Chuyên đề bồi dưỡng học sinh giỏi giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 6)

Chuyfin c i r B D H S G Tojn gia trj I6n nhS't v.i gia tri nh6 nliat - Phan Huy

KITST

Cty TMHH M!V DVVH KhanglZiH"

VAIBAITOAN KHAC VE

G^^yd-.

hay u^ = 2 + 2 V 3 - ( x ^ - 6 x + l l ) .
Ttf do ta di den phU'dng trinh he qua sau:

GIA TRj idiN NHAT VA NHO NHAT CUA HAM SO

= 2 + 2 N / 3 - U <=> U^ - 2 = 2 ^ 3 - u

\J2
§ 1 . LfNG DgNG G I A TRj L6N NHAT V A NHO NHAT DE GIAI

^

PHl/ONG TRINH V A BAT P H J O N G TRJNH K H O N G C O THAM S 6

J

Ne'u nhif ta c6 m a x f ( x ) = m i n g ( x ) = a thi (2) < »
xeD

Ta thu l a i ket qua tren!

f(x) = a

Cac ban c6 nhan xet gi ve tinh hieu qua cua m o i each g i a i t r e n !

g(x) = a

,

, x ^ - 6 x + 15
Pi 7 TT
1. —
= Vx - 6 x + 18,
x^-6x + l l

, , .,

up
ro

'

om

va g(x) = x ^ - 6x + 11.

X-

Xo

3.

2.

,

"

ww

m i n g ( x ) = 2 o x = 3.
xeD

,

.

: rs ,

V a y X = 3 la nghiem duy nha't cua phufdng trinh.
Nhqn xet: X e t each giai khac sau day:
Datu= V x - 2 + V 4 - x

2-6X

+ 11

=1+

x2-6x + l l

(1)

= 1+(x-3)2+2

,

xeR

xeD

fx=3

x^-6x + I l
x^-6x + 15

'

Vay m a x f ( x ) = 3.

D o m a x f ( x ) = m i n g ( x ) = 2, nen phtfdng trinh da cho tifdng di/dng v d i he sau
ff(x) = 2

Ta CO f ( x ) =

x ^ - 6 x + 15
n ~ TT
= Vx - 6 x + 18 .
,

f(x) = 3 o x = 3.

w.

xeD

xeD

1. Xet phifdng trinh

Ro rang V x e R , la c6 f(x) < 1 + - = 3,

fa

V a y m a x f ( x ) - 2 o x = 3.
Ta CO g(x) = (X - 3)^ + 2

HUdng ddn giai

.c

X =

^ ^ < f(x) <

bo

2=4-

<4

ce

2o

ok

Theo bat d^ng thtfc Cosi suy ra 2 < f^(x) < 2 + f(x - 2) + (4 - x ) ]

L a i CO f(x) =

2. V 3 x ^ + 6 x + 7+V5x2+10x + 14 = 4 - 2 x - x ^

/g

Ta tha'y m i e n xac djnh cua phifdng trinh la D = {x: 2 < X < 4 } .

=> 2 < f ' ( x )

J

Bai 2. Giai cac phifdng trinh sau:

,

Ta CO f^(x) = 2 + 2 7 ( x - 2 ) ( 4 - x ) .

- x ^ + 6x - 9 = 0 o X = 3.

duy nha't cua phifdng trinh da cho.

Hiidng ddn giai

Datf(x)= V x - 2 + V 4 - x

•i-.J^j r'^-'~ ^s: ''-it ^- h"^ i
Thuf lai X = 3 vao phifdng trinh da cho ta tha'y thoa man, nen do la nghiem

(2)

B a i 1 . G i a i phiTdng trinh V x - 2 + V 4 - x = x^ - 6 x + 1 L

u = 2.

Ta

X e t cac bai toan m i n h hoa sau day.

o

s/

xeD

"

[u'*-4u^+4u-8 =0

o

.

nghiem.
M e n h de 2: X e t phtfdng trinh f(x) = g(x), v d i x e D

"

Ttf do dan den phifdng trinh V x - 2 + N / 4 - X = 2 o 2 + 2 V ^ t ^ + 6 x - 8 = 4

N e u n h i T v d i phtfdng trinh (1) ta c6 dieu kien m a x f ( x ) < m i n g ( x ) thi (1) v o
X€D

[u'*-4u^+4-12-4u

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(1)

'

J>^
(u-2)(u^ + 2u^+4) =0

va bat phu'dng trinh c6 cau triic dac bict difdc xct den trong muc nay.

xeD

^

>/2
Ta sur dung cac kct qua hicn nhicn sau day de giai mot so Idp phiTdng trinh
M e n h d e 1: X e t phifdng trinh f(x) = g(x), v d i x e D

; ^ .,

=> u^ = 2 + 2 V ( x - 2 ) ( 4 - x ) = 2 + 2 V - x ^ + 6 x - 8

L a i c6 g(x) = V x 2 - 6 x + 18 =

+9 > 3

ai-

g(x) = 3 o x = 3
Vay ming(x) = 3
xeR

Vithe'(l)o

ff(x)-3
g(x) = 3

fx = 3
x =3

o x

= 3.

Do vay X = 3 la n g h i e m duy nha't cua (1)
297


Chuyen de BDHSG Toan gia trj I6n nhjt vk gia tr| nhi nha'l - Phan Huy Khii
Cty TNHH M T V D W H Khang Vigt

2. Xet phurdng trinh Vsx^ + 6x + 7 + Vsx^ + lOx + 14 = 4 - 2x Ta

CO

'

' (2)
p^i 4. Giai phUOng trinh 2^^ ' - 2"

f(x) = V3x^ + 6x + 7 + Vsx^ + lOx +14

,

" = (x -1)^

HUdngddngidi

fv i : ; o ' ;

= V3(x + l ) ^ + 4 + 75(x + l ) ^ + 9 :
Xet phiWng trinh 2"-' - 2 " ^ - " = ( x - l ) ^

V a y V x e R , thi f(x) > 5; f(5) = 5 o x = - 1 .
TCf do ta CO minf(x) = 5 o x = - 1 .

Ta

' '

xeK
CO

g(x) = (x - 1)^ > 0 Vx

Vay ming(x) = 0 <=>x= 1.

g(x) = 4 - 2x - x^ = 5 - (x + 1)^ => g(x) < 5 , Vx e M ;

Do(x - l ) ' > O o x ^ - 2 x + 1 >()

g(x) = 5 c > x = - l .

=> X' -

Do vay maxg(x) = 5 <=>x = - i .

, j,

•

nt,|.,,

Nhir the suy ra (2) o

fffx) = 5
fx = - 1
<^
'<=>i'
o x =- l
lg(x) = 5
[x = - l

f(x) = 0 o

..
- 4 x + 4)
'

'

..x

.,

,.

0.

, ,

om

"

' ' "* *^

o x=1

x=

/

Hudng ddn giai

r
Ap dung bat dang thiJc Cosi, ta c6

.c

'

rx = i

ok

,

o{

m 5. Giai phifdng trinh Sx" - 4x^ = 1 - Vo + x ^

/g

- .

2

Lai c6 4x^ - 4x + 4 = (2x - 1)^ + 3 > 3 => log,(4x^ - 4x + 4) > 1.

\

+ I>3? 1 + ^x2
V
2
)

2

bo

d,

3
2

t

'I <

2

ce

ww

w.

log2(4x - 4 x + 4)

<8;

-i

l +lx^ +
2

3(1 + x ^ ) > 33

fa

^

g(x) = 8 o x = - .

A ' : t • .;.x>1
i

=>

1-

j;

Jil + \^? < - - x ^ Vx e

v.

-.u

u
(1)

2

Vay maxg(x) = 8 o X = - .

Lai ap dung bat dang thtfc Cosi cho 4 so, ta c6:

2

1

{

f(x) = 8

g(x) = 8

X = —

2
1

o
X =

—

2
Vay X = ^ la nghiem duy nhat cua phi/dng trinh da cho.

•
-•

xS2xUx^+ix^>4V^
2

>4x^
(2)

= > 3 x ' * - 4 x ^ > - - x ^ Vx e R .
2

Tir (1) (2) suy ra f(x) = 3x'' - 4x' - (1 f(x) = O o x = 0

298

X <=> X =

Nhir the' (1) CO nghiem duy nhat x = 1.

ro

up

f(x) = 2'"+' + 2^-^^ > l^I^^^Kl'-^' = 24¥ = 8,

Vay minf(x) = 8 o x = - .
xeR
2

g(x) = 0

Ta

.

s/

Theo bat dang thurc Cosi, ta CO

f(x) = 0

Vay (!)<=>

Hitiing ddn gidi

xeR

-

xeR

log:,(4x

V i t h e Vx € R , t a c 6 g ( x ) =

1=

X -

TuTdo suy ra max = 0 <::>x= 1.

Bai 3. Giai phi/Ong trinh 2^"+' + 2^"^^ =

X =

> 2"'

Vay Vx e R, ta c6 f(x) = 2"'' - 2 ' " ' < ( ) ;

Tijr do ta CO X = - 1 la nghiem duy nhat cua (2).

f(x) = 8 o 2x + 1 = 3 - 2x o

1 JI

1

X > X -

2"

<

R va g(x) = 0 <=> x = 1.

6

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Lai

CO

(f) '

^(l + x^f ) > 0 Vx e IR


Chuyen dg BDHSG ToAn gii tr| Idn nhaft va gii trj nh6 nhS't - Phan Huy Kh^i

Cty TNHH MTV DWH Khang Vigt

Vay minr(x) = () o x = 0.

Isin'x + cos'x = 32(sin"x + cos''x)

xel

^ f(x) = g(x)
Ttf cac ket qua tren suy ra

Ro rang phiMng Irinh da cho co the vict duTdi dang f(x) = 0
Ttr do suy ra phu'(tng trinh da cho c6 dang min f(x) = 0 o x = 0

,,

(1)

, '

s

» ,

•K

Vay X = 0 lii nghicm duy nhal can tim.

g(x) = l

^

Dat biet vdi phUdng trjnh dang f(x) = a, x e D
'

xeD

xeD

HUdng dan giai

= n x-(l-x)

Dat f(x) = s i n \ cos^x, x e K .

= n ( 2 x - 1)

sin^x < sin^x
cos'^x < cos^x
' • f t

*

( k 6

up

sin''x = sin^x

Z).

/g

ro

ir
o x =k cos X = cos X
2

1

Ta

•'

s/

=> f(x) = s i n \ cos^x < 1 Vx e R .

.c

.

bo

1

fa

ce

,n-I •

'

ww

I
Dafu bkng xay ra o a = b = —.
Do sin^x + cos^x = 1

X"-2+x"~'(l-x)

+ ... + ( l - x ) " ~ ^

1

0

x

2

1

h'(x)

0

1

h(x)

^•^^'^•^

1
+

1

1

Nhu- vay do 0 < a < 1 => h(a) > h

ok

-

w.

a" + b" >

,

,n-2

3
X "n-2' + ,x "„ n" -- X
l - x ) + ... + ( l - x )

Tijr do CO bang bien thien sau:

om

Tiirdotaco maxf(x) = l o x = k - , k e Z .
"
xeR
2
Ap dung ket qua sau day: Ne'u a, b > 0 va a + b = 1, thi vdi moi n nguyen
> 2 ta c6:

(1) v6 nghiem.

Xet ham so h(x) = x" + (1 - x)" vcti 0 < x < 1
=> h'(x) = nx"~' - n ( l - x ) " " " ' = n x " - ' - ( l - x ) " - '

X€D

Bai 6. Giai phiTdng trinh sin^x + cos^x = 32(sin'^x + c o s ' \

Mat khac f(x) = 1

x = - + k - , k e Z (3)

4
2

Chu y: (*) chtfug minh nhiT sau:

khi do ta c6 r(x) = a <=> max f(x) = a (hoac f(x) = a <=> min I'Cx) = a ) .

Ta c6

(2)

\

Ro rang he (2) (3) v6 nghiem

ma thoa man dicn kicn maxl"(x) = a (hoac minf(x) = a ) ,
xeD

<=>

x-k-,keZ
2

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Chuy:

f(x) = l

sin'^x + cos'^x = (sin^x)' + (cos^)'' > —

oa"+(l-a)">-i^hay a"+b">-i^.

1^

• 5 ' •

•a'u b^ng xay ra<=>a=:^<=>a = b = ^ = > dpcm!

Bai 7. Giai phiTdng trinh cos3x + V2-cos^ 3x = 2(1 + sm^x).
HUdng ddn gi&i

=> g(x) = 32(sin'^x + cos'^x) > 1,
Vxe R.

.

•

Dat f(x) = cos3x + V2-cos^ 3x ; g(x) = 2(1 + sin^x) vdi x e
Khi do phiTdng trinh da cho c6 dang f(x) = g(x).

g(x) = 1 o sin^x = cos^x = <=>x= — + k — , k G
4
2

(1)

De thay g(x) > 2 Vx e R (do sin^x > 0 Vx e R )
Mat khac g(x) = 2 o sinx = 0 o x = kn (k e Z ) .
Vay ta c6 ming(x) = 2 <=> x = kTt.
xeR

Ap dung bat ding thuTc Bunhiacopski, ta c6:

'

(2)


Cty TNMH MTV DVVH Khang Vigt
Chuyen dg BDHSG Toan gia tri Idn nha't va gia tri nh6 nhaft - Phan Huy KhSi

cos^ 3x + (2 - cos^ 3x) (1 +1) > |cos3x + V 2 - c o s ^ 3 x
do suy ra

fir

Vx e

2k7t

ra

\/x' + x

- I +Vx-x^

hay f(x) < 0 Vx e D.

+ 1
=l
| ( x ) = 0 <=> <^X ' + x - l = l <=> s x = - 2
x - X~ + 1 1
X = 1
'x

0 <=> x = I.
l(x) = 0 |x = l
IKct hdp lai suy ra (1) o g(x) = 0 <::> <[x = l
rir do

O

X =

1.

O

X =

1.

di den max f(x) =

' (3)

Bai 9. Giai phU'dng trinh

Tir do suy ra

+1 sin^ X +

COS^ X +

\J
HUdng ddn giai
1 .
Datg(y)=12+ - s i n y . y e R.
COS

ce

bo

ok

.c

om

/g

ro

fa

ww

w.

