luyện thi đại học môn toán

ruyeii chifit & Gi&i thifu dethi Todn hqc - Nguyen Phii Khdnh , Nguyen Tai Thu.
Cau8.a:Tac6 A B = ( l ; - l ; 2 ) , A C = (-l;-l;3)=:> A B , A C

H o n niia: C M

=(-l;-5;-2)

U

nen phuang trinh mat phang ( A B C ) :
Va B N 8 4
3'3

G(?i true tam ciia tam giac A B C la H(a,b,c), khi do ta c6 hf:
a - b + 2c = 3

c=l

Ion nhat khi H = A, VTCP cua A la u^ = U j ; A B

Do duang t h i n g ( A ) nam trong ( A B C ) va vuong goc voi (d) nen:
u.

InABC

nABC,nd

= ( 2 ; l ; l ) va AB = ( l ; 0 ; l ) .

Ggi H la hinh chieu ciia B len A. Ta co: BH < A B , khoang each tir B den

iL
ie
uO
nT
hi
Da
iH
oc
01
/

a + 5b + 2c = 9

u(2;l)=>AC:6x + 3y + l = 0

CauS.brTacoVTCPciiadla:

a=2

C H . A B = 0 ci> a + b-3c==0 <=> b = l = > H ( 2 ; l ; l )
HG(ABC)

n(l;2) la mpt vec to phap tuyen ciia AB nen phuon

3

trinh A B : x + 2y + 2 = 0

- l ( x - l ) - 5 ( y - 2 ) - 2 ( z + l ) = 0 c i . x + 5y + 2 z - 9 = 0

BH.AC = 0

4 8

x=l +t
=(1;-1;-1)=:>A: y = l - t
z = -t

Cau 9.b: Xet z = 0 la nghi|m cua phuong trinh

= (12,2,-11).

Xet z * 0 . Dat z = a + bi (a,b e ]R,a2 + b^ > o), t u gia thiet ta co:

Vay duang t h i n g (A) di qua diem H(2;1;1) vacoVTCP (12;2;-11)

3

+ 1=2

.z + 6.Z.

3.
2
.1 <=> .v3

up

+ 1 = 2a

/g

+ 1 = 2a

- 2b z = 0

b(b + l ) = 12a'

bo

ok

TH2: Lay dugc 8 vien co dung 2 mau: co C j j + C15 + C13 - 2 = 8215

0

3|z|^ = b
a > 0, b > 0

-2b

+I=4a2

a>0,b>0

b(b + ])=:12a2(l)
<=> {

3a2+3b2 =b(2)
a>0,b>0

ww

w.

fa

ce

_,
8215 + 1
316
n^c
xj"
r>
n ooc
=> PT =
=
* 0,065. Vay, PA « 0,935
^
125970
4845
^
B. Theo chUcrng trinh nang cao
Cau7.b: B = B C n d , => B ( 0 ; - 1 ) => B M = (2;2). Do do B M la mgt vec to phap

.^3|z|

6z

<=> \ 3 z ^ - b = 0

-b =0

om
.c

Cl =1 each

z.i

+ 1 =2a
2

ro

gian mau la n(Q) = C^„ = 125970
Gpi A la bien co lay duoc 8 vien c6 du ca 3 mau
A la bien co lay dugc 8 vien khong dii ca 3 mau. Khi do
THl. Lay dugc 8 vien co diing 1 mau (chi xay ra lay dug-c 8 bi vang) v | y co

:4

+ 6 z z.i o z.z^/3

s/

Cau 9.a: Lay ngau nhien 8 vien tu hop gom 20 vien ta c6 so phan t u ciia khong

+ 1=2

^3|z|^ + l = 2 | z | ( a - b i ) + 6

Ta

,
. x-2
y-1
z-1
nen A :
=
=
12
2
-11

tuyen ciia B C = > M B 1 B C
Ke M N / / B C cSt d j tai N , vi tam giac A B C can tai A nen t u giac B C N M la
hinh chCr nhat.
rMN//BC
Do \ , r i > M N : x + y - 3 = 0, N = M N n d 2
lQuaM(2;l)
^
NCIBC
Do

QuaN

3 3^

^

•N

fS

1^

l3'3

Lay (1) tru (2) ve theove taco 9a2 = 4b2 <::> a^ = - b ^ ( 3 )
The ( 3 ) vao ( l ) . ta dugc : ^ b 2

b^ + b « b = 4
13

a = — , (do a > 0,b > 0)
13

2
3 .
=i> z = — + 1
13 13

{1 _ 5
N C : x - y - - = 0,C = N C n d i ^ C
3' 3
3

183

Tuyen chgn b Giai thi?u
Khanli, Nguyen lai

iniu

la so thuc va a + p = 2>/3 .

Cau 9.a: Cho a, p la hai so phuc lien hgp thoa

DETHiTH(jfs628

Tinh a
I . PHAN CHUNG CHO TAT CA CAC T H I S I N H

B. T h e o c h U o r n g t r i n h n a n g c a o

Cau 1: Cho ham so: y = x-* - 3mx^ + 9x +1 c6 do thi {C^)

Cau 7.b: Trong mat phSng tpa dp Oxy, cho duong tron (C): x^+y^-2x-4y-4=0

a) Khao sat su bien thien va ve do thi (CQ) ciia ham so.

CO tam I va diem M ( 3 ; 0 ) . Viet phuong trinh duong thang A, biet A cat (C) tai

b) Gia six duong thing ( d ) : y = x +10 - 3m cat do thj [C^) cua ham so tgi 3

hai diem phan biet A ,

B, C

c6 hoanh do ian lugt

Xi,X2,X3.

sao cho t u giac

iL
ie
uO
nT
hi
Da
iH
oc
01
/

diem phan biet A ,

B

ABIM

la hinh binh hanh.

Cau 8.b: Trong mat phang Oxyz, cho mp ( p ) : x + 2y + z - 3 = 0, duong

Tim m de: xf +X2 +x^ <11.

