Chuyên đề bồi dưỡng học sinh giỏi giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 5)
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Chuyen dg BDHSG Join g\i tr| Ifln nh^t va gia trj nh6 nhat - Phan Huy Khil
min
(x;y)eD
P = min<
min P;
[(x; y)eD|
min
(x; y)eD2
P >.
Cty TNHH MTV DWH Khang Vijt
p 0 i 6. Tim gia tri I6n nhat va nho nha't cua ham so:
(2)
J
'
3 + 4X^+3X^
f(x) = -
Do iha'y D, - {(x; y ) : y = 0 va x^ < 3 } , va P = x^ khi (x; y) e D,.
vdi X e •
(l + x ^ ) '
Tirdo suy ra;
Hudiig dan giai
max P = 3<x>x = ± > / 3 ; y = 0 ,
•>>'
(3)
Goi m la gia trj tuy y cua ham so: f(x) =
(x;y)eD|
P = 0 o x = y==0.(4)
...
. u
, 3 + 4x^ H-Sx''
Khi do phifdng trmh sau (an x):
—- m
l + 2x^ +x'*
Xet khi (x; y) e D j . Liic do: x^ + xy + y^ > 0, vi the:
P=
X -xy-3y / 2
^(x^+xy.y^).
X + xy + y
c6 nghipm. Ta c6:
Xet hai kha nang:
1. Neu m = 3, khi do (2) c6 dang x^ = 0, vay (2) c6 nghiem.
-xy-3y ^^:„2
2
2 khi x"^-xy-Sy"^ < 0 .
x^ + x y + y
Do do m = 3 la mot gi^ trj cua f(x).
yj
X
,
+ - +1
s/
r +t+l
t^-t-3
ce
fa
3
w.
"^"^ < m <
3
y
m-3
= 1 > 0 , d day S va P tiTdng tfng la long va tich hai nghiem
m>—
2
2
Tird6tac6: - < m ^ 3 .
2
.3 khi x^ - xy - 3y^ > 0,
3.^:;^^;^
(4)
S>0
(m-2)^-(m-3)^>0
Ta c6: (4) o 2 - m
>0
m-3
ww
Tilfdosuyra: 0
^
A'>0
• cua (3)
bo
t^+t + l
(3)
m-3
ok
\6i t G R.
Lam tiWng tif nhir tren, ta c6:
c6 nghipm t > 0. Dieu do xay ra khi
DoP =
.c
Dc'n day gpi m la gia tri luy y cua ham so:
f(t) =
2. Neu m ^ 3, khi do (2) c6 nghiem khi va chi khi phiTcfng trinh
(m - 3)t^ + 2(m - 2)t + m - 3 = 0
ro
X N2
/g
/
up
t^-t-3
om
" ' - " y - y ,
X,2• + xy + y 2
(2)
« ( m - 3)x* + 2(m - 2)x^ + m - 3 = 0.
X
Tac6:
'
Ta
P^3^
(1)
(1) o 3 + 4x^ + Sx" = m + 2mx^ + mx"*
p ^ ^ x ^ j c j ^ J j ^ khi x 2 - x y - 3 y 2 > 0
X +xy+ y
Do X + x y + y < 3 , nen
-, X G .
(iH-x^)'
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min
(x;y)eD|
3 + 4x^+3x^
Khim = 3,thi(l)c6dang
khix'-xy-3y'<0.
5
o-
-
HiLJl^ =3ox^
( v i m 9^ 3)
'
=0ox-0.
l + 2x^ + x ' '
,
Nhirvay:
! •
max
P = -3 + 4N/3,
(5)
P = -3-4>/3.
(6)
(x; y)€D2
(x; y)6D2
TO (1), (2). (3), (4). (5). (6) di den:
max P = -3 + 4v'3 va
(x;y)€D
min P = - 3 - 4 N / 3 .
(x;y)eD
K h i m = - , t h i ( l ) c 6 dang 3 + 4x^+3x'^ = - 0 x " - 2 x ^ + 1 = 0 o x = ± 1 .
2
l + 2x2 + x^
2
Vay
maxf(x) = 3 o x = 0,
xeR
1 ''
min f(x) = - o x = ± l
xeR
2
237
ChuySn dg BDHSG Join gia tr| I6n nhSt
Nhdn
g\i tr| nh6
nha't - Phan Huy Khai
Cty TIMHH MTV DWH Khang Vi^t
xet:
f^h^n xet:
Me'u ta CO bai toan sau:
1. Ta da suf dung cac dinh l i ve dau tarn thuTc bac hai de t i m dieu kien sao cho (3)
Co
nghiem khong am.
< .. i ,
Chtfng minh rang v d i m o i x > 0, thi f ( x ) = x + \\ + — > 2 . ' " '
2. Ta CO the giai bai toan tren bang nhieu each khac
-
V
Bang phifdng phap chieu bie'n thien ham so' (xem bai 11, b ^ i 1, chiTdng 4
X
K h i do bai toan ddn gian hPn han. That vay:
cuo'n sach nay)
•
Ne'u X > 2, thi ro rang f(x) > 2 yj ,,
Bang phtfcfng phap lu'dng giac (xem b a i 5, chtfdng 3 cuon sach nay)
•
X e t k h i 0 < X < 2, thi ITx) > 2 ! :
o
J x ^ + - > 2 - x o x ^ + - > 4 - 4 x + x^ ( d o 2 - x > 0 )
X
x
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-
Bang phiTdng phap bat dang thtfc (xem bai 5 phan nhan xet, chtfong 3
cuo'n sdch nay).
B a i 7. T i m gia t r i nho nha't cua ham so: f ( x ) = x +
(!)
x^ + x
O 4x^ - 4x + 1 > 0 <=> (2x -1)^ > 0 .
vdix>0.
la&o suy ra: f(x) > 2 V x > 0 va f(x) = 2 o x = 2
Tuy nhien v d i bai toan da cho, thi lam sao ma tim ra hang so' 2. Ci'i kho cua
HUdngddngidi
G p i m la gia tri tuy
cua ham so': f(x) = x + ^/x^ + — , x > 0.
bai toan la d cho do. V i vay dung phi/dng phap mien gia t r i ham so' giai bai
todn nay la each giai tif nhien va cung kha ddn gian!
+ Jx^ + — = m
(1)
up
s/
x>0
Ta
K h i do he sau day (an x):
X
0
2mx
/g
0
(2)
- m ^ x + l = 0 (3)
om
H? (1) tiTdng diTdng v d i he
ro
c6 nghiem.
ok
(1) c6 nghiem o m ' ' - 8 m > 0 o m > 2 (chu y i\i (2) c6 m > 0).
bo
2m
ce
2m
2
2
4
Do do m > 2 chinh la dieu k i e n de he (2), (3) c6 nghiem
Tiif do suy ra v d i m p i X > 0, ta c6: f(x)
0 <> x2.
<2
K h i m = 2, t h i he (2), (3) trd thanh
4x2-4x +l =0
Tilfd6suyra: minf(x) = 2 o x = - .
x>o
2
1
,
'^''^ir-^-^^'
n
Ham so f(x) xac dinh k h i x - x^ > 0 o 0 < x < 1.
1
^
Gpi m la gia trj tiiy y eua ham so f(x). K h i do phiTdng tiVnh sau day (an x)
ssi!
( I ) e d nghiem.
( l ) o Vx-x^ = m - x
(2)
,
'
:
(2) CO nghiem k h i va chi k h i hai di/dng y = V x - x ^ va > = m - x c^t nhau.
y >0
Ta c6 y = V x - x ^ <=> •
o
x2+y^-x=0
w.
X2 < X| < m. /
ww
m
V i X) + X2 = —
fa
Do do k h i m > 2 thi (3) c6 hai nghiem du'cfng Xj, X2 v d i X| > X2.
•
HUctng ddn giai
Tacd
K h i m > 2, t h i (3) c6 nghiem. M a t khac ta c6: P = — > 0; S - — = — > 0
'
T i m gia tri Idn nha't va nho nha't cua ham so tren mien xac dinh cua no.
x + Vx-x^ = m
.c
X
Bai 8. Cho ham so f(x) = x + V x - x ^ .
'
{
1
X
2
2 1
=4-
I
(1
Vay y = Vx - x^ la nuTa dtfdng tron tam tai d i e m I - ; 0
\2
( p h i n n k m tren true hoanh).
^^rf
va ban kinh R = -
. .
y = m - x o x + y = m.
Tir do de dang suy ra hai dudng noi tren cat nhau k h i va chi k h i
nam giiTa hai difdng
x + y = ()vax + y =
••
x + y = in
Vay
(2)
nghiem
CO
Cty TNHH MTV D W H Khang V i f
gia trj nh6 nhat - Phan Huy Khii
Chuyen dg BDHSG To^n gia trj I6n nha't
f
-
Vay
y
m i n f ( x ) = min{f(()); 1(1)) = min{(); 0} = 0 ,
0
(tuTc
o
0
1-
max r(x) = f
(1) CO nghiem)
0
< m <
/(2 + V 2 ) ( 2 - N / 2 ) ^ 2 + >^ ^ V2 ^ V 2 + I
=
4
do suy ra
min f ( x ) = 0
I
max f ( x ) =
I
2
D o ( ) < x < 1 = > d a t x = sin'(p ( 0 < (p < ^ ).
T f f d 6 x + V x - x ^ =sin^(p + 7sin^(p-sin'*(p = sin^(p + -y/sin^ (p(l-cos^ (p)
1-cos 2(0 sin 2(0
1
yjl
7l'
- +
= - + —cos 2(p2
2
2
2
^ 1. Cach giai tren la sif phoi hitp kheo leo giffa hai phffdng phap m i e n gia iri
s/
V 1 0 < ( p < ^ =^ _ Z ^ < 2 ( p - i ^ < ^ ^
2
4 ^ 4 4
up
2. X e t each giai khac sau day (bang phiTdng phap chieu bien thien ham so)
iyi 'am
x^ + 1 - 2 X
l-2x
Ta CO f ' ( x ) = 1 +
zVx-x^
2Vx-x^
Ta
ham so \k phifc^ng phap do t h j .
•
1
,
,
2Vx-x^ - ( 2 x - l )
X c l khi - < X < 1, luc do Vict l a i f ' ( x ) =
2V X - X
ok
.c
om
/g
V d i O < x < - = ^ l - 2 x > 0 = > f'(X) > 0 .
fa
4(x-x^)-(2x-l)^
w.
ww
+ 2x -1)
D'AU cua f ' ( x ) la diiu cua 8x - 8x^ - 1.
f *
Tff do ta CO bang bien thien sau:
f'(x)
f(x)
0
y
^
1^
+
0
Cue- i»w/2 //?zc/j ciich liicii nao?
r
•
HUdng ddn giai
G o i m la gia t r i tijy y ciia P tren mien ( x ; y ) thoa man dieu k i e n cho tri/(?c.
\^+y^=m
(1)
(x^ - y^ +1)^ - 4x^y^ - x^ - y^ - 0 (2)
1
2 + 72
2
4
0
+
0
+1
min f ( x ) = 0
2
Nhir vay he sau day(an x, y )
Nhir vSy dau cua f ' ( x ) k h i ^ < x < 1 !a dau cua 8x - 8x^ - 1.
X
0
^
T i m gia tri Idn nhat va nho nha't cua bicu thtfc P = x^ + y l
8 x - 8x^-1
l^-x^ilJT- x^ + 2x - 1 ) 2J^- aJ^-
N h f f v a y max f ( x ) = ^ ^ ^ ;
y^ = o.
ce
= > 2 x - 1 > 0 , lufdo t a c o
<2
Bai 9. Cho x va y la cac so thiTc thoa man dieu k i e n (x^ - y ' + 1)^ + 4x^y^ - x^ -
bo
• J:
-:^
0 < - + —cos 2CP-2
2
ro
1
•
. 1
Khi-
(do sincp > 0; cos(p > 0)
= sin^cp + sin(pcos(p
x+y=0
Nhqn xet:
f(x) =
2
'
|. L a i xct each giai sau (bang phu'ctng phap Iffdng giac hoa)
va
a
^ 4 ~
Ta thu l a i ket qua tren!
()
()
4
16
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Tif
2 + V2
Tac6(l)(2)<=>
-
CO nghiem.
x^ + y ^ = m
(x^ + y^)2 -3(\ + y^) + 1 + 4x2 = 0 j^v^,
fx2+y2=m
m 2-3m +l + 4x2=0
(3)
(4)
Cty TNHH IVITV DVVH Khang Vi$t
Chuy§n dS BDHSG Toan gia trj Idn nhat va gia tri nhd nha't - Phan Huy Kh^i
3 + >/5
2
3-N/5
(4) (an X ) CO nghiem <=> m" - 3m + 1 < 0 <=>
2
Nhirvay(5)la dieukicnde (4)c6nghicm.
Ta
CO
Qlum^^ PHIfONG PHAP SO DUNG DO THIHOAC HlNH HOC
(5)
^
*'
(3) <=> 4x^+ 4y^ = 4m <=> -m^ + 3m - 1 + 4y^ = 4m <=> 4y^ = m^ + m +
Dt TIM GIA TRj LON NHAT VA NHO NHAT CUA HAM SO
i
phiTrtng phap nay dSc biet thich help vdi nhffng bai toan tim gia trj Idn nhat,
Do m^ + m + 1 > 0 V m , nen lif do suy ra he (3) (4) c6 nghiem khi va chi khi
'i
nho nhat cua ham so, trong do cac bicu ihifc va dieu kien cua bai toan ban
!•
V a y (5) chinh la dieu kien de he (1) (2) c6 nghiem.