-x" + x + I > 0
Vie't lai phUcJng trinh da cho diTdi dang
\ / x ^ + x - l + V x - x ^ + l - ( x + l) = x ^ - 2 x + l . ( l )
Datg(x) = x ^ - 2 x + l;f(x)= V x ^ + x - l + V x - x ^ + l - ( x + l), vdi x G D .
Tac6g(x) = ( x - l ) ^ > O V x e D
g(x) = O o x = l
TiJf d6 suy ra min g(x) = 0 o X = 1.
xeD

,

1),

T a c 6 g ( y ) < 1 2 - Vy e

I'M

1 ,

g(y) = 1 2 ^ - 0 siny = 1 » y = ^ + k27i. k e Z . ^ ^ ^^^^j ,
Vay maxg(y) = 1 2 - o y = - + k2n, k e Z .
yeR
2
2
Dat f(x) =

2
COS

X+

1
cos z X J

sin.2

X +

1
sin^ X

+ 4=1
1= (cos'* X + sin"* x) + V cos— X+ —Tsin X ;
l - - s i n ' ' 2x
l
2
^
„

.
,
^
+ 4.
= I — s i n ^ 2 x + 16- 2
sin* 2x

y f ( x ) > 1 2 ^ Vx € R .

^^

(1)
.2

/J-iHI /V.?.JU «t> '

J

l--sin^2x

sin^ 2x + — : r
— +4
sin'*xcos'*x
(2)

Tir (2) de d^ng suy ra Vx e R . t a c 6 f ( x ) > 1 - ^ + 1 6 - ^ ^ + 4

1.

Theo ba't d^ng thiJc Cosi Vx e D, ta c6

= 12 + ^ s i n y .

^

sin^ X y

, // . f

Ta
s/

up

x = 2—.keZ
lgW = 2
X = kn, k e Z
3
_ 2kn
^
, X G i •
i
»x = k 2 7 c , k e Z X = k7l,k€Z
j^., '
Vay X = k27t, k e Z la nghiem cua phiitfng trinh da cho.
^'i'H>Bai 8. Giai phtfdng trinh Vx^+ x - l + \ / x - x ^ + 1 = x^ - x + 2.
^,
t.s,.\.:^^:^',:.
HUdngddngiai
Mien xac dinh cua phiTdng trinh la tap hcJp D gom nhiTng phan tijf x thoa man

(1) (chiiy r ^ n g x ^ - x + 2 > 0 Vx)
he x^ + x - l > 0
.2

li >

I vay X = 1 la nghiem duy nha't cua phiTctng trinh d5 cho.

xeR
<=> X =

suy

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f(x) = cos3x + yjl-cos^ 3x < 2
cos3x -Jl-cos^ 3x
f(x) = 2 <=>
1
1
o cos3x = V 2 - ; o s ^ 3x

fcosSx > 0
cos^ 3x = 2 - cos^ 3x
o c o s 3 x = 1.
Vay maxf(x) = 2 o c o s 3 x = 1
Tit


ChuySn de BDHSG Toan gia tri I6n nhat va gia tri nli6 nhat

Cty TNHH IVITV DWH Khang Vl§t

Phan iiuy Kh^i

( 4 s i n \ 2sin^x - Bsinx - 1)^ = 5 - sinx

f(x) =l2-<=> sin'2x = 1 c=> cos2x = ( ) « x = - + n ^ , n € Z .
2
4
2
V i phifdng trinh da cho c6 dang f(x) - g(y)
l-(x) =

<r> 16sinS + 16sin'*x - 20sin''x - 2 0 s i n \ 5 s i n \ 7sinx - 4 = 0

x, y e M

o

(3)

D e n day m d i cac ban giai t i e p ! ! Cac ban thay the' nao?

12-

f

Tijf cac lap luan Iron siiy ra (3) o
g(y) = i

(sinx - l)(16sin' x + 32sin''x + 12sin''x - 8 s i n \ 3sinx + 4) = 0

2 ^

4 x ^ + 1 4 x + 46
9
B a i 1 1 . G i a i phu-dng trinh —
= 2x^ - 8x + 1 3 .
x ^ + 2 x + 10

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Hiidng ddn giai

<::>x = —+ n — ; v = — + k27t, n va k 6 Z
4
2
2

D a l f(x) =

B a i 1 0 . G i a i phiTdng Irinh (sin3x + cos2x)' = 5 - sinx.

x^

+2X

+ 10

G o i m la gia t r i l i j y y . K h i do phU'dng trinh sau (an x )

HUdng dan giai

4x^ +14X + 46
—^
=m

D a l f ( x ) = (sin3x + cos2x)\(x) = 5 - sinx, x e R
D o sinx < 1 V x e M ^

4X^

14V +
4 -46
4(S
+14X

4-

g(x) > 4, V x e R

•

x ^ + 2 x + 10

g(x) = 4 o sinx = 1 <=> x = - + k27i, k G Z .

•

Do x^ + 2x + 10 ^ 0 V x

(VI

•

'

,
^
(l)conghiem

',i

.

'

„• •

x^ + 2x + 10 > 0), nSn

(l)<::>4x^ + I 4 x + 46 = mx^ + 2 m x + 10m

TiJf do suy ra m i n g ( x ) = 4 c:>x = - + k27t.
xeM
2

s/

Ta

(1)

up

L a i CO lsin3x + cos2x|<|sin3x| + |cos2x|<2 V x e

[cos2x = l
sin3x = - l

om

o

K h i m = 4, t h i m - 7 7 t 0 = > ( l ) c 6 n g h i e m .

*

Khim^4,thi(2)c6nghiemo
o

ce
fa

xeR

w.

V i phu-dng irinh da cho c6 dang f(x) = g(x), nen tif cac k c l qua tren suy

ww

phifcing trinh da cho tiTdng diTdng v d i he f f ( x ) = 4 (3)
g(x) = 4 (4)
T i r ( l ) ( 2 ) s u y r a ( 3 ) ( 4 ) « x = - + kn, k e Z

A'>0»m^-8m+15<0
3 < m < 5 ( m ;>t 4)

K h i m = 5, thi (2) CO dang x^ - 4x + 4 = 0 o x = 2
TiJf do ta CO max f ( x ) = 5 o x = 2
(3)

xeR

bo

Tir do ta CO max l"(x) = 4 o x thoa man ( 2 )

, ,

V a y (1) CO n g h i e m <=> 3 < m < 5.

(2)

ok

cos2x = - l

(2)

o3
.c

sin3x + cos2x = 2

/g

fsin3x = 1
f(x) - 4 o

*

ro

=> l"(x) < 4 V x e R

o ( m - 4 ) x ^ + 2(m-7)x +10m-46 =0

Ta

CO

'

g(x) = 2x^ - 8x + 13 = 2(x - 2)^ + 5 '

Nhu" vay m i n g ( x ) = 5 o X = 2

(4)

|\R

Tijf (3) (4) suy ra phU'dng trinh da cho tu'dng dufdng v d i he
ff(x) = 5

fx = 2

g(x) = 5

[x = 2

o x =2

V a y X = 2 la n g h i e m duy nhat can tim.
V a y X = ^ + k n , k e Z la nghiOm ciia phi/dng Irinh da cho.
Nhanxet:
K h o CO each giai nao khac gon gang hdn each giai trcn
Cac ban ciJ thijr ti/dng lu"dng sau khi siir dung cong thuTc sinSx = 3sinx - 4sin
cos2x = 1 - 2sin^x, la difa phiTi^ng trinh da cho ve dang:
304

5 i i iAy

-

A^Aa« jcef: X e t each giai khac sau day
4x^^14x^46 ^^^,_3^^^3

- •
"

'

;
l «

'•• ^ "

x ^ + 2 x + 10
o 4 x ^ + 1 4 x + 46 = ( 2 x ^ - 8 x +13)(x^ + 2 x + 1 0 )

' •

o2x^-4x'+13x^-

. «,

68x + 84 = 0

305


Chuyfin dg BDHSG Toan gia trj I6n nhat vi gia tr| nh6 nhflt - Phan Huy KhAi

Cty TNHH IMIV DVVH Kliang Vi$t

<=>(x - 2)'(2x' + 4x + 21) = 0<:=>x = 2

'

=:>x + y < 6 .

C&ch giai nay hoan loan chap nhan di/dc, neu cac ban doan triTdc d\{a^.

Ap dung bat diing IhuTc Bunhiacopski, ta c6

nghiem x = 2!
Bai 12. Giai phi/dng trinh 4

sinx

ol+sinx

(6)

(75rri.i+7^.i)'< (75^71)'+(7771)'

cos(xy) + 2'^i = 0

HUotng ddn giai

(1^+1^)

=> 7x+T + 7y + 1 ^72(x + y + 2 ) < 4

Viet lai phtfcJng trinh dU'di dang sau:

max ( 7 ^ + 7 7 7 l ) ^ 4 « : / i l L . £ I i v a 7 ^ + 7 ^ = 4

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2l''l-cos^(xy) = 0 .

P = 2 " " " -cos(xy)

(1)

Ta CO 21^1 > 1 Vy € M
cos^(xy) < 1 Vx, y e R .

hit ii.^'f

M

I*til

P = 0 o i 2'^l-cos^(xy) = 0 O '
cos ( x y ) - l

2''"''=cos(xy)

2M = i

[2''"''=1

o x = y = 3J^'^''^S*™i'«-'r'

Nhan xet: Neu khong sit diing phiTdng phap tim gia trj U^n nha't ci'ia hiim so de

"^ly^o

danh gia hai ve', ta c6 the giai ihuan tuy he phtfdng trinh trcn nhU'sau:

T i i r ( 2 ) c 6 x + 1 + y + 1 + 27(x + I)(y + 1) = 16

y =0

y=0

jf

;,;'„ ( 5 .;v v ' •

m id)

f

Ta

X = kn, k e Z ,

t j

s/

sin X = 0

ro

up

Vay minP = 0 o x = k7t;y = 0 , k 6 Z

/g

Tir do suy ra nghiem cua (1) la x = kK, y = 0 vdi k e Z .

om

Bai 13. (De thi tuyen sinh Dai hoc Cao ddn^ khoi A)

ok
ce

t =3
35
t = --

y + l>0

(4)

xy>0

(5)

, ;

De thay neu - 1 < x < 0 va - 1 < y < 0 thi yfx + l+Jy + l<2 vay khong tht"
=>x>0;y>0.
V i the theo bat d i n g thtfc Cosi, ta c6 tijr ( 1 ) : x + y = 3 + 7 x y ^ 3 +

3t^+261-105 = 0

0 1 = 3.
if.'

M

^ f t

Khi do 7 x y = 3 <::> xy = 9

Tir(3)(4)suyrax>-I;y>-1
thoa man (2)

()
T

(3)

ww

x + l>0
Di6u kien de hg (1) (2) c6 nghla la

<=> 27t^ + t + 4 = 11 - 1 <=>
0
w.

7 5 ^ + 7 7 + 1 = 4 (2)

t . ' ^ ff

Thay lai vao (7) va c6 3 + t + 27t^ + 3 + t + l = 14

fa

(1)

,0)'

Dat t = 7 x y > 0, thi tuf (1) CO X + y = 3 + t -

bo

Hudng dan giai
x +y-Vxy=3

<=>2 + x + y + 27xy + (x + y) + ] = 1 6 .

.c

\ y-yfty =3
Vx + i + 7 y + i = 4

306

77^1-2

Vay he (1) (2) c6 nghicMn duy nha'l x = y = 3

2''"" -cos(xy) = 0

X€i he phi/dng trinh

x=y

Tur do suy ra (2) o X = y = 3 va x = y =3 cung thoa man (1).

V i t h e P > 0 Vx,y G R

Giai he phifdng trinh

1

|x+y-6
Vay(l)(2)oj"-_^
" o x = y = 3.
jTa thu lai ket qua trcn!

!nfa^,(1i/
.,(

• • ,'1 f

' i

' <

Chuygn 6& BDHSG ToJin gii tri Idn nh9"t

Cty TNHH IVITV DWH Khang Vigt

gia tri nh6 nhS't - Phan Huy Kh^i

§ 2 . G I A TR! L 6 N N H A T V A N H O N H A T C U A

minf„(x)-2

H A M SO PHg T H U Q C T H A M SO

vfAu-<a5-=>-^2a* ^
4
2 2

v d i x e D. Vcli moi gia t r i cua tham so m, ta xct bai

Cho ho ham so

^4sin^acos^a
^ 2
^=2
-.
= 2tan'a.
4 CDs'* a
(l + c o s 2 a r
sin^2a

toan t i m gia trj Idn nhat, nho nha't cua ham so F^Cx) vdi x e D. NhU vay dT

=>cos2a<0

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la noi dung cua bai loan chiing ta xct trong muc nay.

cos2x = - 1

. „ , , nay mmfoj(x) = 2

A. Bien luan theo tham so' gid tri Idn nhat va nho nhat cua ho ham so
phu thuoc tham so

,
7:
CO nghiem (chang han x = — ) , v i the trong trirCtng h d p

sin2x = 0

V i he

gia tri Idn nhat, nho nhat cua ho ham so Fn,(x) theo mot lieu chi nao do. Do

sin^2a

.4sin^acos^a
.
- =2
:
= 2c()r(x.,
4sin'^a
(-l + cos2ar

Nhi/ vay theo tham so' a, ta c6 ket luan sau:

B a i 1. Cho ho ham so f^Cx) = t a n ^ ( x + a ) + t a n ' ( x - a ) , v d i tham so a e

2tan^a,Ne'uO
minia(x) =

T i m gia tri nho nhat cua faCx) va bien luan theo a .

+•

/g

cos"(x-a)

bo
w.

- =:>()< 2 a < - => cos2a > 0.
4
2

^^f? -

Ta CO Vx e D ^ , ( d day D„ lii mien xac dinh cua ham so f a ( x ) ) /^-j ^, j y,,
0 < sin'2x + sin'2tt < sin^2a

xeR

l
minf,,,(x)

Tac6F;,(l) = - t +^ .

' H

Co ba kha nang sau:

: ;

^Ne'u m > 2, khi do ^ - ^

Do cos2a > 0, nen (cos2x + cos2a)^ < (1 + cos2a)^.

(3)

" ' M

Dau bang trong (3) xay ra <=>'cos2x = 1 .

hdp nay

sin2x = t (-1 < t < 1). K h i do la c6

min 1- ( t ) (2)
i
L

(2)

Dau bang trong (2) xay ra <=> sin2x = 0.

V I he •' ' " " ^
^•\^••^
{^
cos2x = 1

!! I "-f

F rt) = - - t ^ + — 1 + 1, v d i - 1 < t < 1.

ww

Ncu ( ) < a <

-1
HUo'ng ddn gidi

1 .
m
Viet l a i fm(x) di/di dang sau: f,„(x) = 1 - - s i n 2x + — s i n 2 x .