X—1

Cau 2: Giai phuong trinh : cos-* x + sin'' x = cosx + sin2x + sinx

y+1

z—2

thang A: - — = —— = _ _
^/3x^ + x + ^ + j 3 y ^ - y + ^ = m ,
Cau 3: Tim m de he phuong trinh
^
CO nghiem thuc.
N/SX^-X+I +>/3y^ + y + l =m

va diem A ( 4 ; l ; - 3 ) . Viet phuong trinh duon^

thSng d nam trong (P), biet d cat A va khoang each tu A den d bang 42 ,
Cau 9.b: Tim c biet a, b va c la cac so nguyen duong thoa man c=(a+bi)'^ -107i

xe'^+1

Hl/dNG DAN GIAI

Cau 4: Tinh tich phan: I = j (e^+lnx)

Ta

I . PHAN CHUNG CHO TAT CA CAC THI SINH

Cau 5: Cho hinh chop S.ABCD c6 day ABCD la hinh thang vuong tai A, B. Biet

s/

Cau 1:

up

AD = 2AB = 2BC = 2a, SA = SD = SC = 3a. Tinh the tich khoi chop SABC va

a + b + 2c = 8'

-3mx^ + 8x + 3 m - 9 = 0 < : > ( x - l ) r x 2 + ( l - 3 m ) x + 9 - 3 m l = 0 ( l )

om

Cau 6: Cho cac so thuc a,b,c thoa man dieu kien

/g

1
o X = 1 ( gia su X3 = 1) hoac x^ + ( l - 3m)x + 9 - 3m = 0 ( 2 ) .

.c

Tim gia trj Ion nhat cua P = a'^ + b^ + 5c^.

ok

I I . PHAN R I E N G T h i sinh chi dugc chpn lam mpt trong hai phan (phan

De duong thSng (d) cat (C) tai 3 diem phan bi?t thi phuong trinh ( l ) c6 3

bo
ce

A hoac B)

nghiem phan biet <^ phuong trinh (2) c6 2 nghiem phan biet khac 1, tuc la
2

phai c6:

2

w.

fa

A . T h e o chiTofng t r i n h c h u a n

a) Danh cho ban doc.
b) Ham so da cho xac dinh tren R
Phuong trinh hoanh do giao diem ciia (C) voi duong thang (d) la

ro

khoang each giira hai duong thang SB va CD.

Cau 7.a: Trong mat ph5ng tc?a do Oxy, cho duong tron ( C ) : ( x - l ) +(y+2) =1

ww

A = fl-3mf-4(9-3m)>0

va duong thang (A) : 2x - y +1 = 0. Tim diem A thuoc duong thSng (A) sao cho
tu A ke dupe cac tiep tuyen AB, AC (B, C la cac tiep diem) den duong tron (C)
dong thai di^n tich tarn giac ABC bSng 2,7 .

I

m e

l + l-3m+9-3m;.tO

= I va mat phang ( P ) : x - y + 2z + 5 = 0. Viet phuong trinh mat

Voi dieu kien (3), phuong trinh (2) c6

phang ( Q ) d i qua M , song song voi d va tao v6i ( P ) mot goc cp thoa coscp = ^ .

184

x^ +x^

—

'5

3

(3)

11
—
6

Cau 8,a: Trong mat ph^ng Oxyz, cho diem M ( l ; 3; 1), duong thang d:
^ ~ ^ = y±l

7
—00:

Xj + Xj =
XjXj

3m - 1

=9-3m

(theo dinh ly V i - e t )

< l l < ^ l + (3m-l)^ - 2 ( 9 - 3 m ) < l l o m 2 <3omer-N/3;N/3

Doi chieu dieu ki^n, suy ra m e

1^

la gia trj can tim.
185


Cau 2: Phuang trinh du(cosx + sinx)(l-cosx.sinx) = cosx + sin2x + sinx

Di?n tich tam giac AABC: S ^ ^ B C = ^ AB.AC = ^
Gpi K la hinh chie'u vuong goc ciia

<=> i(cosx + sinx)sin2x=sin2xo (cosx + sinx-2)sin2x=0

=> SK 1 (ABCD), SK la duong cao

(keZ)

ciia chop SABC
Hon niia cac tam giac vuong
ASKA=ASKC=ASKD vi
ASKA=ASKC=ASKD chung
va SA = SD = S C = 3a
KA = K C = KD

Vgy, phuong trinh c6 1 ho nghi?m
Cau3:Truvetheove: yjsx^ +x + l-^3x^-x

+ l = ^jsy^ +y+

-^P/-y+

Xet ham so f (u) = Vsu^ + u+ 1 - Jsu^ - u + l,Vu € #
1^
-g u —
6
6;

1

^

u+

V

voi g u + I
6j

1^

Trong tam giac vuong ASHD ta c6 SH = \/sD^-HD^ = \l9a^ - a ^ = 272a

v

2

6,

11

12

3 u
V

f

6>

The Tich khoi chop SABC: V = - S H . S ^ A B C = -2V2a.—
3
3
2

11
12

'

voi t - u + i .

up

Xet g(t) = - p = ^ , V t e #

3 u -

1^

6,

u+

=> K la tam duong tron ngo^i tiep tam giac AACD => K trung voi H

f

1^

>g

ro
/g

om

11
12

.c

.2

3t^4-

. f ' ( u ) > 0 , V u e i ^ nen phuong trinh f (x) = f (y)
.

CD//(SBH)
,

,
,
Mat khac
>0=>g(t) dong bie'n voi V t e # , s u y rd

bo

l^

3t^2 + 11
12
6j

Ta CO

x=y

fa

w.