I
1(
dau da tiem an nhffng ycu to hinh hoc ma thoal lien ta chffa nhin ra no.
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thoa man (5).
. i f
Vdi Idp bai loan nay, ta lhff(:(ng sff dung cac tinh chat hinh hoc sd cii'p sau
„
3 + 75 .
. „ 3-75
Tif do suy ra maxP = — - — va mmP = — - — .
day:
-
Ml •
Nhan xet: B a y gid xet each giai sau day:
Theo iren dieu kien da cho v(3i (x; y) c6 dang
y^f -
+ 1 + 4x^ = 0
3(X^ + y 2 )
x^+y^
O
' x^1+y^2 —3V: = 5- - 4 X ^
4
2,
-2J
(7)
75 _
75
3+
^
2
+y
2
3+ 75^
= — —
X
Cho diem M d ngoai dffdng thang d cho Irffdc. Khi do do dai dffdng vuong
3+
.c
ok
75
It!}
2
+y =
giac CO chu vi vii dicn tich Idn nhat.
Ngoai ra ngffdi la cung dung cac kicn ihffc sau day:
i ' Dung tinh chat ciia do dai vecld, hoac tinh cha't cua ti'ch v6 hffdng cua hai
w-vecld.
. S«^V*1
, . :,
Dfing phffdng Irinh dffdng va mtit va cac cong Ihffc tinh khoang each cua
hinh hoc phSng va hinh hoc khong gian.
,
(U
.
Tim gia tri nho nhat ciia bieu ihffc
p
P= .
n
X"
r
+-r +
r
n
r
• '
1<
HUiiiii' dan viai
i€x cac vecld sau: li = x ; -
3-75«
ertH>'('^"M*-MiM. ^
Cho X, y, z la cac so ihffc dffdng va thda man dieu kien x + y + z < 1.
1* t |/] .IS '
.|J H '
Trong cac lam giac cimg noi ticp mot dffdng Iron, thi lam giac deu la tarn
" a i 1. Be thi tuyen sinh Dai hoc, Cao ddnf' kho'i A)
x=0
x =0
2
/g
x =0
x =0
*
(8)
w.
3+
ww
p
-
I
bo
3-75
., y -•(•• '2
^ 2
2 ^ 3 + 75
< x'' + y^ < — - —
ce
2
—
fa
3-75
75
om
2 2 3
T a laithay(7)<:^ - — < x ^ + y ^ - - <
ro
Dau bkng trong (7) xay ra o x = 0
<=>
Trong mot tarn giac. long hai canh luon luon Idn hdn canh Ihff 3.
Ta
5
<-.
4
3
^ntlc'le-
dffdng thang d.
up
2
'•
-
-
2
'
goc kc tff M xuong d ngan hdn moi dffdng xien ke iff M xuo'ng ciing
(6)
4
Tur (6) suy ra
B la dffdng c6 do dai nho nhat.
s/
(X^ +
Trong ta'l ca cac dffdng noi hai diem A, B cho trffdc Ihi dffdng thing noi A,
V
3-75
V
Fa
CO u + V + w =
,
V =
f
,
f
1'
/;
Xy
1
1
1
N
'i
n
/J
t i i d " , } UtKi.l «J!i!'ft !f''.«
, •
x + y+z; — + — + X
y z
VaymaxP=
Theo tinh chat ve do dai cua vecld long, la c6 u
T a thu lai ket qua tren. Phep giai nay c6n ddn gian hdfn ph6p sur dung mi^"
Tff(l)c6:
+ V +
w
> U+
V +
w .(1)
gia tri ham so.
O/l
1
Cty TNHH MTV DWH Khang Vi?t
Chuyen ai BDHSG To^n gia tr| Ifln nhft v i g\i tr| nh6 nhflt - Phan Huy Khii
(dal t = ( X + y + z)'. Do 0 < x + y + z < 1 => 0 < I < 1).
'I
1
(x + y + z) + — + — + X
v = k2W,k2 > 0 .
'1-.'
Ta c6: f'(t) = 1 - — , va bang bien thien sau:
y
u, v, w la cac vectd cung phiTdng, cung chieu
'
(3)
:\iu>,ii int't ri^'^rhb ;:5iim) ,0,': 'mM' uw-i
1
1
= 81(x + y + z)^ +
-
ri
1
n
y
zj
x
fi',;
Theo bat dang thuTc Cosi, ta c6
VAb
1
'1
1
^1
iV
—+ —+
-
X
z)
y
1
.1 h'un ipj'O
-
1
1^
-
>9.
HUifiig dan giai
ro
Cho ham so f(x) = x + yfi-x^.
nay trcn mien xac dinh cua no.
om
.c
Hiiifng dan giai
ok
bo
ce
fa
4. Tim gia tri nho nhat cua ham so f(x) = Vx^ - x + 1 + Vx^ - V 3 x + 1 , vdi
w.
e R.
ww
X
Hiidng dan giai
f(x)=.
1 1 1
9
Ap dung bat dang thuTc Cosi cd ban, la c 6 : - + - + - >
X y z x+y+z
81
(x + y + z)
2 •
^
X
1
2J
N2
(10)
j.
Viet lai ham so' f(x) diTdi dang sau day
. j
> (x + y + z)' +
Tim gia trj Idn nhat va nho nhat cua ham so
sach nay.
bien thien ham so nhi/ sau:
ZJ
.
mien gia trj ham so de giai bai toan trcn trong bai toan 1, §1, chU'dng I cuon
2. Thay cho phiTdng phap bat dang thtfc ta c6 the sOr dung phiTdng phap chieU
-
„
Bai 3. (De thi tuyen sink Dai hoc, Cao dang khoi B)
phi/dng phap siir dung ba't dang thtfc de giai bai toan dat ra.
—+ — +
X
y
,
>
Xem Idi giai ket hdp giiJa phifdng phap do thi, hmh hoc va phifdng phap
1. Trong bai trcn ta da kct hdp phiTdng phap suf dung vectd trong hinh hoc phang
V i the (x + y + z) +
i tr . ;
(8)
(9)
Nhirthe minP = >/82 o x = y = z = ^ .
1
+ |y - 2| .
•'
bai loan nay. Xem Idi giai trong bai 13, §2, chi/dng 1 cuon sach nay.
<=> dong thcfi co dau bang trong (3), (5) (6)
1
h':
Ta kct hdp phi^dng phap suT dung vectd va chieu bicn ihien ham so' de giai
(7)
> 162.
o x = y = z= - .
(l
.^^.si^'K^'^Ai^-piih
thi tuyen sinh Dai hoc, Caoddnfi khoi B)
V; 1
ZJ
-
Nh4n xet:
^.
P = V ( X - l ) ^ +y^ + ^{X + lf+y^
/g
i
I''
1
Cho x. y la cac so thifc tiiy y. Tim gia tri nho nhat cua bicu thtfc
(6)
, TCr gia thiet 0 < x + y + z < I = > 80(x + y + zf < 80.
T i r ( 2 ) ( 4 ) ( 7 ) ( 8 ) t a c 6 P > >/82.
fiiii2.(De
(5)
TJ
—+ — +
X
y
1
Ta thu lai kct qua trcn.
up
Ttr (5) (6) suy ra 81 ( X + y + /.Y +
1
—+ —+
X
y
Thco bat dang thuTc Cosi cd ban thi (x + y + z)
A Dau bang trong (9) xay ra
.
1 'i0M> (ii
81(x + y + /.)' + [ - + - + - >18(x + y + z)
U
y 7.)
1
Vaytuf(lO) suyraminP= ^^min f(t) =
- 8 0 ( x + y + z)^ (4)
— + — + -
,
Ta
1
—+ — +
s/
(x + y + /.)' +
I'd)
1(1)
Dc thay
,
0
t
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Dau bang trong (2) xay ra o
X.,... ,1 t,
^ I
(2)
+
X -
+
—
Xet he true toa do Oxy, va tren do xet cac diem A
(1)
va
C(x;0),xe K.
81
Xet ham so f(l) = t^ +
y,
Khi do tir (1), ta c6 f(x) = CA + CB.
d day 0 < t < 1.
245
Chuyen dg BDHSG ToAn gia tri I6n nhat va gia tri nh6 nhS't - Phan Huy Khai
Cty TNHH MTV D W H Khang Vi§t
R6 rang ta co C A + CB > A B , trong do
-> O M „ = OC
+
AB =
2
1
V^^'
siin45|^^N/2
I
sin75" • 2
1 +cos 30
,0
73
V2 + 73
V
2
Nhir the ta C O f(x) > N / ^ Vx G E .
Goi C„ = A B n Ox. Ta c6 C„A + C B
= 74-273
(2)
AB.
=J(73-1)"
Ti( do suy ra minICx) = 72 o
Nhir vay ncu dat Xo = OC,, , thi l'{xn) = >/2
=
73-1.
rt>;yrtq
o.^{^U-wwrt,:-;
x = 73 - 1 . Ta thu l a i kc't quii trcn. '
xeR
iL
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nT
hi
Da
iH
oc
01
/
Nhif vay la eo min =
1+
<=:> x = Xd = 73 - 1.
4. Cac ban Ihijr x c m eo the giai bai toan trcn bang each khac ma l a i dOn giiin
hdn hai each tren khong?
Nhqn xet:
\ '
1. X|) C O the linh nhif sau:
,c.
Bai 5. Cho x, y la cac so thiTc thoa man dieu k i c n < x + y + 2 > 0
DifcJng thang qua A B eo phlfdng trinh
y+
]
X
- r^! ,
73
-
i n ! m-^X .^m
, •
nhui
2. R6 rang phu'rtng phap do ihj to ro day uy life eiia no qua RJi giai trcn.
< 0, k h i do ta c6
om
X
••:(*>Y
i " ( - x ) - V x ^ + X + 1 + Vx^ + N/3X + I. "
(**)
bo
V i the do chi quan l a m den m i n f ( x ) , ta chi can xet khi x > 0.
I
S
noi
tren
chinh
la
giac A B C
j>f
K h i do ta eo: I'M ^ MA+ MC > AC = sfl
^
Thco dinh ! i ham so sin trong A M O C , ta c6
ba
max
' ^ i ! n.f
OM,
sin 75"
sin 45"
max
OM'
MeAABC
2
= max O A ^ ; OB^; OC^} = max {20; 16; 4} = 20 = O A ^ ,
minP =
min
O M ^ = O H ^ (d day ta ke O H 1 B C ) "
'"
-'t'
"
MeAABC
Theo cong thi?e linh khoang each tif O den difdng thang y - 2x - 4 = 0, ta c6
oh j i o i :)i.n,} M .
OC
Ro rang
(x + y ) =
(x;y)eAABC
X c l diem M sao cho M O C = 60"; M O A = 30", v i O M = x.
f(x) = N/2 <=> M = M„ e A C .
ca
maxP=
Do do ap dung djnh l i ham .so cosin, xet h m h vuong O A B C vd'\A = OC = '
: r
diem
Ta c(') x ' + y ' = OM^.
T a c o r(x) = J x - - - x . l . c o s 6 0 " + l - + J x 2 - - x . c o s 3 0 " + l ^ .
\„i
"'^
A = ( - 4 ; 2), B = (0; 4) va C = ( - 2 ; 0).
^
-'-a
w.
^
''''Wb ' I \ t . :
^
canh), trong do
ww
I
^'
tap
y)
(kc
ce
fa
Do X < 0, ncn lit (*) va (**) suy ra IXx) > f ( - x ) .
thay
M(x;
lam
ok
| - { x ) - V x ^ - x + 1 + V x ^ V 3 x + l , •.tjiri
'
^• - - -
''^
kien da cho.
hdp cac
.c
Ncu
^
c' -
phang toa do thoa m a n he dieu
De
/g
3. X c l C c i c h gi£u bang phifdng phap hinh hoc sau day
Ta
x„ + — =
~ =>xn= yfi 2
2(1+ V3)
s/
Cho y = 0
+ y^
Hu
Goi M ( x ; y) la d i e m t r c n mat
2
up
2
ro
2
•
. ,1
[y-2x-4>0
vJv»-.
T i m gia t r i Idn nhat va nho nhat cua bieu thiJe P =
' "yf"
2
-x + 2y-8<()
0H =
^^.(OP^
K !
i
247
Chuy6n
BDHSG Toan giA trj I6n nhal
yia In nh6 nhift - Phan Huy Khii
Cty TNHH MTV DWH Khang Vigt
8
V a y maxP = 20 <» M = A
Binh luqn:
x = -4
o
y = 2
16
m i n P = = — < = > M = H<=>
; .
5
;
X
=
Khi (x; y ) thoa man (1), ta viet lai bieu thi?c P drfdi dang
—
5
4
8 x ^ ^ x _ ^ y _ ^ ^ 3 ^ l
-
2
2
,
,
2
^
rkng neu ve he true tpa do Oxy,
K h o n g the c6 phiTdng phap khac hieu qua hdn phu'dng phap do thi
(2)
^
va hinh hoc gian bai toan nay
; ,
,»
2x + y > 2
:
d day M ( x ; y) thoa m a n (1). •
iL
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Da
iH
oc
01
/
A^^rtjcef: X e t bai toan tiTdng ti/sau:
.