Ml

fa

(I)

(cos2x + cos2a)^

tl

ce

2(sin~2x + sin^2a)

j

Bien luan theo m.

m a x L , ( x ) = max f d ) (1)

ok

+(sin2x-sin2a)^

fn,(x) = sin'*x + cos'*x + msinxcosx, x e R

.c

* -'^

(cos2x + cos 2 a ) "

*

om

sin^(x + a)cos^(x - a ) + s i n ^ ( x - a ) c o s ^ ( x + a )
cos"(x + a ) c o s ^ ( x - a )

n

Ta

(J) Kil idi ,C1 i xi\r f

i »f

s/

••

ro

sin'(x-a)

^

Bai 2. T i m gia tri Idn nha't va nho nhii'l cua ham so:

{*'\ >^ 4 / n %

up

B i e n ddi r „ ( x ) ve dang sau day

4

2cot^ a , neu — < a < •
4
2

HUfing ddn gidi

( s i n 2 x + sin2a)^

. ,

p a u bang trong (4) xay ra <=> cos2x = - 1

thuoc tham so m. Tiay theo gia trj cua m hay khao sat cac linh chsi't cua cac

^ ^ sin^x +a)
ia(x) =
cos^(x + a )

(4) r v

p o cos2a < 0, nen (cos2x + c o s 2 a ) ' < (-1 + cos2a)".

nhien gia trj Idn nhat, nho nha't cua ham so F J x ) Ircn D la cac dai lifdng p h u

(.(•, nghiem, ch^ng han x = 0, v i the trong truTdng
'

1

cd bang bien thien sau

tn
t
F„(t)

-1

1

1

+
—

1

2

i

°

1

-

, d dav


Chuyen de BDHSG Join

gia trj I6n nha't va gii trj nhd nhlft - Phan Huy K h j i

TiT do suy ra

max F,,,(t) = F„,(l) =

Cty TIMHH MTV D W H Khang Vi^t

m +I

1+m

2

-i
min F,„(t) = F , „ ( - l ) =

^

l-m

min F,„(l):

2
l-m

, ncu m < 0
,ncu m > 0

-i
2. N e u m < - 2 , k h i do — < - 1 va co bang bicn i h i c n sau '
2

,

,

....

"

:../(P^y<<'.

) Cho ham so f(x) = V a c o s ' x + bsin" x + c + Vasin^ x + bcos^ x + c .

m

2

-1

"

1

Fjt)

Luc do ta

1^

1

^

^

^

F(x) = f(x) + msin2x, vdi x e

1

B i c n luan iheo m .

•

i ^

1. V i e t l a i f(x) difdi dang sau:

I-

1+m

min^F„,(t) = F,„(I) =

2. Gia suf f(x) la ham so xac djnh tren R . T i m gia t r i Idn nha't cija ham so

i

™x_F„,(t) = F,„(-l) =

CO

T i m a, b, c de h a m so' xac dinh vc'li m o i x e R.

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t

f(x) = , a

F,„a) = F^J=

up
ro
om

/g

^ c o s 2 x + i±
2
b —a
,
a+
cos2x +
2

.c

max

i
i

-

1

8

ok

Luc nay

1

0

• " <

b - a ^ ^ a + b + 2c

fa

1+ m

„ UK.'!

T o m lai ta c6

max F,,,(l) =
-l
,ncuO2
•
1+m
, ncu - 2 < m < 0
2

w.

I l - m

F ( l ) ) = min — ^ — - ; — ^

ww

v c , , , ,

1- m

. ncu m < - 2

m- + 8
8
1+ m

:

, ncu - 2 < m < 2 ;

il±^>0Vx€R
2
b+ 2c^„,,
_
>OVxeR
2

a- b
a + b + 2c
X) Vt <1
-l + 2
2

ce

min F,,,(l) = m i n { F ( - i ) ;
-I
1 -cos2x

>OVt <1

• •-^^^ aiii,M
'uc:

min
lmin

a- b

1+

b-a^

,,,
(1)

3xV0!:;(/ii;

a + b + 2c

^ a + b + 2c

>0(2)

>0(3)

Itki

• •
Do

,6!)' W;| S, < m

la hkng so k h i t bicn ihien nen

a + b + 2c ,

a-b
-t > 0 ( 4 )
_
hmin
2
2
|i|......
(2) ( 3 ) . ^
a + b + 2c , . b - a
^ + min
-t + > 0 ( 5 )
2
l.|
,

j f < ' | j,^'»

;*

1 <

I

,ncu m > 2

. l + cos2x
+ b
+ c

.
a + b+ 2 c a - b
. , a+ b+ 2 c b - a
1
cos2x
1
cos2x +
2
2
2
2
Ham so f ( x ) xac djnh v d i m o i x e R k h i va chi k h i

bo

i

F,„(l)

+

s/

7

-1

a

Ta

m
3. N e u - 2 < m < 2 = > - ! < — < 1, va C O bang bien thien sau
.
2
, r a i;t§ f

l + cos2x , l - c o s 2 x
+ b
+ c +

N c u a > b => m i n

,a,

l

i

t

'

£r\

'^jji . I V

J I

a-b.

|i|
311


Chuygn 6i BDHSG Toan gia tri Idn nha't va gia trj nho nhS't - Phan Huy Kh^i

Cty TNHH MTV DVVH Khang Vi?t

b-a

; J 4,Tim gia trj Ic'Jn nhaft va nho nhat cua ham so':

min
|i|
' ' " l " (x) = mx^ + (m + l)x + m + 2, Iren mien D = { x : l < x < 2 } .
1.

Khi do (4) (5)

\

Hiidng dan gidi

i ± ^ ± ^ + ^ _ l > 0 o c + b>0.
2
2 ~

a- b

2

2

c + a > 0, nc'u b < a

> 0 o c + a>0.

ri>minr()(x)

•• *

<=> c + min(a; b) > 0

(6).

2. Dat g(x) = msin2x, ihi F(x) = f(x) + g(x)

(7).

f,„(x)

Vx € K neu m > 0

I

Ta
om

cos^2x

2 J

ba kha nang sau:
a. Neu
2m

ce
w.

X

sin2x = - 1
cos2x = 0

deu C O nghiem, nen suy

ra max F(x) = max f(x) + max g(x) = J2(a + b + 2c) + m
x€R

312

xeR

x€R

I

i

xeD

(12)

xeR

cos2x = 0

0

Neu - H l ± l > 2 ^ 0 > m > - - . Luc nay ta c6 bang bien thien
2m 5

x€R

va

2m

suy ra xet

2

I

2m

xeD

cos2x = 0

= -

m < - - . Liic nay ta c6 bang bien thien
3
m + 1

4(x)

0

T t r d o t a c o m i n L , ( x ) = f„,(2) = 7 m + 4 ; m a x f „ ( x ) = f ^ ( l ) = 3m + 3.

Vay maxf^(x) = 2(a + b + 2c)<^ cos2x = 0

sin2x = 1

-

X

4(x) =

. i ;

= 2(a + b + 2c)

ww

a + b + 2c

If!'

fa

.

r (x) = 2(a + b + 2c) <^ cos2x = 0

V i cac he phiTdng trinh

2

m + 1

(11)

Do thoa man (6) nen a + b + 2c = (c + a) + (c + b) > 0, tiT do tiif (11) ta c6

^ max f(x) = J2(a + b + 2c) ^

,

xeD

• Khi m < 0.1\ ^ ( x ) = 2mx + m + 1 , va

.c

J

[a-b]

bo

2

(10)

ok

a + b + 2c^

/g

ro

(9)

D o f ( x ) > O V x e M . n e n maxf(x)=- /maxf^(x)
xeR
V xeK
2

up

Khi m < 0 thi maxg(x) = - m <=> sin2x = - 1 .

f ( x ) < ( a + b + 2c) + 2

< 0, nen c6

^""""'^^^

xeD

s/

(8)

xeR

,

2m

Tilf do ta CO m i n f^, (x) = L, (1) = 3m + 3; max f„, (x) = f„, (2) = 7m + 4.

Khi m > 0 thi maxg(x) = m ^ sin2x = 1.

Vxe R , t h i

m +1

+

0

/

g(x) < - m Vx € R neu m < 0

I

^

xeD

bang bien thien sau:
, j 4 , 5 . - ! ' f„
m + 1
X
1
2m

< m Vx€;R.

Ta c6f^(x) = (a + b + 2c) + 2,

j

, Khi m > 0. Tif f|„(x) = 2mx + m + 1 , va do m > 0, nen

I'lr!) ft

g(x) < m

= io(l) = 3; maxf,)(x) = f,)(2) = 4.

xeD

Vay (6) la dieu kien can va du de f(x) xac dinh Vxe

Taco g(x)

.

fa ihay l()(x) = 1 > 0, nen f()(x) la ham dong bie'n khi x e f 1; 2]

Tom lai de l'(x) xac dmh Vxe R ta can c6
c 4 - b > 0 , neu a > b

| ..^ g , , .

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(4)(5)

a + b + 2c

,

jChi m = 0, la c6 f;,(x) = x + 2

Nc'u a < b, hoan loan ti/dng ti/ta CO

,

4(x)
fm(x)

1

2
+

m +1
2m
0


Cty TNHH MTV DWH Khang Vi^t

Chuygn di BDHSG loAn gia Iri I6n nhgt va gia Iri nUd nh3't - Phan Huy KhSi

Tfifdotaco

fLijc n a y ( I ) CO d a n g 2 t - - I + at > - 1 hay 2 t ' + at > 0

m i n f „ , ( x ) = f,n(l) = 3m + 3 ; max l„,(x) = L , ( 2 ) = 7m + 4.

^ ^ A

I B a i t o a n t r d i h a n h : Tim a d c ha'l phmJng t r i n h : r.,(l) = 2 t ' + a t > 0 d i i n g v d i

Inioi-1 ^ I ^ 1-Dicu n a y Xiiy ra khi va t h i khi

^ * < 2 <^ —— < m < — —. Luc nay la c6 bang bicn t h i c n
2m
3
3
m +l

I

X

+

max f,„ (x) = f„
xeD

m +l

/
/
/
/

3m2+6m-l

2m

4m

'-^ii'-i^

^

3m + 3 nc'u

1

= m m 3m + 3 ; 7 m + 4

f.(t)

1

4 5 • ,

-••••I

;
\

1<^m < —1 '

1:

Tacd:

..fego'ij

<•••>

rC: ,i\J '

•

KV'.

m i n f^(t) = f j l ) = 2 + a

Khi do tiir (2) suy ra a + 2 > 0 <=> a > - 2 .

minf„(x) =

s/
/g

ro

7m + 4 , ne'u m > - —
. ~ 4
^ ; maxf,„(x) =

up

1
3m + 3, ncu m > —

7 m + 4, neu m < —

3m + 3, neu m < —
4
'.^-.x -"•

Loai kha n5ng nay v i khong thda man a < - 4 .

a
2. N5u — < - 1 (turc la k h i a > 4). Luc nay c6 bang b i c n t h i c n sau:
4
, „.,,
ji^
a
t
1
1
~4

ok

B. Cdc ling dung cua vi^c khdo sdt gid tri Idn nhd't vd nhd nhd't cua cdc

fa(t)

bo

ham so'phif thuQC tham so

/
/
/
/

0

fa(t)

.c

om

4

fa

T i m a, b de ba't phi/dng trinh sau: cos4x + acos2x + bsin2x > - 1 dung v d i mt'i
-;f->^>

ww

Hadngddngidi

V i ba't phiTdng trinh dung v d i m o i x, nen n o i rieng no phai dung k h i
x = — va x = — .
4

4

m i n f.(t) = 4 ( - l ) = 2 - a - :

t

.1 i

'"il'

^

-l
Loai kha nang nay v i khong thoa man a > 4.
a
3- Nc'u - 1 < - - < 1 (turc la k h i - 4 < a < 4). Bay g i d cd bang bien thien:
4

L

-

a

1

~4
0

fa(0

V a y b = 0, va b a i todn trd thanh: T i m a de bat phiTdng trinh sau: j
D a t t = cos2x, k h i do - 1 < t < 1.

ITilfdd:

t

- l +b > - l
b > 0
<^ - <f=>b = 0.
l - b > - l
b<0

(1) dung v d i m o i x.

<

^'Bay g i d tir (2) suy ra 2 - a > 0 o a < 2.

w.

xe R .

i

+

I n' >.

^

ce

Bail.

314

«P! oh §1/ d"j

-i
'

T o n g hdp l a i ta CO :

cos4x + a c o s 2 x > - l

0

Ta

7m + 4. n e u^'

D o do ta CO he:

r

I

f.(t)

0,

«-»r ... /r, 's „

/,;r,;.;fr;MrfJ;.

I N c u — > 1 (ttfc la khi a < - 4 ) . Liic niiy c6 bang bicn thicn f
'•
4

rmnf,„(x) = m i n { f , „ ( l ) ; f „ , ( 2 ) }

X6D

> ^•

' d o xet ba kha nang sau:

0

\
Tir do suy ra

f a c d : l.|(t)= 4t + a v a l,,(t)== 0 « . t = : : ^ -4-'.

2

2m

m i n r . | ( t ) > ( ) . (2)
- Kill '

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c. Nc'u 1 < —

\
1

fa(0

^

^

^

+

^
315


Cty TNHH MTV DVVH Khang Vi?t

cos2x = —1 (5)

b. Neu m = - 2 , Ihi (4) <=> m a x l " ( x ) = m i n g _ , ( x ) = 4
xeR

Tif (2) ta t o : —a

>0^

a- < 0 <=> a = 0.

Ro rang he (5) (6) c6 nghiem (thi du x = — thoa man (5) (6))

Tom lai a = b = 0 la cac gia Iri can tim cua tham so' a va b.

T6m lai phiTdng trinh da cho CO nghiem o m = - 2

'*.

Bai 2. Tim m de phiCdng trinh sau:

2 ,
^/7V
2 ,t

, OA '
1"=
X
+
2(2m
3)x
+
5m
1
6
m
+
20
co
nghiem.
3x^ + 2 X + 1

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20x^+10x4-3

Hudng ddn gidi

HUdng ddn gidi

Dal f(x) - (cos4x - cos2x)\e R
Khi do ro rang do |cos4x — cos2x| < 2 Vx 6 K

Dat f(x) =

\
\

=:>f(x)<4Vxe E

iiiuq

.tit H>ui iU'i r:.,i r

.