Ta CO ( a - l ) ( b - l ) > 0 = > a b - ( a + b) + l>0=>ab>a + b - l = 8 - 2 c - l = 7-2c.
ab(8 - 2c) > (7 - 2c)(8 - 2c) => -3ab(8 - 2c) < -3(7 - 2c)(8 - 2c).
Doc<4=>8-2c>0

Gpi H la trung diem ciia AD => HA = HD = a.
fCH = a

P = -3c^ + 96c^ - 384c + 512 - 3ab (8 - 2c)

Tu gia thie't => ABCH la hinh vuong canh a tam O =>

Trong tam giac AACDco C H la trung tuyen va C H = i A D : ^ A A C D
vuong tai C
186

2

o P = -3c^ + 96c^ - 384c + 512 - 3ab (8 - 2c)

ww

Cau 5: Theo gia thie't ta c6 BC = AB = a.

f CO 1 H B
^
. C O l ( S B H ) ^ C O = d C,(SBH)
[COISH
^
'
L 'V
/J

Cau 6: P = a^ + b^ + 5c^ = (a + hf -3ab(a + b) + 5c^ = (8 - 2c)^ - 3ab(8 -2c) + 5c^

Khiay taco: N/SX^ + x +1 + Sx^ - x +1 = m, tu day tim duoc m > 2 .
Cau4: t^^^xle"+lnx)

, fCD//BHe(SBH)

Ta CO SB va CD la hai duong thSng cheo nhau

ce

^

g "^6

11

ok

Ta c6: g'(t) =

3

Tu giac BCDH la hinh binh hanh (vi HD //BC, HD=BC => CD// BH

Ta

V

s/

f

iL
ie
uO

nT
hi
Da
iH
oc
01
/

<=> sin2x = 0 o 2 x = k7i » x = k ^

S tren mat phSng ( A B C D )

H la tam duong tron ngoai tiep tam giac AACD

P < -3c^ + 96c^ - 384c + 512 - 3(7 - 2c) (8 - 2c)

P < -3c^ + 84c^ - 294c + 344

I

Tugia thietsuy ra 2c<6=>c<3 =>l
P

Xet ham so f (c) = -3c^ + 84c^ - 294c + 344 voi c e [1;3]
f'(c) = -9c2 + 168c-294,
f'(c) = 0 < » - 9 c 2 +168c-294 = 0 o 3 c 2 - 5 6 c + 98 = 0
187

oc =

28 + 7VT0

f(l) = 131, f

r,
«n;3]

0, Theo chirorng trinh nang cao
Cau7.b: (C) cotam l ( l ; 2 ) , b a n k i n h R = 3

u 28-7VT0
hoac c =
e[l;3]

28-7VTo'l

Do A B IM la hinh binh hanh nen AB//MI => M I = (-2; 2 ) la VTGP cua A

=, f(3) = 137

=:>A:x + y + m = 0

Vay gia trj Ion nliat cua P la 137, dat dugc khi c = 3,a = l , b = 1
I I . PHAN R I E N G T h i sinh chi dugfc chgn lam mpt trong hai phan (phan A

Goi H la trung diem AB, ta c6 HB = i AB = - M I = - Vs = sfl
2
2

2

hoac B)

iL
ie
uO
nT
hi
Da
iH
oc
01
/

A. Theo chUorng trinh chuan

=>IH =\/R^ - H B ^ = V 9 ^ = V7 r^d(I;A) = ^ / 7 ^ m = -3±^/l4

Cau 7.a:

Cau8.b:GQi 1 = A n ( p ) , v i I e A =^ l ( l + 2 t ; - l + t; 2 - 3 t )

Duong tron (C) c6 tarn l ( l ; - 2 ) , ban kinh R = l
Ta thay: BIC + BAG = 180°

Vi I 6 ( P ) = > l + 2 t - 2 + 2t + 2 - 3 t - 3 = 0 « t

sin BIC = sin BAG

(1)

T a c 6 m p ( P ) c6 VTPT la: rT = {V, 2; l )

H o n n u a : S^BIC = S A B C + S B , C « I B . A B = i l B 2 s i n B I G + i A B 2 s i n B A G (2)

Gpi u = (a;b;c ) la VTGP cua d, v i d nam trong (P) va cat A nen d d i qua I

Tie (l) va (2), suy ra:

va u l n < = > u . n = 0 <=>a+ 2b +c = 0 o c = - a - 2 b => u = (a;b ; - a - 2b)

2IB.AB
IB^ + A B '
(3)

10

/g

ro

Tir (3) ^ AB - 3 hay lA^ = AB^ + IB^ -10

ok

Me(Q)

w.

fa

(4b + 2 c ) x - b y - c z - b - c = 0. M a : coscp = | ^ 6 ( l 7 b ^ + 16bc + 5c^ j = 6|b

Mat khac

(a + i b f

a3
p2

ww

Cau9.a: Gia s u a = a + ib ( a , b £ K ) t h i p = a - i b

a3-3ab^+3b(a^-b^)i

3b(a2-b2) = 0
b2=3

188

a2=3
b'=3

.Vay a = Va^+V = 3%/2

'

'

z = -4 +1
.\
Cau 9.b: Ta c6: c = (a + bi)^ - 107i = (a^ - 3ab^ j + (sa^b - b^ - 107ji
$• D o c e Z " " nen CO 3a^b-b^ =107 = 0 o b ( 3 a ^ - b ^ ) = 107
|'b = l
fb = 107
< 2 u2 _
hav i o , 2 u2 .
3a2-b2=l
Sa^-b'^ =107
hay

a=

SO nguyen to)

11450

3
b = 107
Ma a, b la cac so nguyen duong nen ta chi nhan a = 6, b = 1
b=l

Tie gia thiet ta suy ra

)

o a = - b =>d: y = l - t

a=6

(ap).(ap)

^

o ( a + b)^=0

x =5+t

dang:

bo

CO

= 7^

0 6b2=2f2a2-.4ab.5b2)

^a2+b2+(a + 2bf

'A

ce

[d = b + c
=i>(d)//(Q)
[a = - 4,b - 2 c , khi do ( Q )

.c

Cau S.a: Gpi phirong trinh mat phSng ( Q ) can tim la ax + by + cz + d = 0

Ta CO |a-p| = 2N/3c^|2bi| = 2 V 3 » b ^ = 3

Vb^.4b^^b^

om

< = > ( a - l f + ( 2 a + 3 f =10 voi A(a;2a + l )

Gia thiet, ta c6:

Ta

27

AI,u

s/

3

AB
1B.AB'
1
Mat khac: S^ec = T I B ^ sin B I G =
2

IB^+AB^ " l + AB^

Vi khoang each tir A den d bSng ^/2 nen

up

2IB.AB = (iB^ + AB^ jsin BAG => sin BAG =

= 2:^1(5 ; 1 ; - 4 )

Vay, c = a3-3ab = 198
189


Viet phuong trinh mat phSng (p) di qua M ( 4 ; 3 ; 4 ) , song song voi duong

DETHITHljrsd29

thang A va tie'p xuc voi mat cau (S).
Cau 9.a: Tim so phuc z thoa man phuong trinh z.z + z^ - ^z - 2zj = 10 + 3i.

I . PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1: Cho ham so y =

- ( m ^ + m - sjx +

B. Theo chUcrng trinh nang cao

- 3 m + 2 ( l ) , c6 do thi ( C ^ )

Cau 7.b: Trong mat phMng tpa dp Oxy, lap phuong trinh chinh t3c cua elip ( E )

a) Khao sat su bien thien va ve do thj ( C Q ) cua ham so.

biet no C O mpt dinh va 2 tieu diem cua ( E ) tao thanh mpt tam giac deu va chu

th5ng y = 2 tai ba diem phan biet c6 hoanh dp Ian

Ixxgt

la

Xi,X2,X3

iL
ie
uO
nT
hi
Da
iH
oc
01
/

b) Tim tat ca cac gia tri thuc cua m sao cho do thi ham so ( l ) cat duong

viciia hinh chii-nhat C O so cua ( E ) la 1 2 ^ 2 + V 3 J .

va dong

Cau 8.b: Trong khong gian tpa dp Oxyz, cho hai duong thSng

thoi thoa man dang thuc X j + X j + X3 = 18
. 3X

sm'^ ~ - cos
Cau 2: Giai phuong trinh:

3X

2 + sin X

- = cosx

Tim M thupc ( A , ) va N thupc A j sao cho M N = 2N/6 va tam giac A M N

vuong tai A.

Cau 3: Giai phuong trinh: \lx^-5\ 6 + N / X ^ + Vx + 21 = V ? + 1 9 x - 4 2
2

Cau 9.b: Tim so phuc z thoa man dieu ki?n z + — = 8 - 6i .
z

Ta

Cau 4: Tinh tich phan: I - J(x + l)lnx.e''dx

ro

BC = aV2,BD = a%/6 . Hinh chieu vuong goc ciia dinh S len mat phang (ABCD)

/g

la trong tam cua tam giac BCD. Tinh theo a the tich khoi chop S.ABCD, biet

om

rang khoang each giua hai duong thSng AC va SB bang a.

ok

a
b
c
c^ b^ c^
— + ^ +^ +—+ — + —
b^c c^a a^b
b
c
a

bo

thuc:

.c

Cau 6: Cho ba so a,b,c > 0 thoa a + b + c < 1,5. Tim gia trj nho nhat cua bleu

Hl/6fNGDilNGlAl

I . P H A N C H U N G CHO T A T CA CAC T H I S I N H
Cau 1 :

up

Cau 5: Cho hinh chop S.ABCD c6 day ABCD la hinh binh hanh thoa man AB = 2a,

s/

1

a) Danh cho ban dpc.
b) Phuong trinh hoanh dp giao diem cua do thi ham so' ( l ) va duong thioig y = 2 .
^'

x3-(m2 + m - 3 ) x + m 2 - 3 m + 2 = 2 < : > x 3 - ( m 2 + m - 3 ) x + m 2 - 3 m = 0

o (x - m)(x^ + mx - m + 3) = 0 o X = m hoac x^ + mx - m + 3 = 0 ( 2 )

hole B)

Do thi ham so ( l ) c^t duong th^ng y = 2 tai 3 diem phan bi?t khi va chi

fa

khi { 2 ) CO hai nghiem phan bi^t khac

w.

A. Theo chuorng trinh chuan

ce

I I . PHAN RIENG T h i sinh chi dvtgc chpn lam mpt trong hai phan (phan A

d i : 2 x - y - 2 = 0, d 2 : 2 x + y - 2 = 0.

ww

Cau 7.a: Trong mat phSng tQa dg Oxy, cho diem 1(2; 4) va hai duong than;.;
m o •

Viet phuong trinh duong tron tam I cat d j t^ii hai diem A, B va cat d 2 t?'
Gia sir
hai diem C, D sao cho AB + CD = — ~ •
o
Cau 8.a: Trong khong gian vai h? tQa dp Oxyz, cho mat cau
(S): ( x - l ) ' + ( y - 2 f + ( z - 3 f = 9 v a d u d n g t h a n g A :
190

m^ + m ^ - m + 3?i0

m>2

A - m ^ - 4 ( - m + 3)>0

m < -6

Xj

= m va

X2,X3

la 2 nghiem cua ( 2 ) .

Khi do theo djnh l i Viet ta dupe:

X2

+ X3

= -m

X2.X3 = —m + 3

^ ^ ^ ^ ^ ^

2

Voi x^ + x^ + x^ = 18 <=> m^ + (x2 + Xg)^ - 2x3X3 = 18

191


<=>m^

- 2 ( - m + 3) = 18

Cau 5: Gpi H la hinh chieu vuong goc ciia S len mat phling ( A B C D ) , M la
trung diem C D va O la tam ciia day A B C D . Do A O la trung tuyen ciia tarn giac
nen

<=>m'^ + m - 1 2 = 0<=>m = -4 hoac m = 3.

ABD

Doi chieu dieu ki^n, ta thay m = 3 thoa man.
Cau 2:

.

X

X

X

.