TCf (2) suy ra maxP = 8 + - m a x O M ^ ,
x + 3y < 9
Cho X , y la hai so' thifc thoa man he dieu k i e n :
x > 0; y > 0.
{ r > > t . . . . ^ f < - . i . . 3,.... t
minP = 8 + - m i n O M ^
T i m gia t r i Idn nha't va nho nhat cua b i c u thtfc P = x^ + y^ - 4x - 8y.
^ t*M'*'-,:xl'-
2 MeQ
Hudiigddngiai
a day Q 1^ difdng tron t a m 1(4; 3)
Cac d i e m M ( x ; y) thoa man he da cho la toan bo ti? giac A B C D v d i A = (1;
9).B = (0;2), •
va ban kinh R = 3.
C = (0; 3), D = (9; 0).
V i e t lai P diTdi dang: P = (x - 2)^ + (y - Af -
De thay m a x O M ^ = 0 M ^ ; m i n O M ^ = 0 M ?
20.
Men
up
'1
1
1
«
om
fa
1
M = H
3
5
<i>x = — ; y = —.
2
2
=
32
24
y =-
minP = 1 0 o M s M ,
=
o
5
\
1
(*)
-
(**)
' = -59
Cac ban c6 the tU" nghiem lai cac ket qua (*) va (**) mot each de dang.
^Hn
xet: Ta co the
SIJT
dung phiTdng phap lifdng giac hoa de giai bai toan tren
" h l 'sau:
Tsan-
Tiif ( I ) suy ra
'''^i
X
-
4 = sin 9
,
•
vdi 0 < (p < 271
y - 3 = 3cos(p
T i m gia trj Idn nha't va nho nha't ciia bieu thuTc P = 4x + 3y.
Tiif do P = 4x + 3y = 4(4 + 3sincp) + 3(3 + 3 coscp) = 25 + 12sin(p + 9cos9
Htidiig ddn giai
V i e t l a i dieu k i e n da cho diTdi dang (x - 4)^ + (y - 3)^ = 9.
2=10
8
B a i 6. Cho cac so thiTc x, y thoa man dieu k i e n : x^ + y^ + 16 = 8x + 6y
*
M = M2 o •
X
bo
B2
= 9 ; y = 0.
minP = - —
Nhir the maxP = 40 o
ce
M = D
I
w.
o x
-A
C3
T i r d o suy ra
maxP = 45 o
4
X
.c
cua I tren C D .
Tir do suy ra maxP = 8^+ 32 = 4 0 ; min P = 8 +
ok
^ d day H la hinh chic'u
/g
m i n M I ^ = H I ^ -' 2 ^'
ww
i
(xem hinh ve)
Do 0 1 = 5 => OM2 = 0 1 + 3 = 8; O M , = 0 1 - 3 = 2.
ro
De thay m a x M I ' = D I ' = 65;
:
d day M | , M2 Ian li/dt la cac giao diem cua 0 1 va difdng tron
s/
ABCD.
Ta
Gpi I la d i e m I = (2; 4), k h i do P = M l ' - 20, d day M ( x ; y) thuoc tiir giac
•
Men
C)
(1)
(3)
Ap dung ba't d^ng thufc quen bie't: V a , ta c6:
Tilf d6 suy ra cac d i e m M ( x ; y) thoa man (1) n ^ m tren diTdng tron tam ta'
~-\/a^ + b^ < a s i n a + b c o s a < Va^ + b^ .
d i e m 1(4; 3) va ban k i n h R = 3.
T a c 6 : - 1 5 < 12sin(p + 9cos(p< 15.
*
^"^^
Cliuyen
BDHSG To^n gia tri I6n nha't va gia trj nhd nha't - Phan Huy KhSi
Cty TMHH MTV DWH Khang Vlgt
Bay gic( liT (3) (4) suy ra maxP = 40; minP = 10.
xet: Ta c6 the giai bai toan tren bang phiTdng phap mien gia trj ham so
Ta thu lai ket qua Ucn (vdi phep giai raft gpn gang).
X + 3y-10>0
Bai 7. Cho x va y la cac so' thifc thoa man dieu kien •
X
+y- 6<0
X
- y + 2 > 0.
nhu^sau:
: ,•„ ,.v,.;;,r
\
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hi
Da
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01
/
'
\.
,^
'
: \•
/
[x-y+2>
CO nghiem.
'
JM.J
'•• \-- • • !'
-y+ 2 = 0
X
•
f^, j j ^ , ^ . j j
~
' ^ V i l M ' ( l ) ( 2 ) (3) ( 4 ) «
,
y+
ra-10>0
-y + m - 6 < 0
o
y>10--m
(6)
y>m-6
(7) ''M-
y ^ i ^ .
(8) «^^-^.'^
Ta
,
'
Ne'um>8thi 1 0 - m < 6 - m = : > ( 6 ) ( 7 ) o y > 6 - m
y>10-m
ro
i ,
/g
,
'
•
y<-
A ^
• '
r^
m + 2 khim<8
.
,
Vay ( 5 ) ( 6 ) ( 7 ) ( 8 ) o
y >m- 6
om
lit;
'
Neu m < 8 thi 10 - m > m - 6 => (6) (7) o y > 10 - m
.c
i'
\r + IS \
•
up
s/
/
1^-t)
(4)
x = m - 2 y (5)
= m-2y
Xet he (6) (7) ta CO
/
(1)
T u f ( l ) s u y r a x = m - 2y.
•
-3y + m + 2 > 0
/
fX + 2y = m
^ ^'.^^^'.1 i
Goi m la gia tri tiiy y cua P, khi do he ,sau day (an x; y) • ^ ^ "^^ 10 > 0 (2)
x +y-6<0
(3)
Tim gia tri Idn nha't cua bieu thiJc P = x + 2y.
.\ dhn giai
;
rn + 2 k h i m > 8
ok
Gia su" M(x; y) la diem, trong do x, y thoa man he dieu kien da cho.
bo
De dang tha'y rang khi do mien D cac diem M(x; y) nhu" vay du"dc bieu dicn
ce
tren mat phang toa do xOy la toan tarn giac ABC, vdi cac dinh A(4; 2);
w.
- Tim gia trj Idn nha't va nho nha't cua tham so m, sao cho diTdng thang x + 2y
= m cat tam giac ABC. Bang each tinh tien difdng thang x + 2y = 0 (theo
chieumuiten)
•'
m +2
^10-mo8>m>7
'•,
ww
Bai loan da cho trcf thanh:
fa
B(2;4)vaC(l;3)
filf d6 suy ra khi m < 8, he (5) (6) (7) (8) c6 nghiem khi
fm«'l---x1
I C6n khi m > 8, he (5) (6) (7) (8) c6 nghiem khi
Nhir vay he (5) (6) (7) (8) c6 nghiem (ttJc Ih he (1) (2) (3) (4) c6 nghiem) khi
chi khi 7 < m < 10.
Dieu dau tien ma difrfng thang x + 2y = m gap tam giac ABC chinh la die'm
>m-6ol0>m>8
/ w
^
;
'
Tir do suy ra maxP = 10 o x = 2; y = 4; minP = 7 o x = : l ; y = 3
^^nh luqn: R6 rang phuTcfng phap giai bang do thi va hinh hoc to ro hieu qua
Khi do m = 1 + 2.3 = 7. Diem cuo'i ciing ma du'dng x + 2y = m gap
chinh la diem B(4; 2). Khi do m = 2 + 2.4 = 10.
Vay maxP =10<=>x = 2;y = 4; min P = 7 < = > x = l ; y = 3.
AAB*-
hcfn h^n so vdi phifdng phap mien gia tri ham so.
j
^ ^ i 8. Cho X, y la cac so' thiTc thoa man dieu kien sinx + siny = ^ .
Tim gia tri Idn nha't va nho nha't cua bieu thtfc P = cos2x + cos2y.
ai BDHSG Toan gia tr| I6n nhS't vi g\& tr| nh6 nhSit -
Phan Huy
Hiidng ddn gidi
Khai
Cty TNHH MTV DWH Khang Vi§t
xet: Ta c6 the suT dung phuTdng phap "chicu bien Ihien ham so " de giai bai
toan tren nhU'sau:
Dat u = sinx; v = siny.
Khi do ta c6 cos2x + cos2y = 1 - 2sin^x + 1 -2sinV = 2 - 2(u^ + v^).
Bai loan da cho trd thanh:
\'y A-.:.,-':!
\ .
Tim gia tri Idn nha't va nho nha't cua bieu thiJc Q = +
.fir::::.
1
.
Ttf ( l ) s u y r a m oui l+i evn=h —
e sau
maxP = 2 - 2minQ, (3)
1
1
Tifu + v = - = > v - — u
(I)
2
; Do do
fa
ce
4
(d day H 1^ hinh chicu cua O ircn AB).
Tiirdd suy ra:
1
1
u
=
—
sinx
=
—
max? = 2 - - = - o M s H < : : >
4
4
:i lit
4 4
1
1
smy = V = —
4
u = —;v = l 4sinx = -—;siny = l
2
M
s
A
m m ? = 2 - - = — <=>
2 2
1
MsB
1
u = 1; V = —
sinx = l;siny = -—
2
w.
8
ww
MeAB
s/
Ta
/g
ok
bo
min OM^ - O H ^ 4=
-1
B
om
2
MeAB
Ta CO ngay
max OM^ - O A ^ = 0 B ^ = 1 + - = - ,
MeAB
u
up
"\
1 O
1
.c
MeAB
u
^
Ta CO f'(u) = -8u + 2, va c6 bang bien thicn .sau
ro
-1
i<:>
T
-^
A
Do do
2
maxQ= max OM ; minQ=: min OM
-' •
|\
-i
= 2u*- - u + 2 - 2(u- + V ' ) = 2 - 4u^ + 2u - - = -4u' + 2u + - .
1
TH do suy ra max? = max f(u); minP = min f(u) vdi l(u) = -4u^ + 2u + 3
'V
2
2
-1
T
,'>
i
•
-1
3 o —1 < u < l .
1
<^
-1<--U<1
—
2.
•
1
— u
KKhi do u- + v' = u +
minP = 2 - 2maxP.
(4)
Ve he toa d p Ouv. Khi do tap cac diem (u; v) thoa man (2) chinh la doan
thang AB (do la phan du'dng thang u + v = ^ nam trong hinh vuong canh 2
trong hinh ve dU'di day).
De thay A = K 2 conB =
t •
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Da
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oc
01
/
Chuyen
u
i1
4
~2
f'(u)
f(u)
+
-—-
'1' = -<=>
7
Vay maxP = f
v4.
4
'il'::i
0
•
^
»
1P
u=—
4
^=4 '
u =— ; v =l
2
= — <=>
minP = min< f . 2 ;f(l)
2
21 21
u~l;v = - - .
2
- min<
: —gang va ddn gian!
Ta thu lai ke't qua trcn. Phep giai
cung—gpn
fiaf 9. Cho bon so thi/c x, y, z. t thoa man dieu kien x + y = 6; + t^ = 1.
""im g'la tri nho nha't cua bieu thtfc P = x' + y^ - 2xz - 2yt
Htidngddngiai
t, ,
,
Xet he true toa do Oxy. Khi do diem M(x; y) vdi x + y = 6 n^m trcn dir5ng
i h ^ n p X + y = 6; con diem N(z; t) vdi r + t^ = 1 nlm trcn diTiIng tron ddn vi
A + y^ =; 1 tam tai O va ban kinh bhng I
.J
Chuyen dg BDHSG T(rfn gi^ tr| I6n nhtft
g\i trj nh6 nhift - Phan Huy Khi\
Cty TNHH MTV DWH Khang Vi^t
d day H la hinh chie'u cua O
Vie'l l a i bieu tMc P diTdi dang
P =
(X -
zf + (y - t ) ' - (z' + r ) =
(X -
zf + (y - t ) ' - 1
(1) „
tren dtfdng thang X + y = 1.
/. ,
72
' ^'^A
Ro rang do O H = — , nen ta c6 t i r ( l )
Tir(l)suyra
P = MN^-1.
(2)
V a y minP = ( m i n M N ^ ) - 1,
P= Vx^+z2+Vy' + t 2 > ^ .
M ( x ; y)
Ihuoc
di/cJng
thang
X+
y t= 6, con N(z;
c> O M , M N cung phu-png, cung chicu va N = H .
.
iron ddn v i . K e ,
.
0M„ 1 di^ng
\g
X + y = ^'^'I'a?'
f;Tac6:
; '
OM = (x;y); MN=:(y;t).
Do vay O M , M N cung phiTPng, ciing chieu va N s H k h i va chi k h i ''^ " '
(3)
y = t.
va gia suf , .
Mat khac N(x + y ; z + t) = H
OMo cat dirc:fng
s/
N„.
up
K h i do CO ngay
(3)
/g
M„N„ = 3 ^ - 1 => MoNf, = 19 - 6^2 .
_ ^
i
^
^
ok
o M = M„;NsNo<=> r
X = z= a
Tir (3) (4) suy ra
2
X = z= a
[z = l = 3
Vay minP = ^
ce
Nhgn xet: LcJi gitii bkng phifdng phap do tfii that trong sang va gon gang!
fa
ww
each giffa hai d i e m tren mat phang tpa do.
w.