20x^ +IOX + 3

y K i ' m xr'/l' '•

Ro rang f(x) xac dinh tren R (do 3x' + 2x + 1 > 0 Vx)

,

Goi a la mot gia tri tiiy y cua f(x). Khi do phifdng trinh sau day (an x)

cos4x = 1

2 0 x ' +10X + 3

mm

cos2x = — 1

£ f

Ri

3x^+2x + l

( i ) Vi; ''

Ta

ocos2x = - l .
cos4x = - 1
^

Ta

CO

(20 - 3a)x^ + 2(5 - a)x + 3 - a = 0 (2)

(1)

up

Ro rang khi

^

thi 5 - a 7^ 0 => (2) co nghiem khi " ~

Khi

* ~

'

thi (2) CO nghiem

o A' > 0 o 2a - 19a + 35 < 0

-

< CY <

ok

+ 2] = (m + if

+(m+lf~l.

ce

- l ] | ( m + if

bo

= l ( m + 2 ) ^ - l ] [ ( m + 2 ) ^ + 2 ] + 7 + sin3x

.c

gm(x) = ( m ^ + 4 m + 3)(m^ + 4m + 6) + 7 + sin3x, x e M

•

om

/g

Tirdo suy ra: maxf(x) = 4 <=>cos2x = - 1 .

ro

(Chu y: khi cos2x = 1 => cos4x = 1; con khi cos2x = - 1 => cos4x = 1)

(1) CO nghiem.

= a

(1) <^ 20x^ + lOx + 3 = a(3x^ + 2x + 1)

s/

cos2x = 1

Chu y rang: A „ = |(m + if

•

3x^+2x + l

V a f ( x ) = 4 <;=>|cos4x-cos2x| = 2

€ir

"

0ai 3. Tim m de phtfctng trinh sau:

(cos4x - cos2x)^ = ( m ^ + 4 m + 3 ) ( m ^ + 4 m + 6) + 7 + sin3x c6 nghiem.

,Dat

s i n 3 x = = - l (6)

xeR

7 a

Vay (2) c6 nghiem <^ — < « < 7.

20
3

f

fa

D o d 6 : - A , „ = - 2 khi m = - 2 (va A m > - 2 k h i m ; t - 2 )

NhiTvay ta di den;

xeR

Dat g„(x) = x^ + 2(2m - 3)x + 5m^ - 16m + 20

ww

7 + sin3x > 6 o-sin3x = - 1 .

Ttr do suy ra: maxf(x) = 7 o -x^ - 4x - 4 = 0 o x = - 2 . (3)

w.

Lai c6: 7 + sin3x > 6 Vxe R

g„, (X) = 2x + 2(2m - 3) va g',,, (x) = 0 o x = 3 - 2 m

^

•

Khi m = - 2 , thi min g,„ (x) = 4 .

(2)

=>mingn,(x) = g n , ( 3 - 2 m ) = m ^ - 4 m + 11 = ( m - 2 ) ^ + 7.

•

Khi m 9^-2, thi ming,,,(x)>4.

(3)

Nhu vay ta c6:

xeR
xeM

Phu'dng trinh da cho c6 dang: l"(x) - gn,(x)
T i r ( l ) ( 2 ) (3)suy ra:

,'6 'I3t

a) Neu m = 2, thi ming_,(x) = 7 (4) '

-:

>:i';

(4)

xeR

b) Neu m ?^ 2, thi mingn, (x) > 7.

(5)

x€R

a. Ncu m ^ - 2 , thi (4) vo nghiem

Tilf(4)(5)suyra:
317


oiiuyuii ue u u i i j u l u j i i L)id in luii i i i u i vd g u ill iiliu ililjl - Hlljll HUy MIUI

Cty TNHH MTV DWH Khan^jyi^

a. Ncu m ?^ 2, Ihi max K x ) < miiig,,,(x) :o I'hiAfng Irinh da cho \

f If do suy ra (1) c6 nghicm

nghicm.
xtK

x>R

.

"

Lap bang sau day:

Vay phiftJiig Iriiih da chi) lifdng diTdng vOi he sau:

g2(x) = 7

>

X

X = -2

(*)

X ==-1

1

0

Hitdng ddn gidi
M,..'Ui

i '{iw fsj £«3

Ukn

m ;

Do do bai toan da trcl thanh:

= mx^ + (m + 1 )x + m + 2 > 0

( 1 ) diing vdi

m

+

+

Lucnaytaco: minfn,(x) = f,„(l) = m + 3 m - 2 .

... m - x £ » : { x >

xeR

Dieu nay xay ra khi va chi khi:

min 1"„ (x) > 0.

(2)

s/

I
ro

up

Theo bai 4, phan A, §2 chu-ctng 8 ta co:

/g

3m + 3, ncu m > - —
"
4
min f,,,(x) =

oh

m^ + 3 m - 2 < 0

2. Neu m < - 1 . Khi do ta co bang bicn thien sau:
-1

m

X

-

1
+

0

fn.(x)

ok

.c

v6 nghiem, nen loai triTdng hdp nay.

R6 rang he:

om

7m 4-4, ncu m < - ( ,. . ~ 4

j

m>l

Ta

moi 1 < X < 2.

bo

m > --

ce

3m f 3 > 0
Tir(2) (3) taco:

ww

4
Vay m > — la cac gia tri can tim cua tham so m.
Bai 5. Tim m dc bat phiTdng trinh sau:
+ 2 | x - m | + m - + m - l < 0 C O nghiem.

-m-2

xeR

Ttjf do di den xet he sau:

W.;.v

w.


m<-l

m<-l

m^-m-2<0

-l
m = -1.

4|.<

i'

^- Ne'u - 1 < m < 1. Khi do ta co bang bien thien sau:
X

'
, -.u!) hh

1

m

-1

-

i-.utwM

.•I."' "

+

f,„(x)

Hif(/ng ddn gidi
DatUx)= x^+2|x-m|+ m ^ + m - l .

Lijc nay taco: minf,„(x) = f „ , ( - l ) = m

« m > - i .

fa

"
™<-i
~ 4
7m+4 X )

'

Khi do ba't phi/dng trinh da cho c6 dang l„,(x) < 0. (1)
318

2x + 2

X

nghicm cua bat phifdng trinh mx' + (m + 1 )x + m + 2 > 0.

Tim m dc ba't phU'dng trinh: ijj.)

2x-2

Neu m > 1. Khi do ta co bang bien thien sau:

Bai 4. Tim m sao cho moi nghicm cua bat phiTdng trinh x' - 3x + 2 < 0 cung i;,

ot, u^-t

X' - 2x + m^ - m - 1

Tirbang bic'n thien, xct cac kha nang sau day:

Tom lai vc'li moi m, phU'dng trinh da cho v6 nghicm.

Vi x' - 3x + 2 < 0 o I < X < 2.

x^ - 2x + m ' + 3m - 1

iL
ie
uO
nT
hi
Da
iH
oc
01
/

Do he (*) M"> nghicm => Phifdng trinh da cho cung v6 nghicm khi m = T

l
xeR

gai toan trcf thanh: Bicn luan theo m gia trj nho nhaft cua ham so fm(x).

b. Ncu m = 2, ihl max f(x) = miiigi(x) -~- 7 .

l(x) = 7

(2)

min i",„ (x) < 0.

Lijcn^y tac6: minf„,(x) = fn,(m)=: 2m^ + m - l
xeR

319


Chuyen

BDHSG Toan gii tri I6n nha't vA gia trj nh6 nhat - Phan Huy Khdi

Cty TNHH MTV DWH Khang Vijt

- 1 < rn<1

-l
1
-1< m< "
~2

TCr do di den x c l he sau
2m^ + m - 1 < 0

m <1

m < —

4~

1

m

-

(1)

2"

-

(2)

3

< X<

~
m i n f , „ ( x ) < ( ) (3)

Ta

Ta c6: f,„(x)= 2x - m va f,„(x) = 0 ^ x = —

bo

ok

.c

+

0

V

ww

w.

m>6
9 Om>6.

9-2m<0

m > -

~2

2. Neu — < - (tiJc la ne'u m < 1). Luc do ta c6 bang bien thien sau:
2
2
1
m

3
X
I

-

^\

+

+

1

ro

= m—

m

Nhu" do dan de'n xet he sau:

lm<0

m

<0
4 ~

<»4
I

m>4

1
Kct hctp lai suy ra: m < - - hoac m > 4 la cac gid trj can tim cua tham so m
Nhqn xet: Hay so sdnh cdch i^idi nay vt'/i cdch fiidi trinh hay tnmi^ bdi 6 phdn
B, §3 chu<rn}> 7 cuon sdch nay. Cdc ban thdy cdch nao thich h(fp \'<)i ban hfn?
Bai 7. Cho ba't phu-cJng trinh: sin3x + msin2x + 3sinx > 0.
Tim m dc ba't phm^ng trinh diing vdi moi x €

fa

ce

min f ^ ( x ) = : f n i ( 3 ) = 9 - 2 m

TiJf do dan den vice xet he sau:

•

l
up

om

2

I -

• it-

/g

1. Neu — > 3 (tuTc la neu m > 6). Liic do ta c6 bang bien thien sau:
2
in

s/

Tif do suy ra xet cac kha nang sau:

m>6

0

L u c n a y t a c o : min f„,(x)=:f,„

m

Luc nay taco:

.i & - i : / ' „ . . ;

A

Tilfd6suyrahe(l)(2)c6nghiemkhivachikhi:

X

'.•

3

7

X

Vie't lai he da cho d\id\g sau:

2

" 2

iL
ie
uO
nT
hi
Da
iH
oc
01
/

u

f (x) = x ^ - m x + m < 0

"

.

3. Neu ^ < ^ < 3 (ttfc la 1 < m < 6). Luc do ta c6 bang bien thien sau:

x^-mx + m<0
HUdngdangi&i

'
1 <»m<

2
. Tim m dc he c6 nghiem.

4

m <1

TiJf do xet he sau:

" • " i ' ••••

Bai 6. Cho hg bat phi/dng trinh:

2

-

2~

K c l hap lai suy ra cac gia trj can tim cua tham so m la: - 1 < m < - .
2x^-7x + 3<0

m

Luc nay ta c6: min f„,(x) = !„

-1 < m < ~ 2

'

Hifdng ddii gidi

DiTa ba't phiTtfng trinh da cho ve dang tU'dng diTcfng sau:
3sinx - 4 s i n \ 2msinxcosx + 3sinx > 0

•0

<=» 2sinx(-2sin^ x + mcosx + 3) > 0
2sinx(2cos^ x + mcosx +1) > 0 .

(1)

K h i x G 0 ; ^ , thi sinx > 0.
2
"$y tren €

, thi (1) <=> 2cos^x + mcosx + 1 > 0.

(2)


Chuyen ai BDHSG Join gii tr| Ifln nhaft

g\& tr| nhd nha't - Phan Huy Khai

m
()<'i_.i
T
NhU the dan den vice xet he sau:

Dat t = cosx. Khi X e 0 ; ^ , t h i l e [0; 1]. J
2
Bai toan da cho trd thanh:
Tim m de bat phiTdng trinh: Im(l) = 2t^ + mt + 1 > 0 dting vdi moi t e [0; 1].
min f„,(t) > 0 .

-

o
Taco: 4(t) = 4 t + m

- 4 < m <()

-4 < m < 0

,<^-2N/2 < m < 0

"11 + 1 X ) I - 2 ^ 2 < m < 2N/2

(3)

£ i

=>4(l) = 0 <^ t =

Ket help lai suy ra: m >

. TOd6 x^t cac kha nang sau:

l

a cac gia tri ciin tim cua m.

iL
ie
uO
nT
hi
Da
iH
oc
01
/

Dieu nay xay ra khi va chi khi:

m +1

8 .- .-.

Tif do ta c6: min L,(t) = 1„,

jV/»fl« xet: Ciich giai trcn di/a vao phUdng phap tim gia trj k'ln nhat va nho nhat
ciia ham so phu thupe lham so m.

1. Ne'u —— > 1 (tu'c la khi m < - 4 ) . Liic nay ta c6 bang bic'n thicn sau:
4

Qic ban hay so sanh each giai tren vdi each giai bai nay bSng phi/dng phap
siir dung gia trj k'ln nhat vii nho nhat ciia mot ham so phan tht?c (khong c6

t

4

C(i)

f,

-

tham so) da Irinh bay trong bai 5 phan B, §3 chi/dng 7 cuo'n sach nay.

0

.

Va tinh hicu qua cua tifng phu'dng phap, xin danh quyen blnh luan cho c^c

r

/

ban.

Bai 8. Cho ham so: r„,(x) = 4x" - 4mx + m ' - 2m. Xct trcn mien - 2 < x < 0.

Tim m dc

m +3> 0

m>-3

he vo nghipm. TiTdo loai kha nang n;iy.

m < 0 (tuTc lii khi m > 0). Liic nay ta c6 bang bicn thicn sau:
,4
r
m
0
I
~ T

&r

/
/
/

0

<

m

<

1

( H J T C la

K

-4

<

m

<

4
Liic nay ta c6 bang bicn thicn sau:
m
t
0
4
/

+
0
4(t)
/

\
/

0

+

/
Vay:

I

0).

-

/
/

w.
khi

•>

y

ww

Ncu

'Jfii'f"

m

0

'

/

TCr do suy ra m > 0 thoa man ycu cau dc bai ra.
3.