X

X

3 sm — cos — = 2 c o s — s m —
2
2

2
2

0

X

X

cos- + s i n 2
2

2
2

X

BM^X

X

tren ta duoc A H l ( S H B )

X

X

——4 j

2N/2

v6 nghi^m.

s/
( t > l ) , t a CO phuong trinh:

ok

• (v/^T^ + l ) = J ( x + 2 l ) ( x - 3 ) , datt = ^x^

.c

x2

om

• Vx-3=0<=>x = 3

bo

- 4t^ +1 + 6 = 0 o t = 2 ho^c t = 3

ce

phuong 2 ve: yjx^-5x + 6 + slx + 2\ Vx^ + 19x - 4 2

2
Ti'nh j l n x.e^dx : Dat

192

ww

du = lnx + - + l dx

w.

fa

«(x-3)(x-6)(x-ll) =0

X

v=e
^2
2,x
2
flnx.e''dx+ [ — d x + f e M x

K h i d o : I - ( x + l)lnx.e

X
Uj

= Inx

d v j =e''dx

up

ro

, v 6 i x > 3 taluonco: 7 x ^ + l > 0

/g

7;7ri|7;^ + l ) = Vx + 2 l ( V ^ - l )

dv = e''dx

43^
2a V3
= 3a^ =>BM = a>/3=:>BH =
4

suy ra

AHIHK

Trong tam giac vuong SHB ta c6

Cdchl:

u = (x + l ) l n x

3

=i> HK la doan vuong goc chung ciia AC va SB
suy ra HK = a .

Ta

<=> cos

^x

Cau 3:

Cau 4: Dat

3

Ke HK vuong goc voi SB, theo chung minh

. x]
TH2: 2 cos—+ sm — = -3
I 2
2j

Cdch2:Binh

C D ^ 6a^+2a^
4
2

2

A H 1 (SHB).

THI: c o s | - s i n - = 0 hay t a n - = l <=>x = - + k27t

f

'

ket hop voi A H vuong goc voi SH ta dugc

X

.

BD^ + BC^
2

2

Ta CO A H ^ + B H 2 =43^ = AB^ r:> A H I B H ,

<r>cos--sin-^0
2
2

hoac 2 cos—+ sin — = -3
2
2
X

4

iL
ie

uO
nT
hi
Da
iH
oc
01
/

Phuang trinh cho <=>

. X
X
s m - - c o s - l + 2- s i n x
2
2
- = cos
2 + sin X

1

1

HK^

SH^

1

•+•

HB^

• SH =

2a

Vs.ABCD =|SH.SABCD =|sH.4.SoAB

| ; C a u 6 : ^ . A , ^ , 3 ^ ,
b^c

c^a

a^b

abc

=|sH.ioA.BH-

4V2a3

^ , ^ ^ , ^ , 3 ^ ,
b

c

a

W

; '

l , 5 > a + b f c>3\/abc
< - . Dat t = ^/abc thi 0 < t < i
2
2

Khido:

P>4-t^ + 3t'' voi 0tim duoc minP = —
2
1
6

khi a = b = c = -!2

I I . PHAN RIENG T h i sinh chi dugc chpn lam mpt trong hai phan (phan A
hoac B )
A. Theo chUorng trinh chuan
Cau 7.a: G ^ i R la ban kinh duong tron can tim va F, G Ian lugt la hinh chieu
vuong goc ciia I tren dj va d j . De thay I F = — ,
5
Laico: F B = VR^ - IF^ =

JR^TI,

GD =

IG =

^ .
5

TR^TJ^ = ^ R ^

-

^6
193


Tuife'u chQH & Giori thifu dethi

Toiin HQC - Nguyen

Theo bai toan: A B + CD =

Phu Khduh

, Nguyen

2(FB + GD) =

Tii't Thu.

M N = 2>/6 » ^ { - 3 a - l)^ + (-3a +1)^ + (3a +1)^ = 2^6

^ R

Cau9.b:Giasu z = a + bi (a,belR, a^ n-b^ >0J

Cau 8.a: Gpi vecto phap tuyen n = (a;b;c) ciia m|t phang ( p ) .

- 25
25
Ta CO : z + — = 8 - 6i <=> a - bi +
= 8 - 6i
z
a + bi
^. 2 5 ( a - b i )
a(a2 + b2+25)
-bi + — ^
--^ = 8 - 6 i < ^ - i
L
a2 , L,2
2 i2
3 +t)
a'^+b^

Mat phang (P) song song v6i duong thang A -3a + 2b + 2c = 0

o-^==^i^i_

= 3 <»(b-2c)(2b-c) = 0

VlSb^+Sbc + lSc^

Cau 9.a: Tim so' phuc z thoa man phuong trinh z.z +

-

iL
ie
uO
nT
hi
Da
iH
oc
01
/

d(l;(P)) = R

- 2zj = 10 + 3i.

a(a2+b2+25)

Gpi z = X + yi (x,y e K),ta c6 z = x - y i va z^ = x^ - y^ +2xyi

b(a2+b2+25)
a^ + b^

(x^ + y ^ ) + (x^ - y ^ + 2 x y i ) - [ ( x + y i ) - 2 ( x - y i ) ] = 10 + 3i

s/

Ta

2

5

3

2

8

Vay, CO hai so' phuc can tim la z = 2 + 3i, z =

a = 0=>b = 0
a = 4=:i.b = 3

up

' = -8

- 8a +16] = 0

Vay, so phuc can tim la z = 4 + 3i

ro

y =3

^

i

DETHITHUfSOSO

/g

y(2x-3)-3

(2)

The (3) vao ( l ) ta c6 a p

^_5
hay

=6

Lay ( l ) chia (2) ve theo ve;ta dupe - = i ^ b = -aC3^
b
b 3
4/ I V /

(2x2 + x) + y ( 2 x - 3 ) i = 10 + 3i
x=2

(1)

a^ + b^

Tugiathiet: z.z + z ^ - ( z - 2 z j = 10 + 3i

2x2 + x = 10

bfa2+b2+25)
\o ...
0
T
l-o-bi
a^+b^

om

B. Theo chUorng trinh nang cao

bo

ok

.c

^2
2
Cau 7.b: Ggi phuong trinh chi'nh tac ciia eli'p ( E ) la : -— + ^ = 1 (a > b > O)
a
b

PHAN CHUNG CHO TAT CA CAC T H I SINK
au 1: Cho ham so: y

x'* - 2mx^ - 3 c6 do thj la ( C ^ )

a) Khao sat sif bie'n thien va ve do thj (C_j) ciia ham so.

tren tryc nho tao thanh mpt tam giac deu.
BF2=FjF2
ABF1F2 deu c:>
BFj = BF2 » c^ + b^ = 4c2 » b^ = 3c2 = 3(a2 - b^) <::> 3a2 = Ah^

b) Tim meM

ww

w.