+ t^ = 1 vii bieu thtfc P gdi ta ngliT den khoiing
T i m gia tri nho nhat cua bieu thiJc P = 7x^ + z ^ +
•
•
- VK!; ...
f^^^,^g
giai
2
X,
y, z la ba so thifc dtfdng va thoa man dieu k i e n : xyz(x + y + z) = 1.
Tim gia trj nho nhat cua bieu thiJc: P = (x + y)(x + z).
* .t
s^' ^-^i"
Hudng ddn giai
T r 6 n mat phang toa dp xOy xet hai diem M ( x ; z) va N(x + y ; z + t)
^ t n tam giac A B C c6 3 canh
Do (x + y ) + (z + t) = 1, nen diem N nhm tren diTdng thang x + y = 1
'^B = X + y ; A C = X + z; BC = y + z.
T a c d O M = Vx^+z^ , M N = ^(x + y - x ) ' + ( z + t - z ) ^ =^|y^+l^
Ta
CO
O M + M N > ON > OH
(1)
I
c6 the giai each khac l a i hay hdn each giai tren khong?
'^^J11. Cho
•
'
1
1
vdi 0 < a < y = t = - - a .
2
^Hn xet: Phu-png phap do thj to ro hieu qua trong viec giai bai l o a n tren. Cac
hn
10. Cho bon so thiTc x, y, z, t thoa man dieu k i e n x + y + z + t = 1.
« •
2
Scr d i CO I d i g i a i nhif vay, v i trong bai toan cai hon " h i n h h o c " da the hicn 10
qua cdc dieu k i c n x + y = 6;
y = t = - - a
vdi 0 < a < - .
2
bo
Tilfdo suy r a m i n P = 1 8 - 6^2
.c
L
nen ta CO x + y = z + t = - . (4) >I nhih itim 1
2
^ ••••
M•• ~
2
(4)
om
Do OMo = 3V2 ; ONo = 1
ro
min(MN-) = M o N f , . -
Bai
U'2j
Ta
tron ddn v j t a i
-
*
Pafu b^ng trong (2) xay ra
t) thuoc dirdng
6,
(2)
iL
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oc
01
/
vdi
•
'
^6
rang A A B C ton t a i v i thoa
" l a n cac tien dc vc dp d a i canh
^"Ja mot tam giac.
A
OB,
Chuy6n ai BDHSG Toan gia tri lan nhS't va gia tri nhd nhat - Phan Huy KhSi
Cty TNHH MTV DVVH Khang Vi§t
K h i do ne'u goi p la niira c h u v i cua tam giac thi p = x + y + z.
IhuTc
M A + A B + B N > M N = V ( 6 - 2 ) ^ + ( 4 + 4 ) ' =AS
Heron, la c6
SABC = V P ( P - B C ) ( P - A C ) ( P - A B )
Gia sur M N cat x + 2y = 9 tai M,, va cat 2x + y = 4 lai N,,.
= 7 ( x + y + z)xyz .
Dd thay M„ = (5; 2); N„ = (4; 0).
T i r g i a l h i e t s u y r a S A B c = 1-
(1)
M a t k h a c SABC = - A B . A C . s i n A = j ( x + y)(x + z ) s i n A .
' '
'
in do
suy ra m i n
? = AS <=>
(2)
HUdng ddn giai
c=> (y + z)^ = (X + y)'+ (X + z)^
V i e t l a i P diTdi dang
(4)
<=> yz = xy + xz + x^.
P= V(x-3)^+(y-5)2 +V(x-5)2+(y-7)2 .
xyz(x + y + z) = 1
Txi do ta c6 minP = 2 <=> <
yz = xy + xz + x^
Ta
toa dp xOy.
s/
up
T i m gia t r i nho nhat cua bieu thiJc
ro
P = Jx^ - 12x + y^ - 8 y + 52 + Jx^ + z^ + y^ +1^ - 2 x z - 2 y t + Vz^ + - 4 z + 8l + 2(1
om
/g
Hudng ddn giai
.c
V i e t lai P difdi dang sau
ok
P = ^ ( x - 6 ) 2 + ( y - 4 ) 2 + V ( x - z ) 2 + ( y - t ) 2 + V ( z - 2 ) ^ + ( t + 4)^ .
Gia sur M ( x ; y ) v d i x, y thoa m a n
X - 2y + 2 = 0, turc la M ( x ; y) nSm
tren diTdng thang x - 2y + 2 = 0. R6
rang tuT (1) suy ra P = M A + M B
Theo m o t b a i loan
(1)
quen bie't ciia h m h
hoc phang, t h i neu
X€i hai d i e m
gpi A ' la d i e m d o i
y)
va B( z; t)
ce
bo
T r e n mat phang toa do Oxy ve hai diTdng thang x + 2y = 9; x + 2y = 4.
w.
fa
+ 2y = 9
ww
Trong d6 A ( x ;
xi^ng
cua
A qua
dirdng thang x - 2y +
2 = 0 va gia sur A A '
ca't dirdng t h i n g X 2y + 2 = 0 tai M„,
tifdng tfng n ^ m
thi M A + M B > M„ + M„B, V M e (d), cl day (d): x - 2y + 2 = 0. , ,
tren hai duf5ng
Co the tinh diTdc ngay Mo = (5; 3,5) => MoA + MoB = A ' B = 6.
thang X + 2y =
Tird6tac6P>6.
(2)
9 v a x + 2y = 4.
H i e n nhien ta
c6
256
(1)
Xet dirdng thang x - 2y + 2 = 0 va hai d i e m A ( 3 ; 5), B ( 5 ; 7) tren mat phang
B a i 12. Cho bon so x, y, z, t thoa man dieu k i e n x + 2y = 9; z + 2t = 4.
P = M A + AB + BN
z = 4 ; l = ()
P = -Jx^ + y^ - 6 x - lOy + 34 + ^ x ^ + y^ - l O x - 14y + 74
o B C ^ = AB^ + AC^
Tir(l)tathay
x = 5;y-2
T i m gia trj nho nhat cua bieu thiJc
a : s r : x } . . i v ' (3)
\
! ' i ! i it '
pai 13. Cho X, y la hai so thi/c thoa man dieu kien: x - 2y + 2 = 0.
Dau bang trong (3) xay ra o sinA = 1 <=> A = 90"
M ( 6 ; 4) va N ( 2 ; - 4 ) .
M
ginh lu4n: K h o n g the c6 IcJi giai nao lai ngan gon va hay hOn lc:fi giai tren.
Do s i n A < l , n e n t i r ( l ) ( 2 ) suy ra ^ ( x + y)(x + z) > 1 => (x + y)(x + z) > 2
VayP>2
(3)
iL
ie
uO
nT
hi
Da
iH
oc
01
/
Thco cong
Dau b ^ n g trong (2) xay ra o M s Mo o
fx = 5
\y; =i 3. , 5^. v,
. V§y minP = 6 c : > x = 5 ; y = 3,5.
257
Chuyfin dg BDHSG Toan g\A trj Idn nhS't va g\i trj nh6 nha't - Phan Huy KhAi
Cty TNHH MTV D W H Khang Vi$t
Nhdnxet:
1. T i r x - 2 y + 2 = ( ) = ^ x : = 2 y - 2 .
B
,5^..,,,^,,
Tur do ta c6:
1 2 ' ~2,
I
vii T(x; x).
2 • 2J
Khi do t i r ( l ) , ta c6
f(x) = TA + TB + T C ;
P = 7(2y - 2 ) ^ + y^ - 6 ( 2 y - 2 ) - lOy + 34 + V(2y - 2)' + y^ - 1 ()(2y - 2) - 14y774
Dc thay ABC la tam giac deu
= Vsy^ - 3 0 y + 50 + ^/5y^ - 4 2 y + 98 . , .
• ,., ,: .
iL
ie
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nT
hi
Da
iH
oc
01
/
vdi tam la go'c toa do O. *'
Bai loan trd thanh: \ rskmisl
Theo ket qua quen bict trong
hinh hoc phang:
Tim gia tri nho nhat cua ham so f(y) = 75y^-30y + 50 + 7 5 y ^ - 4 2 y + 98
„
'JUtU
UM.i
ftiJj.
Do ABC la da giac deu, ncn
!
vdi y e R.
vdi
Cac ban hay thuT dung phifcfng phap chicu bien thicn ham so de giai tiep!
"••
Ta
^
,
p = Jx^ + y ^ + z 2 + 2 x - 6 y + 4z + 14 + V x ^ + y ^ + z ^ + 1 8 x - 8 y - 1 8 z + 178.
'
HU('/ng ddii: Xet mat phang (P): 2x - y + z + 1 = 0 ,
ro
up
Tim gia In nho nhai eija bieu thiJc:
s/
Vay min f(x) = 3 <=> x = 0.
/g
, ^ ij/p
om
ok
bo
S^MP
+
+ -VNH
^ ^KMC •
>
»
.r
(D
Da'u bang trong (1) xay ra khi va chi khi
trong cac tam giac AMP, BMN, CNP cd
w.
tam giac trting vdi AABC. Dieu dd xay
minP = 3x/26 o M = M„ o x = - 2 ; y = 2; z = 3.
ra khi va chi khi Irong ba so' x, y, z cd
ww
Bai 14. Tim giii tri nho nha't cua ham so
xy - yz - jzx,
HiUing d&n giai
Ta cd:
fa
la diem doi xufng cua A qua (P).
Tim gia tri Idn nhat ciia bieu thiye: P = x + y + z
cac diem M, N, P sao cho A M = x; BN = y; CP = z.
ce
dday A ' = ( 3 ; 1;0)
Bai 15. Cho x, y, z la cac so thiTc thupe doan [0; 1 ].
Vc lam giac deu ABC canh bang 1. Tren cac canh AB, BC, CA Ian li/dt lay
.c
Khi do P = M A + M B , d day M(x; y; z) e (P).
Dap so: minP = A ' B = 3>y26 ,
xeK
^ttmh ^
va hai diem A ( - l ; 3; - 2 ) va B(-9; 4; 9). M:+ A M = A
Co the thay AB n (P) = M„(-1; 2; 3).
vdi
Do OA = OB = OC = 1, nen liT (2) (3) cd f(x) > 3 V x e R, ><
/;?
f(x) = 3c=> T = O <=> X = 0.
nfi^h'iit
^ ^^^^^^ ^^^^^.^^^ ^
Cho X, y, z la ba so thiTc thoa man dieu ki6n 2x - y + z + 1 = 0.
ricMig
TA + TB + TC > OA + OB + OC.
2. Ta CO bai toan tiTdng ti/ (giai bang each sijf dung phi/dng trinh difdng thang
, ^
trcn mat
mpi diem T(x; x), ta cd
a r i e b ' i b l J b H'l^si
va mat phdng trong khong gian) ^ ^ ^ ^
moi diem
phang toa dp - ndi
Sau do hay so sanh vdi each giai bang phi/dng phap suT dung do thi va
hinh hoe c( trcn!
1 •c
mpt so bang 1, mot so bang 0 va mpl so
f(x) = ^ / 2 ? - 2x +1 + sjlx^ - ( V 3 - l)x +1 + sjlx^ + (>/3 + l)x + 1 .
ui
iii.ii:) ,j:ii..>
tuyy e [0; l l . T i r ( l ) c d
'
HUofng dan giai
^ [ x ( l - z ) + y(l-x) + z ( l - y ) ] < ^ o x - x z +y - y x + z - z y < l
Vict hji l(x) di/di dang
\
f(x) = Vx- + ( x - l ) ^ +
X - -
if
X + —
2J
Tren mSt phing toa do xet cac diem A(0; 1);
+
x+-
X
2 J
+
—
1)
(1!
Da'u bang trong (2) xay ra <=> cd da'u bang trong (1).
r,r:
Vay maxP = 1 <=> trong ba so x, y, z cd mpl so' bang 1, mpl so' bang 0 va mpt
so'"yye|o;i|.
Chuy6n 6i
BDHSG Join
gii t r i
Idn nha't va gia t r i nh6 nhS'l - Phan Huy KhSi
Cty TNHH MTV DWH Khang Vigl
Nhqnxet:
V
1. Ta CO the sijf dung phiTdng phap "ba't dang thuTc" de giai bai toan trcn nhu
sau:
.j.^^
Do X, y, z € [0; \ } , nen ta c6 ( x - 1)(1- y ) ( l - z) < 0
=> X + y + z - xy - yz - zx + xyz - 1 < 0
=> X + y + z - xy - yz - zx < 1 - xyz.
,
Do X, y, z dcu khong am, ncn xyz > 0.
Tird6kethdpvdi(='0, t a c 6 P < 1.
i
Dau bang irong {**) xay ra <=>
Cong tufng ve' ba bat dang ihtfc tren
;!':••• '.i
, ,
, j, ,
(*)
va P = 1 khi va chi khi trong ba so' x, y, z c6 mot so' bang 1, mot so' bang
0, so con lai tiay y e [0; 1] (ban doc co the nghiem lai dieu nay).
Vay maxP = 1 .
pal 16. Cho X, y, z la ba so' thifc thoa man dieu kien x + 2y + 3z = 4.
(**)
(x-l)(l-y)(l-z)-()
/.j^j,
xy/ = o
,„.,..,
Tim gia tri nho nhat cua bieu thuTc: P = Vl6 + x^ + 2-^16 +y^ + 3 \ / l 6 + ? .