-2

fa

/

/

(XKI

X

ce

/
/

*

Tif do dan den cac kha nang sau day:
m
1. Ncu ~ > ^ ^
m>
Liic nay la c6 bang bicn thicn sau:

ok

/

+

Hiiiln^ ddn giai '' f

Ta c6: f,„(x) = 8x - 4m =» i„,(x) = 0 <^ x = ym .

bo

0

.c

om

/g

2. Ncu

min r|„(x) = 2

s/

m <-4

up

m <-4

ro

Tac6:

Ta

()
min f|„(x) = i',,, (()) = m^ - 2m .
-2
m = l + V3
2-2m = 2
Khi do: m
m>()

m' - 2 m - 2 = 0
m >0

"

„i = l _ 7 ^ 4 ^ m = l + V3
m>()

^

• Ncu — < - 2 (<=> m < - 4 ) .
2
Lijc nay ta c6 bang bie'n thien sau:
323


rgaii y i a u | l u i r i i i i j i vd g i d in iniu i i i i t l l -

m

/
/

0

'in

+

Khi do he:

0

? Ta co: 1 „, (u) = 4u - 2m =^ 1„, (u) - 0 < > u -m

i

m V

Do m > 0, ncn ta co: 0 < m
y < m , vay c(') bang bicn thicn sau:

/

m

u

0

m" - 6 m + 1 6 .

m

y

0

/

m-^ - 1 6 m + 16 = 2
m < -4

.

r

min r,„(x) = 1,,, ( - 2 )
-2
Miai

0

X

Tirdo:

Milan nuy

- 6 m + 14 = 0
m < -4

v6 nghigm.

m
3. Ncu - 2 < — < 0 «:> - 4 < m < 0). Liic nay ta c6 bang bicn thicn sau:

% Tir do suy ra:

/
/

1

m -6m-6

min I„, (u) = i„

()
1)1*'

0

+

iL
ie
uO
nT
hi
Da
iH
oc
01
/

o i i u y K H OB n n n j v i r

max 1'l„,(u) = max {f,„(0); f,„(m)} = max{m^ - 3 m - 3; m - - 3m - .3}

()
.)}•,{

,,„: :,.v;,-' .-f^,

I ••-> •

•,.•,« ( K fUiiH

m^ —6m — 6 ^ ^

= - 2 m . Nhu' vay:

-4 ^ m < 0

•2' .\ (I


Tom lai: m = I + V3 va m = - I la hai gia trj can tim ciia lham so m.'

ok

, ihi u > 0, v > 0 => x + 1 =

va y + 2 = v'.

bo

Dal u = v^x + I ; \ sjyTl

m'

ce

u +v=m

(1)

w.

fa

Bai toan da cho tnt lhanh: Tim m dc he: u" + v^ = 3m (2) CO nghicm.
u > 0; v > 0

(3)

r,„(u) = 2u- - 2mu + m- - 3m - 3 = 0 (4)

,

,

•

Dicu nay xay ra khi va chi khi min r,„(u) < 0 < max f,„(u) (6)
(' II III

-3m-3>0

Tim m dc bat phu'ilng irinh dung vc'^i moi x.
Hiding dan gidi
Viet lai bal phuTttng trinh diftKi dang:
fn,(x) = 2m(sinx + cosx) > - 1 - m'

(1)

>

Tijr do suy ra (1) dung vdi moi X G M khi va chi khi:
(2)

mini" ( x ) > - l - m
X€E

> 1

Xet cac kha nang sau:
min f()(x) = 0. Luc do (2) co dang 0 > - 1

Vay m = 0 thoa man yeu cau de bai

()
' *'

t N e u m > 0 , thi minf„,(x) = - 2 V 2 m (dosinx + cosx <-sl2 Vxe R
X6R

Tir do thay vac (2) CO:

co nghiem.
( 5 ) •::^:-:.::f

fi"«l±:^<„,<,+y,7.

x€R

Tir (1) c6: v = m - u. Do \ ^ 0 < u < m. Thay v = m - u vac (2) ta co:
2u- - 2mu + nr - 3m - 3 = 0. Tir do biii loan da cho lai co dang sau: Tim
m de he sau:
i

( ) < II < m

r—

Bai 10. Cho bal phiAtng Irinh: n r + 2m(sinx + cosx) + 1 > 0.

1. Neu m = 0, thi f,i(x) = 0

ww

Tir(3)va(l)suyram>().

5

,

Do la cac gia Iri can llm ciia m.

om

/g

v ' ^ i +7y + 2 = m „
,
. Tim m dc he co nghicm.
X + y = 3m
HuYfiif/; dan giai

.c

Bai 9. Cho he phifitng Irinh:

s/

r,„(x)=r,„

up

min

ro

Vithc:

:2 •

Ta

Bay gid ket h(1p vrJi (6), la co:

-2m

veil J J. C

= m - 3m - 3.

-2^m>-l-m^

m^ -2V2m + l > 0

m>0

m>0

mm>N/2 + l ^

m>0

(X

m < V2 - 1

m>>y2+1


Cty TNHH MTV DWH Khang Vigt

3. N c u m < 0, thi do sinx + cosx < sjl

Vxe
max

=»minr,,,(x) = 2v/2m.

= max m — 3; m + 8

m -3

xeR

[jsleu m <

Thay viio (2) va c6 he:

8

+ 2>/2m + 1 > 0

m<0

m<0

m>-V2 + l ^

;

m<-V2-i
I-V2

m < 0

max

>N/2

m<-V2-l

Kct hdp l a i la t o :

+ 1 '^^

<^

max

Do la cac gia Iri can tim cua tham so m.
Bai l l . C h o h a m so f ^ ( x ) = | - 2 x ^ + x + m | vc'ii x G [ - 1 ; 1].

liyx^^Uv,

-

= max

ok

fa
w.

max r,,(x) = max g , n ( - l ) | :
-Kx- I
m-3

g,„(l)

'It

m + 8

X e l cac kha nang sau:

I:

t

16
!

1
m + 8

3-m

. . . .

/

25

23 1

16

16

J .

Tif bang bien thien suy ra: m + ^ > m - 1 > m - 3.
T u do suy ra:

= max

-A.

A^Aan jcef; Ta chtfng minh (*) nhiT sau:

ww

ra:

1

23
16

max L ( x ) =
-l
V i ihe m i n max L , ( x )
xeR - I < x < l

bo

0

"•• ^ • •

/

gn,(x)

Tif do suy

+

-

ce

/

'

3 - m , ne'u m <

max f„,(x)
-1
.c

Tir do CO bang bien Ihien sau:
^

vf>f!(

ro

„

max f,„(x)
-1
om

„ •

. ' 1
^23
3 — m neu — < m < —
8
~16

23

/g

x = i.

= max 3 - m; m + 8

m

up

. ,

1. D a l gn,(x) = - 2 x ^ + X + m vc'ti - 1 < X < 1.

-1

8

m + - neu — < m < 3
8
16 ~

Ta

>,fyi'(&iA'iMtm

Hiiihig ddn gidi

3-m.

1
2f;
m + - , neu m > — h . : .
8
I6;V
2. Tur phan 1/ suy ra bang bien thicn sau (iheo m)

s/

max l|„(x) dal gia Iri be nhat.
-l
T a c 6 : g',„(x) = - 4 x + 1 =^ g;„(x) = 0 ^

m+-

m-3

K e l hdp l a i suy ra:

max f,,,(x) va bicn liian ihco m.
-I
2. T m m d e

- =

= max 3 - m ; - m -

g N c u — - < m < 3, i h i m - 3 < 0 va m + - > 0, do do:
8
8

l-V2
1. T i m

m-3

i


• I"

m > 72 + 1

1

iL
ie
uO
nT
hi
Da
iH
oc

01
/

2V2m>-l-ni^

m <-V2-1

, thi m - 3 < 0 va m + - < 0, do do:
8

m —3; m —1 m +

(*).

n'^iW

:

Ne'um+-<0=>m-1 >m-3>0
8

\
,

m-1 < m-3 .

•
(1)

b. N e u m + - > O l h i

1
m + -

•

1
Ne'um+- > m - 1 >0
8

•

Neu m - l < 0 = ^ m - 3 > m - l

a. N c u m > 3, Ihi m - 3 > 0 va m + - > 0, do do:

>m-i

(2)
\'Hv,

(3>
327


"UFiuyen ae BPHSG'iHaw aia trr ion mrarra gra Tri n r o m a r - T n a n Huy Knar

T i i r ( l ) ( 2 ) ( 3 ) s u y r a : max | m - 3 | ; | m - l

max | m - 3

m + -

Kct hdp lai suy ra 0 < m < — va m = - 2 la cac gia Iri can tim cua tham so'm.

m+-

V a y (*) diTdc chuTng minh.

^ Cach giiii tren diTa vao phtfcfng phap tim gia Iri be nhat cda ham so phu

Bai 12. Tim m de ba'l phiTcJng trinh sau: m"x + m(x + 1) - 2(x - 1) > 0 dung V(3j

thupc tham so fn,(x) = (m^ + m - 2)x + m

moi x e [-2; 1].

1^ Xoi each giai sau day:

HU(fng dan gidi

Vi r„,(x) hoac la ham hang so

f^(x)= ( m ^ + m - 2 ) x + m > - 2 c : > U x ) > - 2

nc3n luon luon dong bien (khi m^ + m - 2 > o ) , hoac luon luon nghich bie'n

(1)

De (1) dung vdi mpi x e [-2; IJ dieu kicn can va du la:
_mm/n,W>-2.

,,

„

(khi m^ + m - 2 < o ) . V i vay f j x ) > - 2 V x e [-2; 1] khi va chi khi

.

(2)

,

Xet cac kha nang sau:
1. Neu ra^ + m - 2 > 0 =4> f,„(x) = m' + m - 2 > 0, nen ta c6 bang bien thien sau:
x

-2

1

(khi m^ + m - 2 = o), hoac la ham bac nha't

iL
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oc
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Viet lai bat phtfrtng Irinh da cho difdi dang lu"(ing du^dng sau:

l„,(-2)>-2

-2m^ - m + 4 > - 2

f,na)>-2

m^ + 2 m - 2 > - 2

-2~2

•

m<-2

2m^+m-6<0
4^

m^ + 2 m > 0

0~2
m = -2. -

Tirdotaco:

^

Bai 13. Xet ham so' fjix) = cos2x + mcos x + - , xe
4
ChiJng minh r^ng vdi mpi m ta c6:
\

^

m^ + m - 2 < 0
m^+2m-2>-2
m = l

3. Neum^ + m - 2 = 0<^

m = -2'

maxf„(x)

>2.

V xeR

HUdng ddn giai
IT

Ta c6: fm(0) = 1 + mcos —,
4

(1)
It

fm(7t)

,

+

minf_,(x)

min f„,(x) = L , ( l ) = m ^ + 2 m - 2 .
-2
Vay (2)

nho nha't cua ham so'de giai toan cung dem lai hieu qua!.

bo

I

I

Thi du nay churng to rang khong phai bao gid viec suT dung gia tri Idn nha't va

ok

4 (x) = m^ + m - 2 < 0, nen ta c6 bang bien thien sau:

Ta thu lai ket qua tren. R6 rang each giai nay gon gang hdn!

ce

2. Neu m^ + m - 2 < 0
-2

2

ro

2m2+m-6<0

3

/g

-2m^-m+4>-2

,

om

m^+m-2>0

.c

m'' + m - 2 > 0

fa

^

m \

w.

Vay (2)

up

min f ^ ( x ) = f „ ( - 2 ) = - 2 m ^ - 2 m + 4 + m .
-2
ww

Tirdotaco:

s/

Ta

m>0

= 1 + mcos

4

= 1 - mcos —.
4

T i r ( l ) ( 2 ) suy ra: U O ) + UTC) = 2.
ff„,(0)
(4)

f„,(i;)
(5)

(2)
(3)

Ro rang

Luc n a y f , ( x ) = l ; f . 2 ( x ) = - 2
Vay (2) thoa man khi m = 1 va m = - 2

Tir (3) (4) (5) suy ra:
329


Cty TNHH MlV DVVH Khang Vijt

2 = f^O) + fn,(7r) < 2max 1„, (x) =^ max f,„ (x) > 1
xeR

X6K

: — 1 +m

,2,
IT

2,
/

18;
2;

'

11)

COS

TT

c h i n g han cimg diTrtng thi do x, y G Z =^ x > 1, y > 1 =^ 4x + 5y > 9. D o la

IT

dicu v6 l i .
i i v i A V ' sC'ik)
V i le do D = D , U D j , trong do:

• TT

+ - = _ l + msin^^^'^'"*-7''''^'^
2 4j

I

\

__2

(7)

'-b'mn^ii

2
,/iiiy<

Ro rang: 1„

M a t khac n c u ( x ; y ) G D thi x va y trai dau. That vay neu x va y cung dau,

/

- +4,-= ,2

= - 1 + m cos
TV

\K

•• tn^'f Afn j f l i i /

:|ti6b'.

n'rinf

^aw'.wirt,

D| = {(x; y): x > 0, y < 0; x, y e Z v^ 4x + 5y = 7 } ,

Thco nguyen l i phan ra, ta CO:

061:,

2,

m i n P = min

(x;y)€D

min

(x;y)eD,

Tif 4x + 5y = 7

TCr (7) (8) (9) suy ra: n i i n L , ( x ) < - 1 ^ (min ('...(x)] ^ > 1 . ( 1 0 )
XGM

Do x, y

I

VxeE

Cong lirng vc (6) (10) di den: (min 1„, (x)

+ max f,„ (x))

>2 . t • >

Ta
LLfONG GIAC

/g

HINH HQC,

ro

up

§3. GI6I THIEU MQT SO BAI TOAN GIA TRj L6N NHAT. NHO NHAT

s/

Do

TRONG SO HQC,

om

Ciio'n sach nay diinh dc Irinh bay cac bai loan gia Iri Idn nha't, gia tri nho

.c

nha't thi/ctng gap trong dai so' vii giai lich.

G

P; m i n

(x;y)GD2

P

X>

x=

7-5y

= 2-y +

(2)

x = 3-5t

0; y < 0 nen suy ra:

^ ~ ^ ' > ° = , t < i = > l= 0;-l;-2;...(dotG Z )

41 - 1 < 0
4

Ti^ (3) va do t = 0; - 1 ; - 2 ; . . . . nen suy ra ifng v d i t = 0, ta c6:
min

(4)

P=::12.

Khi (x; y ) e D2 i h i P = -5x - 3y.

khao khac. Tuy nhicn trong muc nay, chung toi muon gicKi thicu vc'Ji cac ban

D o x < 0 ; y > 0 , t i r ( * ) ta c6:

ce

bo

ok

hoc, hinh hoc, liTi.Jng giac so di/dc chiing loi trinh bay trong mot cuon chiiycn

fa

w.

ww

(5)

^"-'^'<"=^1>1=> t = l ; 2 ; 3 ; . . . ( d o t G
4t - 1 > 0
5

Z)

A. Vai bai toan ve gid tri Ida nhat, nho nhat trong so hoc

L u c n a y P = 1 3 t - 12.

Bai 1. Cho P = 5|x| -3|y|,d day x, y thuoc tap hdp D diTdc xac dinh nhiTsau:

Til (6) va do t = 1; 2;... suy ra tfng vdi t = 1, thi:

{ ( x ; y ) : x , y e Z va 4x + 5y = 7)

'

'

'

•

'

>

'

f ^

(tuTc la D la t$p hdp cac nghiem nguyen cua phiTdng trinh 4x + 5y = 7
T m i gia tri be nha't ciia bieu Ihufc P khi (x; y) e D.
llUdng

ddn gidi

-

N c u (x; y) e D thi chac t h a n x 9^ 0; y 9^ 0. That vay ncu trai lai gia siir chaiUhan X = 0 => 5y = 7. D o la d i c u v6 l i VI y 6 Z.

(3)

K h i x = 3 - 5 t ; y = 4 t - 1, la c6: P = 5x + 3y = 12 - 13t

Cac bai loan gia trj k'ln nhat, nho nha't trong cac ITnh vifc khac nhu" trong so

nhSng ITnh viTc nay de cac ban tha'y dufdc tinh da dang eiia kHp biii toan nay.

4

y =4t-l

(x;y)€D|

mot so bai toan lien qiian den vice t i m gia tri Idn nha't, gia trj nho nhii't trong

(1)

Z ^ ^ - i ^ = t, v d i t e Z .