fa

ce

Do cac dinh tren tryc Ion va Fj,F2 thMng hang nen Fj,F2 cung voi dinh B(0;b)

Hinh chii nhat co so c6 chu v i 2(2a + 2b) = 12(2 + N/S) O a + b = 6 + 3\/3
a=6
3a2 = 4b2
Ta CO he:
^ ' 36 27
a + b = 6 + 3N^
CauS.b: M thupc ( A J )

M(a;a;a +1), N thuQC A j = > N ( b ; b + 2;-2b)

Tam giac A M N vuong tai A nen A M 1 A N <=> A M . A N = 0
o ( a + l ) ( b + l ) + a(b + 2 ) - a ( 2 b + l ) = 0 = i > b - - 2 a - l
= > N ( - 2 a - l ; - 2 a + l;4a + 2)

de ban kinh duong tron ngoai tiep tam giac c6 cac dinh la 3

"em cue trj ciia do thj ham so (C,^) dat gia trj nho nhat.
au 2: Giai phuong trinh : sin3x = cosx.cos2x|tan2 x + tan2x
iu 3: Giai phuong trinh: x^ - 3x - 4 = V x ^ J x ^ - 4x - 2)

au 4: Tinh tich phan: I = |- 1
ocos XI

f

x+

1
tan^x-4

dx

Cau 5: Cho hinh chop S.ABCD c6 day ABCD la nua luc giac deu va AB = BC =
CD = a. Hai mat p h i n g (SAC) va (SBD) cung vuong goc voi mat phSng day
195


Tuyeit chgn 6- Gi&i thi?u dethi Todn hqc - Nguyen PM Khdnh , Nguyen Tat Thu.
(ABCD)

Cty TNHH MTV DWH Khang Vift

Hl/dTNGDANGlAr
I . PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1:

. Tinh theo a the ti'ch cua khol chop S . A B C D biet r i n g khoang each

giiia hai duong thang A B va S D b i n g

.

a) Danh cho ban dpc.
Cau 6: Cho cac so thuc a,b,c thoa man (a + b + c)^ = 2^3^ + b^ + c^ j . Tim gia
b) m > 0 t h i h a m s o c 6 3 c u c t r i A ( 0 ; - 3 ) , B ( - V i ^ ; - m 2 - 3 ) , c ( V ^ ; - m 2 - 3 )
tri Ion nhat va gia tri nho nhat cua bieu thuc : Q = -.
;
x-j——;
r
(a + b + c)(ab + bc + ca)
I I . PHAN RIENG T h i sinh chi dugrc chgn lam mpt trong hai phan (phan A

A a A C B C _ ( " ^ + "^'*)-2Vm

A. Theo chUcTng trinh chuan
+

4S

= 13 va

Xet:

phuong trinh duong thSng di qua A va cat ( C j ) , ( C j ) theo 2 day cung c6 do

^

4.m^.>/m

'AABC

( C j ) : (x - 6)^ + y^ = 25. Goi A la giao diem cita ( C j ) va (C2) voi y ^ < 0 . Viet

—1 + m^2
m

f{m) = -

voi m > 0.

Taco: f'(m) = if — L + 2ml va f ( m ) = 0 « m = 3(l.

dai bang nhau.

Ta

(S):

s/

Cau 8.a: Trong khong gian v6i h^ toa do Oxyz, cho mat cau

ok

.c

om

i z - ( l + 3i)z
2
Cau 9.a: Tim so phuc z thoa man dieu ki^n
—= z
1+i
B. Theo chUtfng trinh nang cao

.

/g

A tiep xiic voi mat cau (S) tai A tao voi tryc Ox mot goc a c6 cos a =

up

= 9 va diem A ( 1 ; 0 ; - 2 ) . Viet phuong trinh duong thang

Ban kmh duong tron ngoai tiep tam giac ABC nho nha't

V2

4

Cau 2: Dieu ki^n: cosx

0,cos2x ^ 0

Phuong trinh cho tuong duong voi sin3x = ^"^^^"^'"^ %sin2x.cosx
cosx
<=> sin x cos 2x + sin 2x cos x = cos2x.sin^x
+ sin 2x cosx
cosx

bo

Cau 7.b: Trong mat p h i n g toa dp Oxy, cho hinh binh hanh A B C D voi A(1;1) ,

ce

B(4;5) . Tam 1 cua hinh binh hanh thupc duong thSng ( d ) : x + y + 3 = 0 . Tim

o sinxcos2x =

cos2x.sin^x
cosx

sinx = 0
tan X

=

1

w.

fa

toa do cac dinh C, D biet rang di^n tich hinh binh hanh A B C D bang 9.

ww

Cau 8.b: Trong khong gian voi h^ toa do Oxyz, cho hai duong thang
( d i ) : ^ - ^ ^ = Y,

R=M khi m=3/l

ro

(x + l ) ^ + ( y - l ) ^

= m .Vm

Ban kinh duong tron ngoai tiep tam giac ABC la:

hoac B )

Cau 7.a: Trong mat phang toa dp Oxy, cho 2 duong tron ( C j ) :

=i|y,-y^

8,^3,

iL
ie
uO
nT
hi
Da
iH
oc
01
/

AB.AC.BC = (m.m^).2V;;^,

( d 2 ) : ^ = ^ ^ = Y va mat phSng (P) c6 phuong

X =

kTT

— + krt
«

X =

1

4

Doi chieu dieu kif n, ta tha'y x = k:: thoa man.
Vay, phuong trinh c6 1 hp nghi^m.