Hudng dan giai
vdi X + 2y + 3z = 4, tren mat phang tpa dp Oxy xet cac diem
A(4; x); B(12; x + 2y); C(24; x + 2y + 3z).
if. r K)A
o trong ba so x, y, z c6 mot so' bang 1, mot so bang 0 va so' con lai luy y
e[();l|
TCf do suy ra maxP = I C5> x. y, z ihoa man (**"•)
Do
Ta thu lai kcl qua trcn.
,(
up
ro
/g
Trcn bon canh cua hinh vuong: dal AM = x; DN = y; CP = z; BQ = I (xcni
2
2
K'-^-)^j
2
'
+ y + z + t - (xy + yz + zt + tx) < 2
Cac ban hay tif lam hai dicu sau day:
ww
NhiT vay maxP = 2.
X
-
Giai lai bai loan trcn bang phu'dng phap "bat dang thu'c"!
Tim gia trj Idn nhat ciia bicu thiJc P = x^ + y"' + z^ - x'y - y^z - z^x.
Lcfi giai nhUsau:
x{ 1 - y) > x'( 1 - y)
^
X
_ x + 2y
12
=y=z=
_ X+
2y + 3z
24
2
1 +2+3 3
mien xac dinh cua no.
R6 rang ta c6 x^ - 2x + 2 = (x - 1)^ + 1 > 0 Vx,
+ 4x + 8 = (X + 2)2 +4 > 0 Vx.
Vay f(x) xdc dinh tren R.
< I => X > x'
1- y >0
o O , A , B , Clhdng hang
Hudng ddn giai
Cho X, y, z la ba so ihiTc ihuoc doan fO: 1 ].
CO
Da'u bang trong (1) xay ra
17. Tim gia Iri nho nhat cua ham so f(x) = Vx^ - 2 x + 2 + Vx^ + 4x + 8 trcn
3. Ta CO bai loan sau di/a vao bai loan da cho:
Lai
P > O C = 724^+4^ =4N/37 (1)
M
Khi nao ihi maxP = 2.
X
do do hien nhicn la c6:
Vay minP = 4V37 <=>x = y = z = —.
A
-
Do 0 <
dinh.
Vay P = OA + AB + BC,
<=> X
w.
=^P<2
dicii
BC = A / I ? + 9 ? = 3\/l6 + z^
4
fa
=> X
2
^ABCD
ok
^ x(I-t) ^ y(l-x) ^ z(l-y) ^
-
.c
^H\'Q
bo
CO S^MQ + SDMN + ^CNP +
ce
Ta
om
hinh vc).
C(24; 4) la
Ta
s/
\ y + z + i - xy - yz - zt - tx.
Hifi'liifi dan fiidi
Y^'
2y + 3z = 4
- /i
AB = V8^+4y^ = ijie + y^
2. Ta CO bai loan lifting lif sau: Cho \ y. /. I ia bon so thifc ihuoc doan [ 0 ; 1 [
Ta giai bhng phu'dng phap sif dung hinh hoc nhU' sau:
X +
d i
De thay OA = ^4^ +x^ ;
Cach giai niiy ciing ra'l gon gang va di
c6 x + y + z - xy - yz - zx > P
TiJf do di/a vao bai tren ta c6 P < 1,
iL
ie
uO
nT
hi
Da
iH
oc
01
/
';>..
TiTdng tir CO y - yz > y^ - y^z,
z-zx>z^-z^x.
X
- xy > x^ - x^y
,f
\
Viet lai f(x) durdi dang sau day
261
Chuygn dg BUHSG Toan gia trj I6n nha't va gia Iri nhd nhfl't - Phan Huy Khai
f(x) =
V(x-l)^+l^ +7l'^-(-2)P+(-2)-
.
Cty TNHH MTV DVVH Khang Vi$t
T r c n mat phang toa do Oxy x c t a i c diem A ( x - 1; 1) va B(x + 2; -2)
R6 rang ta luon c6 O A + OB > A B = sjy + 3 ' = 3 V2 . ^
:l,:o;> oofo m^v
(3).
A, O, B thang hang.
o
O A va OB cungphiTdng,
VA BAT PHUONG TBlNH CO THAM SO
,
dung gia tri Idn nhat va nho nhaft cua ham so' la mot trong nhiyng phu'dng
- V M J u.i ( v ' v
Dan hang trong (3) xay ra
o
,
i
S «grtt cnrfi ni>f
phap hffu h i c u dc giai phifdng trinh va bat phufdng trinh co tham so'. Dang toan
%nim v;':
iL
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hi
Da
iH
oc
01
/
V X e R.
Vay Rx) > 3V2
NHD NHAT TRDNG BAI TDAN GIAI PHU0NG TBINH
y ':<^^^^^; :
•^^^'^''^'^-^^J'^^'^i^^*""'"
T i r ( i ) c 6 f ( x ) = OA + OB.
nay cung thi/dng xuyen xua't hicn trong cac dc thi Toan cua cac de thi tuyen
,x osf -
ft{ x
sinh vao D a i hoc, Cao dang trong nhijfng nam gan day.
:J
§ 1 . M O I L I E N H E G I Q A G I A TR! L 6 N N H A T , N H O N H A T C U A
ngiTcfc chicu.
o(x<=>
H A M SO V A Si; C O N G H I E M C U A P H U O N G TRINH V A BA'T
1; l ) = k(x + 2 ; - 2 ) v d i k < ( )
x-1
liNG DUNG CUA GIA TBI LlN NHAT VA GIA TRj
P^^Mif^
(1)'-"
PHUdNGTRiNH
= k(x + 2)
Ta thu'dng xuyen sii" dung ket qua sau day:
l = -2k
Gia
< x > x - 1 = - ^ ( x + 2) . .
f
su" f(x) la ham so lien tuc trcn mien D , va gia suf ton tai M - m a x f ( x ) ;
m = min f(x).
x=0
. ' • • --r-- - ' - i
x = 0.
iuI ilnjm^kisii'j
•.
CO nghiera k h i va chi k h i m < a < M .
chi k h i m > a.
om
ce
fa
5. Bat phu'dng Irinh r(x) < a diing v d i m o i x e D khi va chi k h i M < a.
w.
Chiing minh:
,
!• Gia suf he phufdng trinh da cho co nghiem, tufc la ton tai Xd e D sao cho f(X(i) = a.
Theo dinh nghla ta co
^
m i n f ( x ) < f(X(,) < max f(x),
"eD
H :
4V
(MUV'
fl"(x)
••^niniHi"'
4. He bat phiTdng trinh <
co nghiem khi va chi k h i m < a. ^
[xeD
bo
'If
ww
•
ijrf'i \n
3. Bat phu'dng trinh f(x) > a dung v d i m o i x e D khi
ok
„
$
2. He baft phuTdng trinh \ nghiem k h i va chi k h i M > a.
[xeD
.c
...1
t
r(x) = a
ro
/g
va 17. Cac ban c6 each giai nao khac ngan gon hdn khong?
...
3-i;sjq|> ^a.ji,} ylvtfriH
,
f
1 . He phiTdng trinh
Nhgn xet: Ro rang phi/dng phap do thi to ro qua hicu qua trong hai bai toiin 16
^
_.
. ^,
,
Kin do ta co:
up
Vay m m l ( x ) = 3V2 o
' •
xeD
Ta
2x - 2 = - x - 2 o
s/
o
u
^'^-^
xeD
\r
=; ,!
minf(x)
H © a o lai gia siir
minf(x)
• | '
X€D
BVi
tiJc la
..f^TI
X€D
f(x) la ham lien tuc nen no nhan m o i gia tri tuf m i n f ( x ) de'n max f ( x )
xeD
xeD
K
N o i rieng no nhan gia tri a, tiJc la ton tai Xo e D sao cho f(xo) = a.
1
D i e u do co nghla la phu'dng trinh da cho co nghiem tren D => dpcm. -
•
263
Cty TNHH MTV DWH Khang Vigt
Chuy6n dg BOHSG Toati gia tii Idn nha't va gia trj nho nha't - Phan Huy KhJIi
2.
Gia
sur he da cho
Ro rang
c6 n g h i c m , ttfc la ion tai x„ e D sao cho
f(X(,) >
X
a.
m a x f ( x ) > IXx,,) > a
r'(x)
xeD
Dao l a i gia sur m a x f ( x ) > a
i
xeD
Gia thie't phan chi?ng he da cho vo nghiem, ttfc la f(x) < a V x e D
•a'
TO
T
i j rdo
d o suy
s u y rra
a
',
maxf(x)
,
,
(21
,
|||,,.
nghiem => dpcm.
3. Gia sur m > a
•
Lay X(, luy y e D .
,J'AHfl M O J M A W A i i l © tSH H : M 'tOM
Ta c6 f(x„) > min f ( x ) = m > a -Q
^4
Dao
''
HWmi
Ta
1=
I
up
s/
:o'} iiJ dh '
TCfdo
o
"
D i e u k i e n de phiTcfng trinh c6 nghla la <^
bo
w.
••
' ^ ^
f4-x>0
ww
i»
» '
o
1 < x < 4.
^
•
sfe
Tiir
264
do suy ra bang bien thien sau:
I
+
- ^ 0 ^
he (1) (2) c6 nghiem
g(t)oO
x/3
so f(x) khi 1 < X < 4.
4,,
nha't ciia mot ham so dc bien iuan mot phiTrtng trinh c6 tham so.
l
16-4x-2x-l
\M Id'
2- Day lii mot v i du mau mi/c minh hoa cho vice ap dung gia trj Idtn nha't va nho
4
+ 276-x = m
Tim m dc phu'dng trinh co hai nghiem phan biet.
^
^ ^ i 2. (De (hi tuyen sink Dai hoc Cao ddn}i khoi A)
Cho phifdng trinh Hlx + 72x + 2 ^ 6 - x
^
- ^ & ^ + 2^/43^
V2x-2 ~
2V4-X.V2X-2
T a t h a y f ' ( x ) = 0 <=> 2 A / 4 ^ = V 2 X - 2 O
0
ciia bien mc'Ji t, ta da giai mot bai toan t i m gia tri Idn nha't, nho nha't ciia ham
D e tim m i e n \ic dinh cfia t, x e t h a m so f(x) = V 4 - x + V 2 x - 2 v d i 1 < x <
2V4-X
1
1
1- K h i doi bien ta phai t i m m i e n xac dinh cua bien m d i . De t i m m i e n xac djnh
Datt= V4-X+V2X-2
Tac6f'(x)=
2
Nhaiixet:
fa
HUdngddngiai
CO nghiem.
7-473
min g ( t ) < m < max
v'5
ce
nghiem.
(2)
Phifrtng trinh da cho co nghicm o
*'
Cho phiTdng trinh 6 + x + 2 V ( 4 - x ) ( 2 x - 2 ) = m + 4 f V 4 - x + V 2 x - 2 ) .
T i m m de phU'dng trinh c6
N/3
min g(t) = g(2) = 0;
/g
'
(1)
max g(t) = max{g(73);g(3)} = m a x { ? - 4 7 3 ; 1 j = 1,
om
.c
ok
'
g(l):=l2 _4t + 4 = m
N/3
ro
'
CAC Kl THI TUYEN SINHVAO DAI HQC, CAO DANG
Bai 1. (De thi tuyen sink Cao ddtifi khdi A, B-2011)
I.
+ 7 2 x - 2 => t ' = X + 2 + 2 V ( 4 - x ) ( 2 x - 2 )
i
g(t)
VA BAT PHOCfNG TRINH CO THAM SO CO MAT TRONG
.
N/6
^'
73
g'(t)
„...U>ljiwii
§2. CAC BAI TOAN BIEN LUAN PHOONG TRINH
V
" ^ " ^ ^ ^
do suy ra k h i : 1 < x < 4, thi 73 < t < 3 .
t
xeD
4,5. Chu-ng minh tu-cfng tu" nhu" 2,3.
, '
N/3
Ta CO g'(t) = 2l - 4, va c6 biing bien thien sau:
K h i do do m = m i n f ( x ) , nen theo djnh nghTa ta c6 ton t a i Xo e D ma m = f(x
,.....,.4
3
l
,
Tim m de he sau
%(M£:9,ii&fU,.
Ttr f(X|,) > a => m > a. Nhu^vay ta CO dpcm.
-
V i the bai toan da cho t r d thanh:
miOim-
l a i gia sur f(x) > a Vx G D.
I
Tir t = V 4 - X
i
Vay IXx) > a Vx e D.
0
iL
ie
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Da
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oc
01
/
Ttf
Tiir (1) (2) suy ra v6 l i , vay gia t h i c l phan chtfng la sai, ttfc la he da cho c
4
m i n 1"(X) = N/3 va max f ( x ) = 3.
Vay
xeD
+
i
l(x)
(1)
3
1
, .
HUifng ddn giai
I ' D a t l X x ) = 72x + 2 7 6 ^ ; g ( x ) = 72^ +
276^
Luc nay phiTdng trinh da cho co dang h(x) = f(x) + g(x) = m
265
Cty TMHH MTV DWH Khang Vigt
BDHSG Toan gia tri Idn nhit va gia lii nh6 nhS't - Phan Huy Khii
Chuygn
PhiTdng trinh (1) xac dinh trong m i e n 0 < x < 6.
Ta
CO
f
(X) =
_76^-N/2^
2
2
2>/2x
276-X
72x(6-x)
0
X
+
0
'">'^"
^"'^
^^"^
f i m m de phiTOng trinh S V x - l + mVx + 1 = 4\/x^ - 1 c6 nghic
icm.