Tif do ta co:

( I •< •

D 6 la d i c u phai chiyng minh.

330

J:

Khi (x; y ) e D, t h i P = 5x + 3y.
xeK

D=

-i i t ; / w ; ; ,

D2= { ( x ; y ) : x < { ) , y > 0 ; x , y e Z v a 4 x + 5y = 7 } .

d i . ^ - . ,

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L a i c6: f„

f max f„, ( x ) f > 1 ( 6 )

min

(6)

•- ^ I

(7)

P = I 3 . 1 - 1 2 = 1.

(x;y)eD2

T i ) f ( l ) ; (6); (7) di den:
min
(x;y)eD

P = min{l2;l} = l ^
^

^

I '''if,

x = 3-5.1
[y = 4 . 1 - 1

'

»

x =-2
y =3

II xet: Bhi toan la sir k c t hcJp giffa nguyen l i phan ra trong bai toan t i m gia tri
Idn nha't, nho nha't va phep giai phi/dng trinh nghiem nguydn trong so hoc!.
331


Cty

Chuy8n d l BDHSG Toan gii t r j Idn nhat va gJA t r j nh6 nhaft - Phan Huy Khjii

Trifdc h c l la chiJng minh rang irong each phan tich da chpn n h t f t r e n thi tich

B a i 2. Cho m , n la so nguyen diTtJng.
Tim

gia tri nho nhat ciia bicu thiJc: P = |l2"' - 5 " | .

'

'a,a: ••• ak chi' g o m nhCTng ihiTa so nguyen to 2 va 3, va khong eo qua 2 ihiTa so

' ' '

nguyen to 2. T h a i vay:

HUifiig dan gidi

. . r , i J.y<

,{} > i, (ij).;t/

, V i 12"'" la so chan, con 5"" la so tan cung hang 5, nen suy ra:

lhay lich aiaj ... a^ tang Icn. D i c u nay mau ihuan v d i u'eh da cho la Idn

r,*; <

nhii'l.

12'"" - 5 " " la so Ic : ^ 1 2 ' " " - 5 " " / 2 .
D o 12"'"/5,con 5"" 15
Lai

-

12"""-5""/5.

thay 12'"" 13, con 5""/3

lhay hai so 1, a bnng so I + a va de y rang do 1. a < 1 + a, nen sau khi

5

•

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:S.

,

12"'" - 5 " " / 3 .

x - * f^^^fc

4 <=> 2b - 4 = b <=> b > 4). V a y long chi ehiVa cac so hang 2 va 3.

x'

-

(3)

.

ojljir

Vdi tinh chat nhuf vay ihi de dam bao aiaj... a^ idn nha'l, ta can phan ti'eh

(mod 13)

(4)

s/

R o r a n g t a c o : 12'"" = ( - 1 ) " ' "

Ta

hoac la 12'"" - 5"" chia cho 13 thi diT 12.

(5)

Tim

gia trj Idn nhat va nho nhat cua P.

fa

w.

rkngXi

1. Xet bai loan tim min P.
X e t bp so

l < X , < X 2 < . . . < X 3 o

I

K h i d d t a c d : P = x,

Tim

Bay g i d xet bp so

gia trj Idn nhat cua bieu thuTc: P = aiaj... ak.
HUt^ng dan gidi

de'n cich chpn ma tich cua cac so nguyen dufdng la Idn nhaft. N e u c6 nhidu
each phan tich nhuf vay t h i ta se chpn each phan tich ed nhieu so hang nha't.

:

(x|,X2,...,X3„)

ak = 100.

Trong cac phan tich 100 ra tong cac so nguyen diTdng a i , aa,... ak ta quan tam

"? i

Do vai tro binh dang giDTa x,, x , , x , , ) la cd the khong giam long quat ma gia sif

ww

Bai 3. X e t tap hdp ta't ck cac so' nguyen di/dng a i , a2,... ak sao cho aj + 3 3 + .. +

x.,,, sao cho

HUdng ddn gidi

Mat khac neu chpn m = 1 ; n = 1, thi P = 7.

Theo dinh nghla ve gia trj Idn nhat suy ra: m i n P = 7.

I''

X i + X2 +... + x,„i = 2011. X e l d a i lu'ilng P = X | X : . . . x,„.

Tir (4) (5) suy ra k h i dem 12'"" - 5 " " chia cho 13 t h i khong the dir 1 hoac 12.
D i e u nay mau thuan v d i (3). V a y khong the c6 (2), «?c la (1) dung

.

"

Do3".2<3''.2-(vi3<4)suyraaia.... ak<4.3".

Bai 4. X e l lap hdp ta'l ea cac each chpn 30 so nguyen Jififng X|, x^

ce

(tifdng i?ng v d i z,, = 0; 1 ; 2; hoSc 3).

332

32 so hang

bo

Tir (5) ta CO k h i 5"" chia 13 thi so di/ se la 1, 5, 15 hoSc 8

33 so' hang

„^

Vay max P - 4.3^'.

ok

Tir do suy ra: 5"" = 625''".5'" = l''".5'" (mod 13)

100 = 3 + 3 + . . . + 3 + 2 hoac 1 ()() = 3 + 3 + ... + 3 - I 2 + 2 .

Da'u " = " xiiy ra k h i va ehi khi k = 34; ai = a^ = ... = a,: = 3 ; a,, = a u - 2,, f,*

.c

om

khac ta luon luon c6 the bieu dien:
n,, = 4k„ + z„ v d i k<, e Z va z,, e {0; 1 ; 2; 3}

,,j

;

ro

^ ^ . . j ,

/g

13)

up

Nhir vay neu m„ la so chan thi 12""" = 1 (mod 13), c6n neu m,, le thi

Mat

ifing Icn,

licli aia,.

dd la d i c u vd l i . Do vay thiTa so' 2 trong lich se khong \l q u i i 2.

hoiicla 1 2 ' " " - 5 " " c h i a c h o l 3 l h i d i r l

(mod

Gia suf long ed nhieu hdn hai so hang 2. Chii y rhng iicii lhay ba so 2 hang

hai .so 3 Ihi 2 + 2 + 2 = 3 + 3, nhirng 3.3 > 2.2.2, vay

Tif(3)tac6:

=12

Trong each phan tich da cho khong cd so hang b > 4, vi la ed ihe lhay b
bang hai so hang 2 va b - 2. Rd rang 2(b - 2) > 4 ( T h a i vay, vl 2(b - 2) ^-

K e t hdp iai cac d i c u tren, tuf (2) suy ra: |l2"'" - 5 " " 1=1

12'""

,

la lay mot so hang a tiiy y khiie, a > 1 ( d l nhien no ton tai). Luc nay la

That vay gia s i l r ( l ) khong diing, ttfc la ton tai hai so nguyen diTtJng m,,, n„ s,^,
" =-V:«' "''•.•.tfl- m

,

Trong each phan lich da chpn, khong cd so hang 1 v i neu eo so hang 1 i h i

TrU'c'Jc hot la chufng m i n h rang v d i m p i m, n la so nguyen duftJng thi P > 7. (| j

cho | l 2 ' " " - 5"" I < 7.

TIM HH MTV DVVH Khang Vigt

.

trong d d :

•'/«':;.
, _ ^

v a x , + X 2 + . . . + X3„ =

2011

X2...X3(, .

(x|-1,X2,...,X29,

x^o+1).

Vf/V

Khidddeyrangdo xj'>l:^x, > 2 = i > x ^ - l > l .
•Mat
•

khac:

- 1 ) + ^ + ... + x ^ + ( x ^ + 1 ) = 5^ + x^ +... + X j , , = 2011, nen

bp so m d i nay cung thoa man y e u cau de b a i . Cfng v d i bp so nay ta cd:

•L...

333


Chuyfin

BDHSG Toan gia Iri I6n nha't va gii tr| nh6 nhS't - Phan Huy KhSi

P = (^-l)5^...X^(x3o +l) = X2...X29(x3o-X,

Do x ^ > ( ) Vi = 2r29, va x 7 < x ^ ^ P > P .

P = 67^'".68.

(1)

T o m lai la co: max P = 67^*^68
K c t hop l a i ta co: m i n P = 1982 va max P = 67'''.68.

'"''^ '

p a i 5. Cho k la so nguyen di/Ong > 3.

nha't. V a y mot dieu k i e n can de P dat gia t r i nho nha't la X| = 1.

''* '

Tir do lap luan hoan toan ti/rfng tif suy ra X2 = X3 = ... = X29 = 1 cung la dieu

Tim gia tri Idn nha't ciia ham so f(x, y, z) = xyz trcn mien

k i e n can de P dat gia t r i nho nha't.

D = {(x; y; z) : x, y, z nguyen dU'Ong va x + y + z = k } .

1 (2011 - 29) = 1982.

\'

^

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1.1

trong do:

,

C3„-X,>2

<

,

max

X j ...

Ta se chu'ng minh rang: X o - Zo < 1

• v - ^
"

"

' 'r?

(2)

That vay, neu trai lai ta co: Xo - Zo > 1.

X3,, . Bay g i d xet bo so m d i sau day:

^^

K h i do chi co the xay ra 3 triTdng hdp sau day:

(x, + I ; x 2 ; . . . ; x 2 y ; x 3 0 - l )

'

'

'*

s/

a. NcII x„ = y„ > z,, + 1. V i x„ + y„ + z„ = k nen ta co: x„ + (y„ - 1) + (z,, + 1) = k

up

Chu y rang do X3(, - x, > 2 ma x, > I => Xj,, - x, > 1

/g

ro

khac de thay: (x7 + l ) + x^ + ... + x ^ + ( x 3 o - l ) = x , + . . . + X30 = 2011,

om

nen bp so m d i cung thoa man yeu cau de bai. l?ng v d i bp so'n^y ta c6:

.c

P = (x, + l ) x 2 . . . X 2 y ( x 3 „ - l )

bo

ok

+ l)(x3(, - 1)- X, X3(,] = X2 ... X29 (X3„ - X, - 1).

ce

>2=>P>P

w.

fa

V a y bat dang thuTc nay chtfug to r^ng bp so (x, .....Xjojchifa lam cho P dat gi;i

ww

tri nho nha't. V i the dieu kien can de P dat gia tri Idn nhat la X30 - Xi < 2 hay
xw - X i < 1 => X30 - Xi e {0; 1} tdrc la:

"

Ta

C n g v d i bo so nay ta c6: P = x,

f(x,y,z) = r ( x , „ y , „ z , ) ) = : x „ y „ z „ .

(Chu y d day x„ > y„ > zo)

X, + X 2 + ... + X 3 o = 2 0 1 1 .

Tir X 3 „ - x ,

gidi

xyz phai dat gia tri Idn nha't tren D . Gia suT:
X,
.

X > y S z. V I D la tap hilu han phan tuT (x, y, z) ncn dl nhien ham so f(x, y, z) =

2. X e t bai toan t i m max P

P- P =

0gi>]

Do vai tro binh dang giffa x, y, z nen khong giam tdng quat co the cho rang

V a y m i n P = 1982 <=> c6 29 thuTa so bang 1, va m o t thifa so bang 1982.

Mat

^

,f'

HUdng dan

29 thiira so'

Xet bo so' ( x , , X 2 , X 3 , , )

K i i i n g Vi^t

Nhu" vay chon day chting han 29 so' bang 67 vi mot so bang 68 t h i :

+l).

(1) chuTng to rang bo so (x,, X j , . . . , X 3 „ ) khong lam cho tich P dat gia trj nho

Tir do suy ra: m i n P =

[ivvil

Cty TNHH M I V

^

hoac X3(, = X|
hoac X3(, = X | + 1

Nhir the dieu kien can de P dat gia tri Idn nhat la trong 30 so thi khong diTdc

CO hai so bat ki nao trong chung lai chenh nhau qua 1.
D i e u nay co nghla la phai c6 I so bang a va 30 - t so b i n g a + 1 (1 < t < 30)
sao cho:
t a + ( 3 0 - t ) ( a + 1) = 2011 => 30a + ( 3 0 - t ) = 2011 (*)

Do X() = y,, > Z|, + 1 va Zo > 0 => Xo, yo - 1, Zo + 1 cung nguyen di/Ong, turc la:

,

(Xo, y , , - l , z „ + 1 ) e D .
Mat khac: f(Xo, y,) - 1, z,, + 1) = X|,(y„ - l)(z„ + 1) = x„y(,z„ + x„(y(, - z,, - 1).
Do X(i = y,, > z,i + 1, nen tiTtren suy ra: f(x,i, y,) - 1, Z(, + 1) > X(,y„Z|, = f(x„, y,,, z,,).
Bat dang thijTc thu diTOc mau thuan vdi (1).
Vay trong trufOng hOp a. khong the xay ra.
0. Neu

X(,

:

> jin^'y

> y„ > z„.

X e t bp ba nguyen duTdng ( x „ - l , y o , zL a i c6: f(x„ - 1, y„, z<, + 1) = y „ ( X ( , - l)(z„ + 1) = x„y„Zo + y,)(x„ - z^, - 1)
V i x„ -

Z|, -

1 > 0 (do

X(, > Z() +

1) nen ta co:

l'(X() - 1, yo, z<, + 1) > x„y„z„ => f(x„ - 1, y„, z„ + 1) > f(x„, y,,, z,,).
Ba't dang thiirc nay cung mau thuan vdi (1).

,1 i ,

Vay trong trU'dng hdp b . khong the xay ra.
Ne'u X() - 1 > y,, > Zo. L a p luan nhiT tren cung suy ra mau thuan

' '

T o m lai gia thiet Xo - z,, > 1 la sai, vay (2) dung
Tir (2) suy ra chi cd the xay ra hai kha nang sau:

De thoa man (*) c6 the chon a = 67, t = 29 (vi 30.67 + 1 = 2011)
335


Chuygn d l BDHSG Toan gia tri Idn nha't va gia tri nho nhat - Phan Huy Khii

i. Neu x„ -

Z|) =

Cty TNHH MTV DWH Khang Vigt

0. Ket hdp vdi x„ > y„ > z„, Ihi x,, = y„ = z„ = — .

Dieu nay xay ra khi va chi khi

x=y=z=

^-f::<:.::-.-K,

.-r,.:

k = 0 (mod 3). (Chu y: Xo, yo, z,, nguycn difring).

,, ,

ii. NSu X() - Z() = 1. Luc nay lai c6 hai triTdng help nho sau:
k+2
•

Neu Xo = y o + 1 = Z o + 1 - T O d o ta c6: X o =-—;

•

Neuxo = y() = z,)-i-l.Tijfdotaco: X o = y ( ) =

3'3'3j

Vi the chiia c6 the ke't luan gi ve

y,, = z , , = — ~

k +1

max f(x,y,z) neu sijf dung baft dang thtfc

(x.y,/)eD

,v,

Cosi.