:au3: x 2 - 3 x - 4 = V ; r ^ ( x 2 - 3 x - 4 ) - ( x - 2 ) V ^

trinh x + y - 2 z + 3 = 0 . Viet phuong trinh duong thSng A song song voi (P)
va cat d , , d 2 Ian luot tai hai diem A , B sao cho A B = \/29
Cau 9.b: Tir cac so 1, 2, 3, 4, 5 c6 the lap dupe bao nhieu so t u nhien c6 nam chu

o(x-2)

^x^ - 3 x - 4
1

+

=0 o
(x-5)

x + 1-

=0
2+ Vx^.

so, trong do chii so' 3 c6 mat dung ba Ian, cac chir so'con lai c6 mat khong qu^i
mot Ian. Trong cac so' t u nhien noi tren, chpn ngau nhien mpt so', t i m xac suat

N / X ^

x=2

Ta c6: x +1 > 2,

5==

< - nen x +1 -

7 = = 0 v6 nghi^m.

de so dupe chon chia he't cho 3.
196

197


4

\

X

Cau4: I = J

D|it x = a + b , suy ra (a + b)^ + = 4ab + 2cx => (x - c)^ = 4ab < (a + b)^ = x^

dx

—dx+

—

=> 0 < c < 2x . Khi do: Q =

COS^-•
X
It

| _ ^ d x = |xd(tanx) = x.tanxg71 - 4Jtanx.dx = ^ + (ln|cosx|),4lo = 4- + l n —
2
0 cos^ x
0
0

iL
ie
uO
nT
hi
Da
iH
oc
01
/

n - ,I n t a n x - 2r
4 tanx + 2

d(tanx)

Xet f(t) = (t^* + l^\f voi t > i , 2taco f'(t) = O o t = l hoac t = 5
II. PHAN RIENG Thi sinh chi dugrc chpn lam mpt trong hai phan (phan A
hoac B)

1, 1
=-ln0
4 3
4

A. Theo chUcrng trinh chuan

Cau7.a: A (2;-3) la giao diem (C,) va ( C j ) .

/g

ro

Vgy, I = —+ ln — + - l n ^
4
2 4 3
Cau 5: Gpi H la giao diem cua AC va BD. Do (SAC) va ( S B D ) cimg vuong goc
voi mat ph^ng ( A B C D ) nen SH vuong goc voi ( A B C D ) . Coi K la hinh chieu
vuong goc cua B len duong thang SD.
Do A B C D la nua luc giac deu nen AB vuong
goc voi BD, ket hop voi AB vuong goc voi S H
suy ra AB 1 ( S B D )

AD

.2SH =
Honnua:

3

.c
ok
SH.aV3 ^.VsH^
^

3

1

.SH =

1

+ HB^

1 2a 33^73 a^^f3
1

VABCD = 3SH.SABCD " 3 y -

Cau
198 6: Gia thiet suy ra: a^ +

4

5

+ c^ = 2(ab + be + ca).

5

17
I
,
i(a + b i ) - ( l + 3i)(a-bi) , ,
C a u 9.a: Theo gia thiet, ta c6-^
^ .
-=a +b
W

2a
^0= 3V3a'
—

S^BCD = S A B D + S B C D = ^ A B . B D . - B C . C D . s i n l 2 0

1I
Vay,

w.

2SSBD = S H . B D = B K . S D ^

ww

Mat khac:

^ a= 5 c
- = = a|= = = = VTo o a = -1c .ho|c

4J^b^

fa

2^33

b = 2a - 2c . A tao voi tryc Ox mpt goc a c6 cosa =

bo

HD

1
2

BC

Phuong trinh duong thang A di qua A c6 dang: a ( x - 2 ) + b(y + 3) = 0.
Duong tron (Cj) c6 tam O(0;0), ban kinh Rj =>/l3
I Duong Iron (Cj) c6 tam l(6;0), ban kinh = 5
Theogiathiet, suyra: R ^ - d ^ ( 0 , A ) = R ^ - d ^ ( l A ) =>x + 3y + 7 = 0
Cau 8.a:. Mat cau c6 tam l(-l;l;0), ban kinh R = 3
GQ\ = (a;b;c) la vecto chi phuong cua A , tu gia thiet suy ra lA.n = 0

ce

cua AB va S D suy ra BK =
HB
Do BC//AD suy ra

om

=> AB 1 BK => BK la doan vuong goc chung

•

Ta

X

Neu o O , bang each datt = - > ^ t h i Q = ^ * ,3
(t.l)

s/

i t a n ^ x - 4 cos

t a n x - 2 tanx + 2J

Neu c = 0 thi Q = 0

up

1 ' .

(x.cf

30

^-a-4b.(b-2a)i^^,^^,^[-a-4b.(b-2a)i](l-i)^^,^^,
1+i

-3a-3b + (5b-a)i

= 2(a2

- 3 a - 3 b = 2(a2+b2)

,5b,-a =

fl

„

+ b2)

2

45
26b2-9b = 0
' = 2-6
o a = b = 0 hay;
a = 5b
199


Vav, C O hai so phuc can tim la
•y

45
9

= 0, z, = — + — i .
2 26 26

1

^

DiTHITHIJfs631

B. Theo chuorng trinh nang cao

x-1
4a-3b-l

4a-3b-l

Tir ( l ) va (2), suy ra

o 4x - 3y - 1 = 0

5

b) Tim tat ca cac gia trj tham so m de duong thSng d: y = - x + m - 1 cat do

iL
ie
uO
nT
hi
Da

iH
oc
01
/

= lAB.h = l

=

thi (C) ham so tai hai diem A , B sao cho tam giac OAB npi tiep trong duong
5=

a = -2
b = -6

2 o 4 a - 3 b - 1 = 9 o 4a - 3 b a=

hoac

Cau 2: Giai phuong trinh: cos 3x. tan 5x = sin 7x

32
7

X''

C2

[7

r
I

-24]
' 7 J

53

e

52^

7'

7

Cau 4: Tinh tich phan: J = J In x - Vx dx
1

J

Cau 5: Cho hinh chop tu giac S.ABCD co day ABCD la hinh chCr nhat va AB = a,

A ( l + 2 t ; - l + t;t),

Ta

Cau 8.b: (P) c6 n ( l ; l ; - 2 ) A e d ,

(-32

BC = aVs . Mat phang (SAC) va mat phang (SBD) vuong goc voi day, I thupc

up

• B e d j =>B(l + a;2 + 2a;a) => A B ( a - 2 t ; 3 + 2 a - t ; a - t ) .

ro

r:> A B ( - t - 3 ; t - 3 ; - 3 )

/g

AB//(P)=> AB.n = 0<r>a = t - 3

om

AB^ = 2 9 < » ( t + 3)^ + ( t - 3 ) ^ + 9 = 2 9 0 t = ±l

.c

Cau 9.b: Gpi aja2a3a4a5 la so tu nhien c6 nam chO so, trong do chii' so 3 c6 mat

ok

diing ba Ian, cac chix so con lai c6 mat khong qua mpt Ian voi ai,a2,a3,a4,a5

bo

{1; 2; 3; 4; 5]

ce

6

+ v'' = 5x — V

Cau 3: Giai hf phuang trinh: \

7
Vay, C i ( - 2 ; - 6 ) = i . D i ( - 5 ; - 1 0 ) ,

tron CO ban kinh R = 2\/2 .