, nen C O bang bie'n thien sau-
HUdng dan gidi
6
2
f'(x)
P»i 3-
phiTdng trinh 3Vx - 1 + mVx + 1 = 2 V x ^ ^
-
p i c u k i e n de (1) c6 nghla la x > 1. Do x > 1, nen V x T T > 0, vay
f(x)
(1) <^ 3 J — +
x +1
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4/(5
- ^(2x)''
TuTdng tif ta c6 g'(x) = ^^^^—,
= - , nen cijng c6 bang bie'n thien sau
2^(2x0-^6-x)-^
/X —1
x —1
E)3t t = ^
. Do
=1
Vx+I
x+1
r a v d i t l a O < t < I)
'•"•2
0
g'(x)
6
0
-
^
0
^
^
max h(x) = h(2) = ^
^
m
+ V 4 + 2 ^ + 2V4 = 6 + 3>/2 ,
ok
()
bo
min h(x) = m i n { h ( 0 ) ; h ( 6 ) } = m i n | 2 7 6 + 2 V 6 ; ^ + 2V3| = h ( 6 ) = ^
2
2
< 0 ( v i X > 1 nen
> 0 nen dieu k i e n dat
x+1
i
x+1 ,f
£
^
m
nghiem.
0
f'(x)
^•""•?
.if :ji>
'•••'-^'•^'i'0
-
f(t)
i
+ 2^3
ny m a x f ( t ) = f
()
it''
',:!•(r
= - ; c 6 n l i m f(t) = - l (chu y d day khong ton tai minf(t)).
3
t ^ r
i)
^
ce
()
^ ,
Ta
s/
-
ro
h(x)
TCf do suy ra phu'Ong trinh da cho co hai nghiem phan biet k h i va chi k h i
r;
2 ( ^+ V 6 ) < m < 3 V 2 + 6.
TiJT do suy ra phU'dng trinh da cho c6 nghiem khi - 1 < m < ^ . ' " t J , Cf-)
w.
fa
,1,
ww
Chiiy:
1. Ne'u dau bai chi d o i h o i : T i m m de phiTdng trinh c6 nghiem t h i dap so cua ba'
toan la ^
;:^r"i^
1
t
/g
+
.c
h'(x)
6
up
2
om
0
X
Taco
Vx + 1
Ta c6 f ' ( t ) = - 6 t + 2, nen c6 bang bie'n thien sau:
V i the ta C O bang bie'n thien sau doi v6i ham so h(x), 0 < x < 6
i'-
ir
B^i toan da cho trd thanh: T i m m de he: - ^^^^
3t + 2t
• [0
g(x)
„
x^-l
(x +
Vx + 1
(vdi chu y : a^ - b ' = (a - b)(a^ + ab + b'), d day a^ + ab + b^ > 0.
X
m = 24)
+ 2^3 < m < 6 + 3V2 .
2. T u y nhien d day b a i toan d o i hoi: T i m m de phiTOng trinh c6 hai nghic''^
Cac ban nen lifu y rSng khi suf dung dieu k i e n c6 nghiem cua he phifdng trinh
•
" t a can lu\ y gia thie't phai ton tai mini"(x), m a x f ( x )
XeD
•
xeD
xeD
CJ day v i xet khi 0 < t < 1, nen khong Ion tai m i n f ( t ) nhu^g ton tai
()
phan biet, thi de y k h i m = max h(x) k h i do phtfclng trinh da cho chi c6 m'?'
()
|m f ( t ) = - l . D o do dieu kien theo ly thuyel - 1 < m < - phai thay bang - 1 <
nghiem duy nha't.
K e t hcJp v d i bang bie'n thien, ta suy ra ket qua tren. K h i l a m bai ta can liA" ^
dieu nay.
266
- (ttJTc la ta da thay dieu kien m > min 1(1) thanh m > l i m 1(1))
3
" S K I
_
,
•
d ji;
,
,,
267
Cty TNHH MTV DWH Khang Vi^t
Chuy6n 6i BDHSG To^n gia tri Icin nhSt va gia tri nh6 nhS't - Phan Huy Khii
f^' tuyen sink Dai hoc Cao danf^ khdi B)
p^i 4.
Cac ban can lifu y d i c u do (d day ta cung dfi ket hOp sit dung chieu b i f
thien cua ham so' da xet trong bang bicn thien ham so' l'(t) trong bai).
Tim m de phirong trinh Vx^ + m x + 2 = 2x +1 c6 hai nghiem thiTc phan biet.
2. Ta CO each giai khiic sau day:
- 3 l - + 2 t - m = 0 (3)
0
I
xV'T' thmi
pifa phifdng trinh da cho ve he sau:
'^rf'^
(4) ,
f2x + l > 0
De he (3) (4) c6 nghiem, triTdc h e t (3) can c6 nghiem, tiJc la phai
x 2 + m x + 2 = (2x + l)^
Co
<=> s
!•;')'>; -
x> —
2
•' •
•
<• ,
"
x^ + m x + 2 = 4x^ + 4x + l
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X e t he
HU
A' = l - 3 m > ( ) < ^ m < ^ . K h i do (3) <^ t = ^ " ^ ^ ^ ' ^ ^
' 3 x ^ + 4 x - l = mx (1)
TiTdo suy ra he (3) (4) CO nghiem k h i va chi k h i
^, „ • i . ,
1
2
3
0
-l
Vl - 3 n i <;2
0 < l - V l - 3 m <3
-2 < V l - 3 m < 1
Vr^-3m SI
3x^ + 4 x - I
<2<::>-l
Ta thu l a i ket qua tren!
y1
'
*
3x^+1
f'Cx)
om
a. hoac (3) v 6 nghiem, tiJc la A' = 1 - 3m < 0 o m > ^ .
f(x)
.c
b. hoac (3) c6 nghiem, nhifng m o i nghiem deu khong thoa man (4)
bo
ok
2
K h i (3) CO nghiem t i , t2 thi tj + t2 = - , do do m o i nghiem t i , I2 khong thoa
ce
man (4) k h i va chi k h i ti < 0 < 1 < t2
V a y he (3) (4) v6 nghiem k h i
fa
w.
De tha'y (5) xay ra k h i va chi k h i
ww
A'>0
(5)
m<0
f(l)<0
m +l<0
, va CO bang bien thien sau:
1
0
2
i
1
+
+
+00
4
=1
i':'
.
+00
9
2-"
-00
Tiir do suy ra he (3) (4) cd hai nghiem phan biet <=> m > —.
^hqnxet:
3
f(0)<0o
n •
(4)
X
/g
D i e u nay xay ra trong hai IriTdng hdp sau:
up
'
Ta CO f ' ( x ) =
ro
Tru'dc hot ta tim dieu kien de he (3) (4) v6 nghiem
s/
3. L a i con each giai khac sau day
'
= m (3)
x> —
2
Ta
Vl-3m
y i 11
Do X = 0 khong phai la nghiem cua ( I ) V m , ncn (1) (2)
3
o
i
(2)
X> —
o m < - l .
1
m >3
1- Ro rang trong b a i nay ta c6
m i n f ( x ) = — , v a da k e t hdp v d i chieu bien
-'-
thien ham 86"de g i a i bai todn.
2- X e t each giai khdc ciia b ^ i toan tren:
m<-l
Do do he (3) (4) c6 nghiem k h i va chi khi - 1
3 x ^ + ( 4 - m ) x - l = 0 (5)
B a i toan da cho cd dang: T i m m de he
cd h a i
x> —
2
(6)
I n g h i e m phan biet.
ham so v d i cac phi/dng phap vijfa trinh bay! Cac ban thich each giai nao?
268
269
cnuyen de B D H S G loan gia tr| lan nhat va gia trj nh6 nhat - Phan Huy'Khai
Cty TNHH MTV DWH Khang Vigt
D i e u d o x a y r a k h i v a c h i k h i ( 5 ) c 6 h a i n g h i e m p h a n b i e t X j , X 2 sao c h o
TCf do suy ra
m a x ^ f ( t ) = 1(0) = 1,
0
X2>X,
f ( t ) = f(N/2) = V 2 - 1 .
min
1
R6 rang (5) c6 hai nghiem phan biet ( v i - = — < 0 )
a
3
C
. i „
X | + -
[ X | + X2
>0
X-, + •
> -1
U
X|X2+-(x,+X2) + - > 0
2
3
m-4
m > -
>-l
2 <=>m> —.
m >1
:
yjfKy,
^!
Cho
Tim
J ' f . i ^ " * n n o c y Q ' * ^ ,,
Ta t h u l a i kc't q u a t r e n .
Bai l o a n da c h o In't t h a n h :
0, v a y (1) o
Ta
s/
/g
t+2
w.
Ta
CO f'(t) =
,? . .
t
f'(t)
f(t)
—t
0
(3)
,
s
—
4t
(1 + 2)-^
1
vafcui
'
va cd bang bien thien sau:
, ^
i M isij^jVfidi* mfirf tits'•• ,
'(tSnMM'A i i i i f rf'J^b J A
1
^
i
—
.
1
1
—
'
1
i
i
i
x^/: X e t each giai khac nhau:
Bai loan da cho C O dang:
T i m m de he sau
0
1
+
1
'nfrfBimbt,
CO nghiem.
t +2
nghicMii
2
(1)
ww
Timmdehe:
to
2
no
fa
B a i loan da cho trd thanh:
ce
t+ 2
(2)
I
bo
Luc d o la di/a phuTdng u-inh v e d a n g m(t + 2) = 2 - t^ - t .
Do 0 < t <
1<1<2
.c
2 - 2V1 - x"* < 2, v a y 0 < t < V2 .
(1)
Ta C O f ' ( t ) = 2t + 1 va c6 biing bien thien sau:
om
,
ok
CO t ' =
_
up
c6 nghiem. .
r(t) = t ^ + t - 2 = 2m
ro
+ 2 ) = 2 V l - x ' * + V l + x2 - V l - x ^
Hudng dan gidi
Ta
Tim m de he •
f y V I 6:.
r,^nti\M <;<'»^>^ h-i
.Khi d6t>0.
I s i y
Liic nay t h e o b i e n m d i t , phU'dng t r i n h da c h o c6 d a n g I " + t - 2 = 2m.
S y.
B a i 5. (De thi tuyen sink Dai hoc Cao dang kho'i B)
D a t t = V l + x^ - V l - x ^
/n)e\y''
D a t l = , / l o g ] T + 7 . K h i 1 < x < 3"^ => 1 < t < 2 .
B a n doc h a y tuT b i n h l u a n v e t i n h h i e u q u a cua m o i phiTcfng phdp.
m ( V l + x^ - V l - x ^
I—7—
phiTdng t r i n h l o g ^ x + J l o g ^ x + l - 2 m - l = 0
m d e p h i / d n g t r i n h c6 i t n h a t m o t n g h i e m t h u p c d o a n [ I ; 3 ^ 1
HU(ing dan gidi
"
Tim m de phiWng Irinh
:.'
Bai 6. (De thi tu\en sinh Dai hoc Cao dami kho'i A)
y
9
-.Z':::
nhat c u a h a m so' de g i a i b a i t o a n n a y .
.. in = i --.J<l->
1 1 ^
- -6+ -4> 0
%
h a i de tha'y ro t i n h h i e u q u a c u a phU'dng p h a p suf d u n g g i a I n I d n n h a t va n h o
(9)
l^x, + X 2 > - l
(
''
glith luan: Cac b a n h a y g i a i l a i b a i l o a n t r c n b a n g phu'dng p h a p t a m thtfc bcic
(8)
4
A p dung djnh H V i e t vdi (5), ta c6 (8) (9)
m-4
Vayhc(2)(3)c6nghiem<=> V 2 - l < m < l .
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V i vay (7) <=> <
()
s,h cmi,} • 1 v,
t =
t=
Jl^ + l - 2 - 2 m = 0
l
- l + V8m
(3)o
1
j
(3)
C O nghiem.
(4)
+9
V8m + 9
^ Q ^j^^.
^^^^^ ^^^^
^^^^
271
ChuySn
BDHSG Toan gi^ trj I6n nhat va gia trj nh6 nhjit - Phan Huy KhSI
Cty TNHH MTV DWH Khang ViQt
V a y he (3) (4) c 6 n g h i c m khi va c h i khi
m >
Do do he (2) (3) c6 nghiem <=> min f(t) < m < max <=> - — < m < - 2 .
•
()
9
—
8
<=>
<=> 0
8
< m <
3 < V8m + 9 < 5
E)6 chinh 1^ cac gia trj can tim cua tham so'm.
2.
plnh luan:
, CHI n;/.:,<,;
-*
...^
J R6 rang suf dung phiTctng phap gia tri Idn nhat
Ta ihu lai ke'l qua trcn.
> '
,:>•..
nh6 nhat ciaa ham so de
gi^i b^i toan nay day hieu qua: Ldi giai gon gang, trong sang.
Ban tha'y the nao vc tinh hieu qua cua hai each giai vijfa tnnh bay!
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2. Ta hay x6t them cdc each giai khac dc nhan thay cdch giai tren la thich hdp
nhS't.