:

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;:

k k k

k—1

2
3
Chii S rang triTdng hdp nay xay ra khi va chi khi k s 1 (mod 3).

J

'

Tuy nhien do k / 3, nen

Vai tro cung lap luan so hoc to ro hieu life trong triTdng hdp nay.
i i 6. Cho X = { 1 ; 2; 3;... ; 2011}. Tim so nguycn diTdng n (n < 2011) nho nhat

k—2

- y - ; z „ = —

sao cho ton tai mot each lay di mot tap hdp con gom n phan tur cua X sao cho
tap hdp con lai c6 tinh chat: khong c6 phan tur nao b^ng tich hai phan tu*

Trurdng hdp n^y xay ra khi va chi khi k = 2 (mod 3).
Tom lai ta di den kel luan sau:

khac.

,

e

Hudng dan gidi

khi k = 0(mod3)

Khi do phan tur con lai cua X bao gom cac phan tuT 1, 45, 46,..., 2011. '

(k + 2 ) ( k - l ) '

khi k = l(mod3)

Vdi cac phan tuT con lai nay chi xay ra hai kha nang sau:

Ta

27

(x.y,/.)eD

a. Xet tich hai phan

(k + l ) ^ ( k - 2 )

khi k = 2 (mod 3)

s/

max f(x,y,z) =

up

27

ro

Nhdn xet:

.c

giai cua M i toan c6 the suy infc tic'p mot each ddn gian nhif sau;

max
(x,y,/)6D

k

k k

3 ' 3' 3

€D.

fa

k-^
f(x,y,z) = — , k h i k > 0 va k : 3.

CO phan tuT nao bhng tich mot phan tur khac vdi 1.
^b. Xet tich hai phan tur bat ki khac 1.
Ro rang tich ay Idn hdn: 45' = 2025 > 2011.
Vi moi phan tur con lai deu < 2011, suy ra khong c6 phan tur nao bang tich
tiir

khac.

thoa man btft d^ng thurc:

n>43

(1)

That vijy, gia s u r ( l ) khong dung, tuTc la ton tai so nguyen du'dng no < 43 ma

ww

Do k > 0 va k 13 nen

la 1. Tich ay chinh bang

Ta se chiJng minh rang moi so' nguyen duTdng n thoa man yeu cau de bai

w.

Dau " = " xay ra khi va chi khi x = y = z = - .

tiif

hdp con thoa man yeu cau de ra.

ce

27

trong do eo mot phan

Nhif vay ta da chi ra mot each loai bo di 43 phan tur, de thu diTdc mot icip

bo

z+ y+ z
3

tur

phan tur thur hai. Do moi phan tur cua phan con lai deu khac nhau, nen khong

cua hai phan

ok

Theo ba'l dang thitc Cosi Ihi vdi moi (x; y; z) e D, ta eo:

Viledotaco:

om

/g

1. DI nhien neu k = 0 (mod 3), titc la k nguycn diTdng va chia het cho 3, thi

xyz <

si

Loai di 43 phan tur sau: 2, 3, 4 , 4 4 .

27

van thoa man yeu cau dau bai.
Xet 43 bo ba sau: (2, 87, 2.87); (3, 86, 3.86);... (44, 45,44.45).
'

27

2. Tuy nhien neu k / 3 thi khong the iip dung difde ba't dang thufc Co-Si de g ' ^
bai toan nay.
L i do d cho da'u " = " xay ra trong bat dang thuTc x y z < — khi va chi khi:

Dat l'(x) = x(89 - x) vdi 2 < X < 44. Ta eo: f (x) = 89 - 2x > 0 khi 2 < x < 44
=> f(x) la ham dong bien tren [2; 44].

' ~ •

Tir do ta eo: 2.87 < 3.86 <... < 44.25 < 2011

»+

Nhif the 43 bo tren gom cac phan tur doi mot khac nhau thuoc tap X. - ' *
Vi ta rut ra no phan tu" (no < 43), nen phan con lai cua X sau khi rut no phan
tur luon chtfa it nhat mot trong 43 bo ba noi tren.

336

,

3137:


Chuygn 6i BDHSG Join gia tri I6n nhat

glA tri nh6 nhat - Phan Huy Khii

Cty TWHH MTV UVVH Khang Vi§t

R 6 r a n g b o b a d o chufa b a p h a n luf c u a d a y c o n , U-ong d o c 6 m o t p h a n tff

p a j 8 . H a m so' f ( x ) x a c d i n h t r e n t a p h d p c a c so' n g u y e n diTdng v a n h a n g i a t r i

b a n g t i c h h a i p h a n tuT c o n l a i . D o l a d i e u v 6 l i v i no la so' n g u y e n difcfng thf^
m a n y e u c a u d e b a i . V a y g i a t h i e t p h a n chiJng l a s a i

Il"(l) = l

(1) diing

c u n g t r e n d o v a diTdc x a c d j n h n h u sau:

K e t h d p l a i t h c o d i n h n g h i a g i a t r i n h o nha't s u y r a so n g u y e n di/dng n nho

r(2n)=:r(n)
f(2n + l) = f(2n) + l

nhat thoa m a n y e u cau dau bai la: m i n n = 43.
2,... T i m

B a i 7 , X e t t a p h d p la't ca 7 so n g u y e n t o k h a c n h a u c 6 c a c t i n h c h a t s a u :

m a x i"(n)
l
1. C h i i n g CO d a n g a, b , c, a + b + c, a + b - c, a - b + c, b + c - a.

Hudiig

a ^! - '

G o i d l a k h o a n g e a c h giffa so I d n n h a t

B6^ d e : f ( n ) c h i n h b a n g s o ' c a c chiT so' 1 t r o n g b i e u d i e n n h i p h a n c i i a s o n . B o

"
HUdng

de du'dc chiirng m i n h b a n g q u y n a p nhvt .sau:

ddn gidi

V d i n = 1. i h e o d i n h n g h l a ta c o : f( I ) = 1

D o v a i t r o b i n h d a n g giiJa a, b , c n e n c 6 t h e g i a suf ( m a k h o n g l a m m a t tinh

so 2 )

•

R 6 r a n g c < 8 0 0 . T h a t v a y n e u c > 8 0 0 => a + b - c < 0 , d i e u n a y m a u thuan

V a y b d d e d u n g k h i n = 1.

v d i v i c e a + b - c l a so' n g u y e n to'.

•

s/
up

ycji i n . :

ro

(1)

/g

R 6 r a n g a, b , c p h a i l a c a c so l e . T h a t v a y n e u t r a i l a i t o n t a i i t nha't m o t

om

t r o n g b a so t r e n l a so c h a n , t h i d o 2 l a so n g u y e n t o c h l ' n d u y nha't n e n c h i co

ok

ce

bo

c a c so c h 5 n , m a c h i i n g l a i k h d c n h a u . NhiT v a y t a c o n h i e u h d n 2 so nguyen

fa

w.

ww

'

,j

X e t k h i n = k + 1. C o h a i tru'dng h d p x a y r a :

-

H o a c la k c h i n ( k = 2 m ) . K h i d o : l ( k + 1) = f ( 2 m + 1) = l(m) + 1.
D o m < k , n e n t h e o g i a thic't q u y n a p suy r a l ( k + 1) c h i n h b a n g so c a c chu"
sd^ 1 t r o n g b i e u d i e n n h i p h a n c u a m c o n g t h e m 1.
M a t khac, k+ 1 = 2 m + 1 m i i :

2 m + 1 2_
1

TiJfc l a : 2 m + 1 = (\|a2...(Vpl

m

, d d a y m = a|tt2...ap la b i e u d i e n c i i a m

k + 1 c h i n h bang so chiir so 1 t r o n g b i e u d i e n n h i p h a n ciia m c o n g t h e m 1.

N o i r i e n g so n h o nha't t r o n g 7 so d a c h o I d n h d n hoSc b i n g 3.
.
(2)

B a y g i d la chi ra d a u " = " trong (2) c6 the x a y ra.
•

t

t r o n g h e n h i p h a n . T i f d o suy ra s d c h i J so 1 t r o n g b i e u d i e n n h j p h a n c u a so

V a y c a 7 so n g u y e n t o d a c h o d e u l a c a c so n g u y e n t o l e .

d < l 5 9 7 - 3 = > d < 1594.

'

•

k h a c n h a u . T i r d o suy r a c a c s o a + b + c, a + b - c , a - b + c, b + c - a d e u la

K e t h d p v d i ( 1 ) suy r a :

*

G i a su" b d d e d a d u n g d e n n = k > i , i i k la v d i m o i / < k , ("(/) c h i n h b a n g

.c

d i j n g m o t t r o n g b a so a, b , c l a 2 , c o n h a i so' c o n l a i l a h a i so n g u y e n to Ic

to chS^n k h a c n h a u t r o n g 7 so n o i t r e n . D o l a d i e u v 6 l i .

'

sd^cac chi? so' 1 t r o n g b i c u d i e n n h i p h a n c u a s o / .

R o r a n g a + b + c la so Idtn nha't t r o n g 7 so n g u y e n t o n o i t r e n

a + b + c = 8 0 0 + c < 8 0 0 + 7 9 7 => a + b + c < 1 5 9 7 .

'

Ta

ij i ,

So n g u y e n t o kUn nha't d i r d i 8 0 0 l a 7 9 7 , v i t h e :

' '

Mat k h a c : 1 = I / 2 d d a y ta d u n g k i h i e u a,a2...a„ 12 d e c h i so g h i t h e o h e c d

tdng quat)

a +b = 800va a < b .

dan giai

T a CO b o d e sau d a y :

so n h o nha't t r o n g 7 s 6 ' d 6 .

T i m giA t r j Idn nha't c u a d .

iL
ie
uO
nT
hi
Da
iH
oc
01
/

2. H a i t r o n g b a so a, b , c c 6 t o n g l a 8 0 0 .

vdimoin=l,

. . ^ p ;
• ;

V a y bd d e d u n g t r o n g tru'dng h d p k = 2 m .
H o a c l a k Ic ( k = 2 m + 1). K h i d o : f ( k + 1) = f ( 2 m + 2 ) = l"(m + 1)

D o m + 1 < k , n e n t h e o g i a thie'l q u y n t i p suy r a f ( k + 1) c h i n h b a n g so c a c

V i dtj c h p n 7 so sau:

chff so 1 t r o n g b i c u d i e n c i i a m + 1 t r o n g h e n h j p h a n .

a = 13; b = 7 8 7 ; c = 7 9 7 ; a + b + c = 1 5 9 7 ,

M a t khac, k + 1 = 2 m + 2 m a :

R o r a n g d i i n g 1^ 7 so n g u y e n t o k h a c n h a u t h o a m a n c a c t i n h c h a t 1. ; 2.
N g o a i r a : d = 1 5 9 7 - 3 = 1594.
Tuf d o t h e o d i n h n g h l a v e g i a t r j n h o n h a t suy r a : m i n d = 1 5 9 4 .

2m+ 2
.0

a + b - c = 3, a - b + c = 2 3 , b + c - a = 1 5 7 1 .
T u - c l a : 2 m + 2 = 3|[32-3pO

m + 1

d d a y m + 1 = (3,32...i

m + 1 trong he nhj phan.
,(^,||

fsJ ,?

D e ' n d a y ta tha'y b d d e d u n g k h i k = 2 m + 1.

la b i e u d i e n c u a

i &/ i
. stS,

'* >


Cty TNHH MTV DWH Khang Vi^t

Chuyen 66 BDHSG Toan gii tr| Mn nhS't va gi^ trj nh6 nhat - Phan Huy Kh^i

Tom

l a i , bo de dung k h i n = k + 1.

\i the: a-^ + 3 = f25a2 + {5r2 + 3) ^ + 3

Theo nguyen l i quy nap suy ra bd dc dung v d i m o i n (dpcm).
,1

>a, + 3 = ((5r2 + 3 ) ' ( m o d 2 5 ) .

Ta lha'y so' nho nhii't c6 11 chiy so' 1 trong bicu d i c n d\idi he nhi phan

1_1__1

(5r2 + 3 ^ + 3 j ; 2 5

2"-l

= 2 ' " + 2'^+ ... + 2' + 2 " =

, ,-,,^1, ,> r, ,.

W TiJf (1) suy ra, noi rieng (a"* + 3): 25

so 2) la so:
N=

(4)

= 2027.

- /'

rj = 2 (do rj e {0; 2; 3; 4 } )

Vay a = 25a: + 13. L a i dat a. = 5a., + r, vdi r, e { 0 ; 1; 2; 3; 4 } .

2-1

=^r(n)< 10.

(1)

r, = 1 = > a = 125a,+ 38.

Do phat hicn ra da'u bang trong (1). la lict kc ra cac so la nho nha't, g;1n nh6

&'

Ihco thi? tir lis nho den kitn la:

Lai dat a = 16t + r v d i r khong am va 0 < r < 15
10

=1023;

lOchCTsd 1
= 1791;

Vay a = 1 6 1 +

IJLJ

1110

1_KJ
= 1983;
6 chu* so' 1 2

urn

i_Lj

= 2015 > 2011.

0
5 chas6' 1 5 cliffsd' 1

lai thu djnh nghla vc gia trj be nhat ta c6:

max

t'(n) = 10 o

n thda man (2).

T i r ( l ) s u y ra, n o i r i e n g ( a - V 3 ) ; 5 . V l the tif (2) c6
Do r,

G

| 0 ; 2; 3; 4} nen suy ra: r, = 3 => a = 5ai + 3.

D;)t a, = 5a: + f:, v d i rj e {0; 2; 3; 4 } , ta c6:
a = 5(5a:

+ r . ) + 3 = 25a2 + Srj + 3.

V

Ta lha'y a cd: 9 + 2.51 = 111 chCT so. Do vay sau khi xda di 100 chu" so tuy y
(1)

< 4, ta c6:

(mod 5) => (a^ + 3) =rl+5 (mod 5)

16

HUdfiig ddii gidi

Do 2000 = 2 .5 , nen ne'u dilt a = 5ai + ri v d i ri la so' nguyen thoa man 0 ^ fi

a' =

Do t nguyen du'cJng nen — — — = h (h = 0; 1; 2;...) suy ra a, = 16h + 1 1 .

Tim min n va max n.

ce

X = a' = 1997 (mod l O ' ) => a' + 3 = 2000 (mod 10^).

16

Xet tap hdp cac so n thu du'dc iCr A bSng each xda di 1(X) ehOT so luy y cua A.

bo

ww

Ta l i m so nguyen dufdng X thoa man:

16

^

du-dng lir 1 den 60 Iheo thiir lit tir nho den Idn, lu-c la: A = 1234...5960

w.