§ 3 . sCr D g N G G I A TR! LCJN NHAT V A NHO NHAT C U A HAM SO
2: Bai toan da cho CO dang:
D E BIEN LUAN PHUCING TRJNH V A BAT PHl/ONG TRJNH C O
Tim m de he
3t - 2 t - 3 - m = 0 (4)
0
THAM SO
'
c6 nghiem
(5)
He (4) (5) CO nghiem trong hai triTcJng hPp sau:
A. Bi^n ludn phiCdng trinh c6 tham so
Bai
1.
Cho
phiTcJng trinh 2(sin"'x
phi/dng
trinh
+ cos'x) + cos4x + 2sin2x +
c6 it nhat
m
= 0.
cho diTcHi dang
s/
j
tiTPng difdng sau:
.
t 'ifh i r ' '
(t,-l)(t2-l)>0
t|
+t2
<2
ro
trinh da
t,i2>0
Dieu nay xay ra khi va chi khi -{t, +12 > 0
up
Hitting ddn giai
Vie't lai phiTdng
fA'>0
'--y'iU.
mot n g h i c m thupc d o a n
Ta
T i m m de
a. (4) CO nghiem t|, t2 m^ 0 < ti < t2 < I
Bai l o a n da cho trd thanh:
w.
ww
Ta CO f'(0 = 6l - 2 va CO bang b i e n thien sau:
0
3
•
.c
3m + 10>0
:
Tur do ta c6
ce
f(t) = 3 t ^ - 2 t - 3 = m (2) c6 nghiem.
0
fa
Tim m de he
bo
ok
t)5t t = sin2x. K h i O < x < - = > 0 < 2 x < 7 t = : > 0 < sin2x < 1.
2
Dira vao djnh Ii Viet, thi t, + ta = | ; t,t2 =
om
/g
1
^
->
l - - s i n ^ 2 x + l - 2 s i n ' ^ 2 x + 2sin2x + m o 3 s i n ^ 2 x - 2 s i n 2 x - 3 = m (
.
2
)
-3-m>0
o
10
-Y
(6)
1 .1
. ••
- 3 - m — + 1>0
3
b. (4) CO nghiem t i , t2 trong do mot nghiem 6 [0; 1] \h mpt nghiem khong
thupc [0; 1]
Dieu do xay ra o f(0)f(l) < 0, d day f(t) = 3t^ - 2t - 3 - m
o (-3 - m)(-2 - m) < 0 o (3 + m)(2 + m) < 0
<i>-3
"
(7),
Ket hpp (6) (7) suy ra - ^ ^ m < - 2 .
Vay
min f(t) = f
()
o
Ta thu lai ket qua tren.
^ch 3: Tn/dc hct tim dieu kien dc he (4) (5) v6 nghiem
i
He (4) (5) v6 nghiem trong hai triTcing hPp sau:
07^
Cty TIMHH MTV DVVH Khang Vi$t
ChuySn 6i BDHSG Toan gJA t r j \dn nhaft vA g\& M nh6 nhSt - Phan Huy KhSi
fx>3
U)
3m + 10 < 0 <=> m < -
a. Ban than (4) v6 nghiem, tiJc la can c6 A ' < 0
x
b. (4) CO nghiem nhiTng ca hai nghiem deu « |(); 1 ].
;;
'
2
Khi (4) CO nghiem l , ,
la c6 t, + = - , nen dieu nay xay ra k h i v a chi
X>3
<:> (2) (3)
I
f ( ( ) ) < 0 <=> - 2 .
C:> i 1(1) < ( )
-2-m<()
Vay he (4) (5) v6 nghiem o
m <
10
3
m >-2
Do do he (4) (5) c6 nghiem o - 10
y < m < - 2 . T a thu lai ket qua Iren.
Cddi
4: Trirdc h e l (4) c6 nghiem
»
0, nhU" vay (1)
• < v>. . i ^ ; ; , ^ .
7
X
1
f'(x)
0
1
oA'>0o3m+10>0c:>m>-^.
Tir do suy ra
1 - V 3 m + 10
/g
Vay he (4) (5) c6 nghiem
Nhanxet:
Tim m de he sau
.
s
•
f ;;(.,,
l i m f(x) = + « .
ok
bo
fa
- 2 ( m + l)x + 5m + l = 0 (6) ,
C O nghiem
x>3
•
'
(7)
X
ww
He (6) (7) v6 nghiem trong hai tru"dng help sau:
a. (6) v6 nghiem <=> A ' < 0 o m ' - 3m < 0 o O < m < 3
B a i 2. T i m m de phiTdng trinh 72x^ - 2 ( m + 4)x + 5m + 1 0 + 3 - x = 0 c6 nghiem
b. (6) C O nghiem X i , X2 va X| < X2 < 3.
>f.
,^
*
<,
Trirdc h e l t i m m de he (6) (7) v6 nghiem.
nhat va nho nhat trong tri/cfng hdp nay la help l i nhat.
TCr d o s u y r a ( l ) o
lim f(x) = + c »
ma ta van sir dung)
Bay gicf cha'c cac ban dong y vdi chiing t o i : Suf dung phtfcfng phap gia t n U^"
V i e t lai phi/dng trinh da cho di/(3i dang sau:
•
2. Xct each giai khac sau day (de cac ban thay no kem hieu qua hcJn phiftJng phap
Bai loan da cho C O dang:
Hitffng dan gidi
:A, ,
•
•
10 ^,
<=>
3
w.
Ta Ihu lai ket qua Iren.
' • •
1. O day d i c u k i e n m < max f ( x ) hien nhicn thoa man do
.c
0<3m + 10
- l < V 3 m + 10<2
V2x^ - 2 ( m + 4)x + 5m + 10 = x - 3 .
•
Vay he (4) (5) c6 nghiem <=> min r(x) < m <=> m > 3.
ce
- 2 < V3m + 10 < 1
om
10
2
/
f
x>3
Ta
ro
up
s/
l + N/3m + 10
+
m i n f ( x ) = 1(4) = 3
(con m a x f ( x ) = +oo hieu theo nghla
2
"•'
+C0
0
1
f(x)
:
4
1
x>3
K h i do (4) C O nghiem
(4)
,
X -2x + !
l(x) = —
- — = m (5)
2x-5
,
^ 2x^ - l O x + 8 . , ^,
Ta C O I (x) = —
— — va co bang bien thien sau:
(2x-5)^
JiSji;" ••s/ .
khil| ^ 0
3, nen 2x - 5
- 2 x + l = m ( 2 x - 5 ) (3)
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'fij 6".. f ' i u . «
po X >
(2)
, ^ ,,
[A'>0
(1)
X|+X2<6
x-3>0
Do
X | + X2
= 2(m + 1);
X|X2
"
(8)
(9)
(8)xayrac::> j ( x , - 3 ) ( X 2 - 3 ) > 0
"
(10)
lit
hi
t \ ,1
(11).
= 5m + 1, nen
2x^ - 2(m + 4)x + 5m + 1 0 = (x - 3)^
4
275
Cty TrjUII MTV DVVH Khang Vi?t
(9)(10)(11)«
m^-3m>0
0 < m hoac m > 3
5m + l - 6 ( m + l) + 9 o
m<4
2(m + l ) < 6
m<2
HUdng ddn giai
<=>m<0.
Ipat t = N/X - 9 > 0 => X = t' + 9. Khi do phu'dng trinh da cho c6 dang
Vay he (6) (7) v6 nghiem khi m < 3 => he (6) (7) c6 nghicm khi m > 3.
Ta thu lai ket qua tren. Ban thay the nao?
.
;
V(t + 3)' +V(i--^)^ =
Bai 3. Cho phiTdng trinh V 2 - x + -Jl + x - 7(2-x)(2 + x) = m .';
t+3+
.,
HUdng ddn giai
^ f|
'
D a t t = V 2 - X +V2 + X .
.
^2
t-3
Tim m dc he •
2^.
r(t) = - t ^ +6t + 9 + 6 t - 3 = m (1)
t>()
Ta
=m
up
<=>-r+ 2l + 4 = 2m.
ro
Bai toan da cho tn'nhanh:
om
ok
bo
0
1
Id)
w.
-
.
,
ft'.
•
i<\
0
I'd)
i"'(t)
i
1
V/.
f(i)
i
3
+00
6
-t^+12t- 9
- I - + 27
-2t
- 2 t + 12
+
27--^.^^^^^^^
0
-
27-^,^^^
=^ 1 8
^
Ta CO max 1(1) = max {1(0); 1(6)} = max {27; 27} = 27.
' ' "
-co
^ * ''^
IL'O
O day la coi nhU' min I'd) = -00 vi iim 1(1) =
-00.
I>0
Vay he (1) (2) CO nghicMii o m < m a x H D o m < 27.
i
Nhqn xet: Cac ban hay giai lai bkng phiTcfng phap khac, de thay ro hieu U^'^
cua phiTdng phap ma ta vuTa suT dung de giai bai toan nay!
Bai 4. Cho phi/dng trinh Vx + 6 V x - 9 + A / X - 6 V X - 9 = ^ll^.
6
Tim m de phtfdng trinh c6 nghiem.
-jd-.U in
phir(tng phap nay.
,
i/r
'^ai 5. Cho phirong irinh 9 ' ^ " ^ - ( m
m + 22)3'
) 3 " " ^ + 2m + 1 = 0
Tini m dc phU'dng trinh c6 nghicm.
o 4V2-4 < 2m < 4 o 2>/2 - 2 < m < 2
276
(2)
^hqii xet: Thco chung toi khong the c6 phUcJug phiip niio khac lai ddn gian hdn
max f(t) = f(2) = 4 ;
min f(t) - f(272) = 4 V 2 - 4
i<\
m i n ^ f ( t ) < 2 m < max^
ww
Vay
1
ce
f'd)
2
fa
1
.c
Ta CO f'(l) = -2t + 2, va c6 bang bien thien sau
t
f(t)
/g
ff(t) = - t ^ + 2 t + 4 = 2m (1)
CO nghicm.
Tim m de he <^
[2
,
CO nghicm.
Lap biing sau day:
, vay difa phifcfng trinh da cho ve
s/
dang t -
• (
J^i toan lrc"< thanh:
",,•
t
t^-4
(do I > 0)
- r + 6t + 9 + 6 t - 3 = m.
;,
Ta CO I ' = 4 + 2V(2-x)(2 + x) =^ 4 < r < 4 + (2 - x) + (2 + x)
f +9 + m
iL
ie
uO
nT
hi
Da
iH
oc
01
/
Tim m de phiTcfng trinh CO nghiem.
t^ + 9 + m
''
Hifdiig ddn giai
) a t 3 ' ^ ^ = t . Tac6 3 < t < 9 .
thi do dU'a phu'cfng trinh da cho ve dang tu"(Jng du'iJng sau:
t' - (m + 2)1 + 2m + 1 = 0
1--2t+1 =m(t-2).
(1)
277
Chuy§n 6i BDHSd loan qui tn Ion rihfi! jz (jia in ntio t'ih.ti
-2t + l
D o 3 < I < 9 => t - 2 ?t 0, v a y (!)<=>
V i the b a i l o a n da cho t r d t h a n h :
1-2
^ = m (2)
3 < t< 9
T a CO r ' ( t ) =
,
laco
=m.
,
.
3
I'd)
1
...
nghiem.
64
V a y m a x 1(1) = 1(9) - — ;
3ii
TiC d o he ( 1 ) ( 2 ) co n g h i e m o
^j/i/i /Hfl/i." X e m c
%
T i m m d e phi/dng I r i n h t^ + I + m = 0
i
, ,
m i n 1(1) = 1(3) = 4
3
a m . V i I h e (3) c o n g h i e m a m k h i va c h i k h i (3)
\'VJ%H
khi A = I - 4 m > 0 o
.c
2
0
'
\
X
(I)
2
^
f"
'
+mx + 2m + — + — =
X +
.
CO n s i h i c n i
—
+ m
X +
V
^
0
t=
X
1
+ -,
X
k h i do
—
X/
t
=
1
x + —
X
X
+
1
X
o
m ( t + 2) = 2 - t ^
\d)
(t)
O-r;
(2)
D o t = - 2 k h o n g p h a i la n g h i e m c u a ( 2 ) vcti m p i m , n e n ( 2 ) <=>
l(t)
1
4
^
1„
,;
( l ) c 6 dangt^ + m t + 2 m - 2 = 0
+
0
dan giai
CO t h e du'a phiTiIng t r i n h b a n d a u v c d a n g ti/dng duTdng sau d a y : |
Dal
1
278
Hitdng
N2
(2)
T a CO r ' ( t ) = 2t + 1, va c 6 b a n g bie'n I h i e n sail
f'(t)
T i m m d e phiTdng t r i n h c o n g h i e m .
D o v d i m o i m t h i x = 0 k h o n g p h a i la n g h i e m c i i a phiTdng t r i n h d a c h o , v i the'
ok
ce
t<0
-co
B a i 7. C h o p h i f d n g t r i n h x'' + mx"* + 2 m x ^ + m x + 1 = 0
fa
r(t):=:r + l = - m
ww
B i i i l o a n da c h o t n J l h a n h : T i m m d e he
w.
Dal t - log.x. K h i 0 < x < I ihi t < 0
t
gian nhat!
t;H)ji!/fc3i#ini*:»m
TiT d o phiAJng I r i n h da c h o c 6 d a n g san: l o g ; x + log-, x = - m .
.
C o n g b a n g m a n o i t r o n g b a i l o a n n a y , d a y m d i la e a c h l a m h a y nhaft, d d n
om
/g
|
^-logjX.
'
isf-.i!'.
Ta
s/
up
ro
1
bo
l o g , >/x
m < i/'S"
c o n g h i e m , luTc la k h i va c h i
Ta i h u l a i k e l qua I r e n .
dan i>iai
log, x - - l o g , x ,
( 3 ) co n g h i e m a m .
Ro r a n g n e u ( 3 ) co n g h i e m t h i d o S = - 1 < 0, n e n c h a c c h a n ( 3 ) co n g h i e m
1 I'
11 d e phiTcJng I r i n h c 6 n g h i e m i h i i o c ( 0 ; I )
Vdi dicu kien x > 0 ihi
(...
V a y m < — la c a c g i a I r j c a n t i m c u a l h a m so m .
4
1
Uiiiin^
,
1
1
o - m > — < = > m < —.
4
4
B a i 6. ( h o phiTctng I r i n h 4 ( l o g , \/x | - l o g , x + m = 0 .
„ - r ! " ' i
- m > m i n 1(1)
i<()
64
D o do 4 < m < — la c a c g i a I r i c a n t i m c u a l h a m s o m .
7
Tmi
l i m ( r + l ) = ().
(3)
1
f(t)
+ l ) = +oo va
n o d o m i n f ( l ) = - - va c o i n h i r m a x f ( l ) = +oo.
^
4
1-^0
-4t + 3 _,
,
va CO b a n g bie'n t h i e n sail
(1-2)^
t
l i m (I
iL
ie
uO
nT
hi
Da
iH
oc
01
/
T i m m d c he
-21 + 1
f(t) =
t-2
.
Cty TNHH MTV DWFTKliang VigT
"iim Huy Khtii
2-t^
— t+ ^
•
B a i l o a n da cho t r d t h a n h :
Li.
279
Cty TNHH MTV
t(l) = ^—^
T i m m de he
1+2
>2
f'(t) =
CO
,
. (4)
CO nghiem.
(t,
.
f'(l)
+00
+2)(t2+2)>0
t, + t 2 > - 4
(7)<^
£.."i
f
(2-t2)(2-t,)>0
t|+t2<4
va CO bang bien thien sau
-2-V2
0
+
Do ti + t2 = - m va tit2 = 2m - 2, nen
- 2 + V2
m ^ - 8 m + 8>()
[m^-8m + 8>0
2 m - 2 + 2(-m) + 4 > 0
+00
(7)ci>'-m>-4
n,<4
<»•
1
4-2(-m) + 2m-2>0
, . , . maxr(t) = — ;
,^2
2
He (3) (4) CO n g h i c m
t>2
o
h o a c la phiTdng I r i n h 1(1) = m c6 n g h i e m I r e n D,
( d d a y D i = { t : t < - 2 } c6n D2 = { l : t > 2 } )
2V2
m . - l .
*
om
•
/g
2
A^/ia/i xet:
v d i D = D, u D 2
fa
, ft^+mt + 2m-2=-0
T i m m de he
^2
w.
(6)
CO nghiem.
Trirdc het ta t i m m de he (5) (6) vo nghiem.
—
2
m>4^-2^/2
.
, j .
Ta thu l a i ket qua tren. Ban doc tiT danh gia ve tinh hieu qua cua tijrng
j
,=
»ft ' ^' '
3. X6t mot bai toan tiftfng tiT sau:
Cho phi/dng trinh — —
sin
^ ^ B ^ + m(tan x + cot x) - 1 = 0 .
X
T i m m de phu'dng trinh c6 nghiem.
3(tan^ + col^x) + m(tanx + cotx) + 2 = 0
«
•'•/ft.
(8)
3(tanx + cotx)^ + m(tanx + cotx) - 4 = 0.
)at t = tanx + cotx => t
tan x + c o t x
tanx
cotx
(do tanxcotx = 1 > 0)
>2.
•
rtr do
(8) <=> 3t^ + mt - 4 = 0 <=> mt = 4 - 3 t \
c>
4-3t^
t
He (5) (6) v6 nghiem trong hai triTdng hdp sau:
a. (5) v6 n g h i c m <=>A = m ^ - 8 m + 8 < 0 o 4 - lyfl
m <
<=> 3[(tanx + cotx)^ - 2] + m(tanx + cotx) + 2 = 0
ww
(5)
(iVI'*^'?"
[m>-4
DiTa phifdng trinh da cho ve dang:
nghiem tren D , , hoSc la phiTdng trinh f(x) = m c6 nghiem tren D2.
B a i toan c6 dang m d i sau day:
o
phiTdng phap t r e n !
R6 rang he nay c6 nghiem k h i va chi k h i hoac la phiTdng trinh f(x) = m c6
2. X e t each g i a i khac cho bai toan tren
! f I S t =v
Lc(i gidi nhU sau:
bo
xeD
ce
cho tru'dng hdp he
ok
.c
1. Trong bai tap tren ta da mcl rong ke't qua suT dung gia t r i Idn nhat va nho nhat
'f(x)-m
I'
up
<=>
,'
ro
1
nghiem
h o a c la phu"dng t r i n h f ( t ) = m c6 n g h i e m t r e n Dj
m >4+
*fe ,!
V a y he (5) (6) vo n g h i c m khi - ^ < m < 4 + 2>/2, ttfc la he (5) (6) co
min f ( t ) = - c o
Ta
;:iYVrt»
o--
"^^"2
-m<4
m i n f ( l ) = 4 + 2N/2; m a x f ( t ) = + 0 0 ;
t<-2
t<-2
s/
Ta CO
Khang'Vigt
A>0
^.^
-t^ - 4 t - 2
I
f(l)
(3)
iL
ie
uO
nT
hi
Da
iH
oc
01
/
Ta
= m
PWH
>2
nen (8) <=>
t
= m .
< m < 4 + 272 .
b. (5) CO n g h i c m t| < Ij va khong thoa man (5), tiJc la - 2 < l | < tj < 2.
(7)
281
Chuyfin dfi BDHSG Join gia tri Idn nhat va g\i trj nhd nhat - Phan Huy KhSi
l(t) =
B ^ i loan da cbo Irc'J thanh: T i m m do he <
Cty TMiiH MIV i i v v i i Khang Vigt
4-3t"
- m (9)
I
>2
,
p o - 1 ^ V < 1 Ci> - 1 < - - u < 1
CO nji
(10)
'
^ - - < u < - .
Ta CO f ' ( t ) =
-3t^-4
'
(6)
, nen c6 bang bien thien sau
Ttf (5) (6) suy ra —^< u < 1. L u c nay (4) c6 dang 8 u ' - 4u - 3 = - 2 m .
f'(t)
2
1
f(t)
T i m m de he
111
> 2 } , thi D = D| u D2, d day D, = {I: t < - 2 } va D2 = {I: t > 2 )
m a x f ( t ) = +oo;
maxf(l) = f(2) = - 4 ;
teD2
minf(t) =-00.
\h)
teD|
u
s/
.c
bo
w.
1
Dat u = sinx; v = siny. K h i do (1) (2) o
2
'
"1 l
1
4
1
i
+
{)
7
4
^ r; f ( i )
= max{l;l} = I .
--
Bi/i/i
- - < 2m < 1 o
1 '
- -
2 •i&j^f^h nt>o\
X e t mot each g i a i doc dao bai toan tren nhiT sau:
i1-
(1)
(2)
^ \y.
/.^
(3) va (5) chinh la doan thang A B
„J(.iii,i;
vdi A
(3)
(5)
I
vaB
2 >
1;
V
A
r
1
i\
2.
diTcfng tron t a m O ban kinh
' • ;•
R =
2-m
I He (3) (4) (5) CO nghiem
1/2
\
K h i 2 - m > 0, t h i (4) \k
u 2 + v 2 = l z i l l (4)
u <1; V 41.
Tir (3) suy ra V =
CO nghiem.
De tha'y cac d i e m (u; v) thoa man
sin^x + sin^y = ^ — ^
ww
o
u +v=-
1
2~""
. u,t\,tt Vfoi'.
fa
sinx + siny = —
ce
1
1
Vay he (7) (8) c6 nghiem o
ok
HUdng dan gidi
min f(u) = f
max f ( u ) = m a x O
/g
^
T i m m de he c6 nghiem.
V i e t l a i he da cho diTdi dang
1P
1
-'
ro
1
smx + smy = —
^ 2
cos2x + cos2y = m. ^j. , •
smx + smy = —
s
2
l-2sin^x +I-2sin^y = m
Ta CO
up
s t - « ^ »>'»F
om
B a i 8. Cho he phi/dng tnnh
2
l"(u)
m <-4
leD2
282
•>if
m>4
<=>
hoacla m < m a x f ( t ) = - 4
D o la cac gia t r i can t i m cua tham so m.
(8)
f'(u)
hoacla m > m i n f ( t ) = 4
S
--
1
leD]
He (4) (10) CO n g h i e m o
(7)
Ta CO f'(u) = 16u - 4, va c6 bang bien thien sau
*>.ta i
minf(t) = f(-2) = 4 ,
f(u) = 8 u ^ - 4 u - 3 - 2 m
Ta
G o i D = {t:
Taco
Bai loiin da cho trcf thanh:
-
iL
ie
uO
nT
hi
Da
iH
oc
01
/
-2
t
-1
-1/2
-1/2
° \
I
B
-1
O di/ctng iron tren c^t doan A B
283
Cty TNHH MTV DWH Khang Vi$t
Chuyfln 06 BDHSG Join g\i trj Ifln nha't va g\i trj nh6 nhat - Phan Huy KhSi
Ttfd6tac6
<=> O H < R < O A (cf day H la hinh chieu ciia O tren A B )
oOU'
c^--
2
4
2
max f ( t ) = m a x { f ( - l ) ; f ( l ) ) = max{4; 4} = 4 ;
-i
,
-l
4
••
:•>
Ta thu l a i k e l qua tren. Ban doc lif binh luan ve tinh IAJ viet cua tuTng phucjfjg
Thay v a o ( 4 ) t a c 6 0 < 2 m < 4 < = > 0 < m < 2 .
phap da trinh bay.
E)6 la cac gia t r i can t i m cua tham so m .
9. T i m m de phiTdng trinh sau V l + 2cosx + V l + 2 s i n x = m co nghiem v d i
B a i 9. T i m m de phufdng Irinh 2 + 2sin2x = m ( l + cosx)' co nghiem tren doat)
iL
ie
uO
nT
hi
Da
iH
oc
01
/
71 K
HUdng ddn gidi
fji({ingddn
.mir'^iuM:.'
gidi
Gpi D la m i e n xac dinh cua phi/Png trinh, t h i
R6 rang v d i m o i m thi 1 + cosx ^ 0 ( v i neu 1 + cosx = 0 => cosx = - 1
0 => V T = 2 vo l i ) . '•SI-J'"
sinx = 0 => sin2x
K"?
i
Do do dU'a phufdng trinh da cho ve diing tuTdng du'dng sau:
2 ( l + sin2x)
(1 + cosx)^
= m.
D = {x: 1 + 2cosx > 0, 1 + 2sinx > 0}
\
1\i sinx > - — va cosx > -—
2
2
1
n
27t
suy ra — < x < — .
^
6
3
(1)
Vay D = {x: X Ihoa man (1))
Ta
X
'
2t
1 — t^
D a t tan — = t. A p dung cong thuTc sinx =
vk cosx =
k h i do
2
l + l^
1+t•
1-t^^
1
= m o -({^ - 4 f ^ + 2 t ^ + 4 t + l)==m.
2
^
/g
n6n ( 1 ) o
1-t^
om
1+
up
2t
-l
(2) (3) CO nghiem o
fa
-i
(2)
xeD
V i f ( x ) > 0 V x e D , n c n m a x f ( x ) = /maxl"^(x)
xeD
7t
271
K h i — < x < — =>
6
3
w.
m i n f ( t ) < 2m < max f ( t ) .
-i
xeD
.
= > t = t a n - e [ - 1 ; 1].
ww
n n
bo
,
CO nghiem.
(3)
2'2
1;'
(2)
ce
f f ( t ) - t ' * - 4 l ' ' + 2 t ^ + 4 t + l = 2m
minl(x)
D a t t = sinx + cosx = \/2 cos X
ok
B a i loan da cho t r d thanh: T i m m dc he
Khi do phiTdng trinh da cho co nghiem k h i va chi k h i
xeD
/minf^(x)
Y xeD
(4)
Luc do m a x f ( x ) =
57t
—
4,
..
5n
_..
...
max
F(t) ;
minf (x)=
ct day F(t) = 2 + 2t + 2 V2t^ + 2 t - l .
T a c o F'(t) = 2 +
71
< x — < — => c o s — < c o s
12
4 12
12
>/6-N/2
< cos X — < 1
4j
Ta c6 f ' ( t ) = 4t^ - 12t' + 4t + 4 = 4(t - l ) ( t ' - 2t - 1).
V i the'CO bang b i c n i h i e n sau:
; minf(x)=
Ta c6 f"^(x) = 2 + 2(sinx + cosx) + 2^1 + 2(sinx + cosx) + 4 s i n x c o s x
.c
l+t^
(chu y khi X e
ro
'
s/
1 + sin2x = (sinx + cosx)'
Dat f(x) = V l + 2cosx + V l + 2sinx , x e D.
4t + 2
,
V2l^+2l-l
V3-1
>QVte
X
<1
4j
max
F(t), »