Hii(ing dan gidi

,11-a,

Bai 10. So nguyen A du'dc tao lhanh bling each viet lien nhau cac so' nguyen

fa

b. n viet trong he thap phan c6 4 so tan cung la 1997.

„

^^..^

dau bai.

IJai 9. T u n n so' nguyen du'dng n nho nhat thoa man cac tinh chat sau:
a. n la lap phu'dng cua mot so' nguyen dtfdng.

i^^A"^.wMf^

Nhir vay n = 1413 = 2821151997 la so tir nhiC-n nho nhat thda man yeu cau

ok

l

(6)

Tir do a = 125( 16h + 11) + 38 =^ a = 2()()()h + 1413; h = 0; 1; 2;...

/g

(2)

om

Tom

{1023; 1535; 1791; 1919; 1983).

ro

da'u bilng trong (1) xay ra

o n e

5.

33 f 125a,

7 chCf so' 1

2

(mod 16) = ^ ( r V 3 ) ; i 6 = > r = 5.

Tir (5) (6) di den: 125a, + 38 = 16l + 5 => 33 = 16l - 125a3

= 1919

.c

Vay

9chiIso' 1

Ta

11110

Tir (a- + 3 ) = { r U 3 )

= 1535

s/

0 ( . JjciVf!

1_LJ

up

OLJ
9 chiJsiY

11

(5)

' V i 2000 : 16, nen tir (1) cd (a"* + 3): 16.

nhat,... nen trong b i c u d i c n dxiiVi he nhi phan c6 chiJa 10 chiJ so 1. Cac so ay

II...1

iL
ie
uO
nT
hi
Da
iH
oc
01
/

De y rang: (a^ + 3): 125 va lap luan nhu'tren suy ra:
Tirbddc vado 1 < n < 2 0 1 1

(2)
(r|V3);5.
(3)

ciia A thi con lai so n cd 11 chff so (chi'i y rhng n cd the cd cac chuT so' 0 d^ng

i

dau).

f

Ta lha'y A cd 6 chi? so 0 phan b d nhir sau:
12... 10... 20... 30... 40... 50 51 52 53 54 55 56 57 58 59 60,
De n la nho nha'l, ta chon 5 chiT so dau lien cho n deu bang 0 (nghla la phai
xda bdt cac chiT so khac 0 trong cac so tir 1 den 50). Ta se xac dinh them 6 chOf
. so con lai cho n lijr day .so:

, ,

51 52 53 54 55 56 57 58 59 60
j D e t h a y 6 s d d d l a : 123450.

^»s;tj

..

Vay m i n n = (KX)(X) 123450 = 123450.
341


Cty TNHH MTV DVVH Khang Vigt

ChuySn 6e BDHSG Toan gia t r i I6n nhat vi gia t r i nh6 nhat - Phan Huy KhJi

D c n la \iln nha'l, ta c h o n 5 chu" s o ' d a u t i e n c h o n d e u b S n g 9 ( n g h l a l a p h a i xo-y

(2)

b d t c a c chcr so k h a c 9 i r o n g c a c so l i f 1 d e n 4 9 ) . T a se x a c d i n h t h e m 6 c h i j sg-

f

c o n l a i c i i a n tiT d a y so:

Tir(l)(2) suyra:

5 0 51 52 53 54 55 5 6 57 58 5 9 6 0
Dcthay6sod6

'

X

;

<3)

-•

1a:785960

V a y m o i h i n h v u o n g csinh a t h o a m a n y e u c a u d e b a i , ta d e u c 6 ( 3 ) .

. „ . ' , •

V a y max n::: 99999785960.

.

,

B a y gif* x e t h i n h v u o n g A B C D c 6 a = l-Jl + 2 . X e t n a m h i n h i r o n c 6 t a m l a

•

luan:

O, A|, B|, C|, D i ( x e m h i n h v e ) , t h i m o i y e u c a u d e b a i t h o a m a n . T o m l a i ,

iL
ie
uO
nT
hi
Da
iH
oc
01
/

Binh

V2 > 2 ^ a > 2 V ^ + 2

,1

Q u a 10 h a i l o a n t r c n , l a l h a y d e g i a i c a c b a i t o a n l i e n q u a n d e n v i e c x a c dinh

hinh vuong

gia t r i k i n n h a t , n h o nha't I r o n g ITnh viTc S O H O C ta c a n k e t h d p k h e o l e o giija

272 +2.

cac d a c t h u so h o c c i i a b a i l o a n v d i c a c phifcfng p h a p t r u y e n t h o n g n o i chung

c 6 k i c h Ihu'dc b e n h a t c a n t i m l a h i n h v u o n g

v d i canh

bang

•

B a i 2 . D u n g n m a u d c t o ta'l ca c a c c a n h c i i a m o t h i n h l a p phuTdng sao c h o m o i

c u a b a i l o a n gia t r i h'in n h a t , n h o nhat.

d i n h d e u c 6 b a m a u l i e n t h u o c , d o l a b a m a u c i i a b a c a n h chiJa d i n h d o . T i m
so' n n h o n h a t d e h a i d i c u k i c n sau d a y d o n g thcJi d i f d c t h o a m a n .

B. Diem qua hai todn gid tri Idn nhd't, nho nhat trong hinh hoc to

a. K h o n g c 6 m a t n a o c 6 h a i c i i n h c u n g m i i u .

hdp

b. K h o n g CO h a i d i n h n a o c o c u n g b a m i i u l i e n t h u o c .

B a i 1. T i m h i n h v u o n g c 6 k i c h Ihifdc b e nhift, d e t r o n g h i n h v u o n g d o c 6 the sap

Hiidng

xe'p n a m h i n h t r o n b a n k i n h 1, sao c h o k h o n g c 6 h a i h i n h t r o n n a o t r o n g chiing

giai

Ta

up

Hiti'ing dan

gidi

la g i a t h i c t a . k h o n g c 6 m a t n a o c o h a i c a n h c i j n g m a u , m a m o i m a t c 6 b o n

s/

CO d i e m t r o n g c h u n g .

ddn

c a n h , v a y so m a u p h a i d i i n g i t n h a t la b o n , tiJc l a n > 4 .
M a t k h a c , n e u n = 4 , t h i so b o b a m a u t a o ra t i f b o n m a u l a

.

ro

G i a sur h i n h v u o n g A B C D c 6 l a m O va c a n h a, chufa n a m h i n h t r o n k h o n g cat

C ^ 4 .

/g

n h a u v a d e u c 6 b a n k i n h b a n g 1. V I ca n a m h i n h t r o n n a y n a m t r o n t r o n g hinh

Co b o n b o ba m a u khac nhau d e t o cho 8 b o ba c a n h t a i

c a n h b a n g a - 2 , d d a y A , B | // A B .

tam

.c

difdng t h a n g n o i cac trung d i e m

ciia cac canh d o i d i e n ciia hinh vuong

nguyen

bo
ce

nho. Theo

lhanh bon
li

Dirichlet t o n t a i m o t trong b o n hinh
vuong nho, m a trong hinh vuong nay

A,

A

fa

vuong

AiBiCD,

\ V):

2x/2

V a y n > 4 v a n 91^ 4 , n e n n > 5.

B|

D,

chiJa ft n h a t h a i t r o n g so n a m l a m h i n h

X e t k h i n = 5. V i h i n h l a p p h i f d n g c o sau m a t ( m o i mat b o n c a n h ) , v a l a i c o 12
c a n h , n e n v d i ba't ci? b o n c a n h n a o l u o n l u o n c h p n diTdc i t n h a t h a i t r o n g c h i i n g
t h u o c c u n g m o t m a t c u a h i n h l a p phi/dng. D o d o , d e t h o a m a n g i a t h i e t a . m o i

(

m a u c h i d u n g d e t o c h o t o i d a b a c a n h ( v i n e u k h o n g , tir n h a n x e t t r e n n e u c o

D

tron n o i t r e n ( k h o n g g i a m l o n g quat, gia

mau d o chang h a n d u n g de to it nhat b o n canh, t h i i t nhat m o t m a t c o hai canh

siJfdolaOi.Oj).

d o , t r a i v d i g i a t h i c ' l a.)

D e y r a n g v i k h o n g c 6 h a i h i n h t r o n n a o ( t r o n g so n a m h i n h I r o n ) c i l t n h a u , n t i
O A > 2 .
Mat

khac

Oi,

O2

n e n l a i t h e o n g u y e n l i D i r i c h l e t suy r a c o i t nha't h a i m a u ( g i a suf d o l a m a u
ciing

canh ciia hinh vuong nho bang

342

D i i n g n = 5 m a u d e t o c h o 12 c a n h , m a m o i m a u d u n g d e t o t o i da c h o ba c a n h ,

(1)
do

l a p phi/dng, v a y t h e o n g u y e n l i

t h u o c , tuTc l a g i a t h i e l b . k h o n g t h o a m a n .

B

2N/2 + 2

w.

hmh

chia

ww

A,B,C,D|

A

dinh cua hinh

D i r i c h l e t c o i t nhat hai d i n h co cung m o t b o ba m a u l i e n

ok

Cac

om

v u o n g , n e n c a c l a m c i i a c h i i n g n ; l m t r o n g h i n h v u o n g A | B | C i D | c 6 t a r n 0 vii

a-2

n h m trong

mot

hinh

vuong

nhc

i

xanh ( X ) , va m a u d o (D)), ma m o i m a u d i i n g de to d i i n g ba canh.

, nen lai c6:
343

C h u y S n dS

B D H S G T o a n g i a tri I6n n h g t v a

gia

P h a n Huy

C t y T N H H f/iTV D V V I ! K i i a n g V i j t

Khii

V I moi canh chuTa hai dinh va lai de thoa man gia t h i c l a.

y

ncn hai mau X. D se hen thuoc tdng cong 12 dinh (moi mau

\

f

i
1

lien Ihuoc sau dinh). Ma hlnh hinh lap phUctng chi c6 tarn

di/dng net diirt (

\3
^

c2

(5)

<13

)

'^^ - S < 156 =o
13

) chi mau do,

) dam chi mau xanh).

1

S-i ,

a

!

KV,f {

5

10

13

i

- 13S - 2()2S < 0

V/.

%

S < 52.

(6)

Dang thijfc trong (6) xay ra khi va

X , D. C h i C O kha nang sau xay ra:

chi k h i X| = X2 = ... = X|, - 4, ti'fc la

k

iL
ie
uO
nT
hi

Da
iH
oc
01
/

V d i m o i dinh trong so nay, xet canh thi? ba xuat phat tiif do va khong t o ma,,

moi dong c6 4 6 to do.

10

mau con lai (ngoai X , D ) , nen phai c6 it nhat hai canh ciing mau, va nghla la

Bay gi(:f ta chi ra each to miui nhi/ sau:

dieu kien b. bj v i pham vi c6 hai dinh ciing c6 bo ba miiu lien thuoc gion^

V a y maxS = 52.

nhau.

:

i=l

Tir do, ta thay ( I ) va (5) vao (4), di den:

B o n canh dang xet doi mot khac nhau. Do bon canh nay chi du'dc diing ha

13

f
i

Bai 4.

M o t lam giac deu diftfc chia thanh n^ tam giac deu b ^ n g nhau. M o t so' tam

V d i n = 6, xet each to mau sau day: Do (D), Xanh ( X ) ,

giac do di/dc dcinh so bc'li cac so' 1, 2, 3,... , m sao cho cac tam giac v d i cdc so'

Vang ( V ) , T i m ( T ) , Nau ( N ) , L a m ( L ) .

hSn tie'p thi phai c6 canh chung.

Cach to mau nay thoa man m o i yeu cau dc ra.

T i m gia trj Idn nhat c6 the c6 ciia m.

Ta

T 6 m l a i , neu n = 5, ta luon di den mau thuan. Do do n > 5 1 / ,

s/

Hi((ing din gidi

Chia cac canh tam giac deu thanh n phan bang nhau. Tir cac d i e m chia ve cac

vuong cua bang sao cho khong c6 bo'n 6 do nao nam d bon goc cua mot hlnh

dirdng thang song song vclti cac canh cua lam giac. K h i do so' tam giac deu con

chO" nhat. H o i gid trj Idn nha't cua S c6 the la bao nhieu.

la: 1 + 3 + 5 + ... + ( 2 n - l ) = n '

'^ '
i=i

2

ok
bo
ce

fa

.

Irang xen ke nhau nhir hinh ve. K h i do so' cac 6
n(n + l )
, con .so cac 6
den la: 1 + 2 + 3 + ... + n =
trang la: 1 + 2 + 3 + ... + (n - 1) =

n(n-l)

dirdc danh so' l i e n tie'p phai c6 ctinh chung, do do

Chie'u cac cSp 6 do xuong mot hang ngang nao do. V I khong cd bon 6 do
nao nam d bon goc cua mot hlnh chff nhat nen khong c6 cSp 6 do nao co
hinh chie'u trung nhau.

no phai C O mau khac nhau.

13

^•
i=l

^i(xi-l)

1.3

13

i=l

i=l

<78 hay: X^^-S^i ^^^6.

Theo ba't dang thtfc Bu-nhi-a-c6'p-xki, ta c6:

'

V i Ic do, trong so' cac tam giac diTdc danh so', so' cac tam giac den chi co the

nhieu htfn so' cac tam giac tr^ng la 1. V a y to'ng so' cac tam giac diTdc danh so
m phai thoa man ba't dSng thu'c:

V i the: C f j = 78 > A .
Tir (2) va (3) suy ra:

To mau cac tam giac thanh cac tam giac den,

Theo each danh s6' tam giac thi hai tam giac

>.

ww

V a y tong so'cac cap 6 do l a : A =

=

••\: ,

w.

6 hang thi? i , so cac cSp 6 de c6 the c6 la: C^

.c

^ ,•
, ,
HUdngddngidi
'

'
•
n
GoiXilaso'odetodhangthuri. Taco: S = ^ X i .
i=l

om

B a i 3. Cho bang hinh vuong gom 1 3 x 1 3

/g

6 vuong. NgiTdi ta to mau do d S 6

ro

up

V a y gia t r i nho nhat phai t i m la n = 6.

344

13
1^13

2

i= l

dinh nen phai c6 it nhii't bon dinh c6 ciing hai mau X , D lien

thuoc (xem hinh ve, du'c^ng net lien dam ( —

13

m<2"^"^"^^l,

i, - ^

haym
Ta chi ra mot each danh so' dat dau dang thuTc. M u o n vay, each danh so' nay
[ p h a i danh so diroc het tat ca cac 6 ir^ng. T h i du each danh so nhir hinh ve
345