2

s/

'ABC

Cau 1: Cho ham so y = — ^ co do thj la (C)
x-1
^ '
a) Khao sat sy bien thien va ve do thj (C) ciia ham so'.

v - l

Phuang trinh duong thSng AB
Lai CO d(C;AB) = h =

I. PHAN C H U N G C H O T A T CA C A C THI SINH

.Doled=i>a + b + 8 = 0 (l)

Cau7.b:Giasu C(a;b) =>I

fa

S3p chix so 3 vao 3 vj tri, c6 C5 = 10 each

canh SC sao cho SI = 2CI va thoa man A I vuong goc voi SC. Tinh the tich cua
khoi chop S.ABCD theo a.
Cau 6: Cho cac so thuc x, y, z thoa man x^ + y^ + z^ = 9 . Tim gia trj Ion nhat
ciia bieu thuc: P = (9 + 2yz)(y^z^ - 4yz + s]
II. PHAN R I E N G Thi sinh chi dugc chpn lam mpt trong hai phan (phan A
hoac B)
A. Theo chUorng trinh chuan
Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC co dinh A(1;1), true

Vay khong gian mau c6 10.12 -120 phan tir

tam H ( - 1 ; 3 ) , tam duong tron ngoai tiep l ( 3 ; - 3 ) . Xac djnh tpa dp cac dinh

ww

Co ( l + 5);3;(2 + 4);3

w.

Con lai 2 vj tri, 4 chii so. Chpn 2 chii so xep vao 2 vj tri do, c6 C 4 = 12 each

B, C, bie't rang X g < x,^.

Gpi A bien co: "so dupe chpn chia het cho 3" c6 2 phuong an

Cau 8.a: Trong mat phSng Oxyz, cho hai diem A(3; 1; 2) va B(l; 2; 0). Lap

2 c h i f s o c o n l a i l a 1,5 c6 C5.2! = 20 so

phuong trinh mat ph^ng (P) chua A, B va tao voi mat phSng (Oxy) mpt goc

2 chu so con lai la 2,4

9 sao cho eos(p = - .
3

CO

C5.2! = 20 so

Vay bien co A c6 40 phan tu.
Xac suat ciia bien co' A la P =
200

40__1
120 ~ 3

Cau 9.a: Tim modun ciia so'phuc z, bie't z = ^ + 2z + 3
z+1
201

B. Theo chUcrng trinh nang cao

1
,
, Vm^ - 6 m + 5. m - 1
Taco: S^OAB = ^ A B . d ( 0 ; A B ) =
—

Cau 7.b: Trong mat p h i n g toa dp Oxyz cho hinh bmh hanh ABCD c6 D(-6;-6)
Duong trung true cua doan DC c6 phuong trinh ( d ) : 2x + 3y + 17 = 0 va

^_OA.OB.AB

duong phan giac goc BAC c6 phuong trinh ( d ' ) : 5x + y - 3 = 0 . Xac djnh toa

4SoAB

dp cac dinh con lai cua hinh binh hanh.
Cau 8.b: Trong khong gian tpa dp Oxyz, cho bon duong thang
x-1 _ y - 2 _ z
. x-2 _ y-2 _ z
•
, d, :
1
-2
4
-4
.x_y_z-Jl

.xzl-l z - 1
-1
Viet phuong trinh duong thSng A cat dupe ca bon duong th3ng da cho.

o s i n 8 x = sinl2x<=>x = ^

I. PHAN CHUNG CHO TAT CA CAC THI SINH

I

ro

/g

trinh (1) c6 hai nghiem phan biet, khac 1

om
ok

bo

-4x1X2

X j + X2 = m - l

(l)

( x H - y ) ( x - y ) = 3 (2)

hoac x . 2 y

Voi X = 2y thay vao phuong trinh (2) ta dupe: y^ - 1 = 0

Xj.X2 = m - 1

Cau 4: Xetham so f(x) = l n x - \ lien tuc tren doan [ l ; e ] .

g('^i) = 8(x2) = 0

= 2 m -6m + 5

(m,k6Z)

Voi X = - i y thay vao phuong trinh (2) ta dupe: y^ + 4 = 0 ( v6 nghifm )

ce
fa

ww

w.

B ( X 2 ; - X 2 + m - l ) , gpi X j , X2 la cac nghifm ciia (1), ta c6:

(x + y ) ( x 2 - x y + y 2 ) = 5 x - y

^'-^y^y^^<^x^-3xy-2y^=0ox^-ly
x-y
3
2

.c

o m < 1 hoac m > 5 ( 2 ) .

Cau 3: H? da cho tuong duong voi:

+

Nhan thay \^±y nen lay ( l ) chia cho (2) theo veta dupe:

Vdi (2) thi d cat do thj (C) ciia ham so tai hai diem A { X J ; - X J + m - l ) ,

=2(x2+Xj)

Vay, phuong trinh cho CO nghiem la: X = m7t, , X =

s/
up

x^l.

Duong thSng d cat do thj tai hai diem phan bif t khi va chi khi phuong

AB^=2(x2-Xj)

Vol x = — + — thi cos5x = cos ^ l 3 U 0 , ( V k 6 Z )
4 2
20 10

Ta

I

- ^ ^ = - x + m - l <=>g(x) = x^ - ( m - l ) . x + m - l = 0 ( l ) vai

g ( l ) = 1^0

hoac x = - ^ + Y ^ , ( k e Z )

Vdi x = — thi cos5x = cos — ^ 0 o k = 2m ( m e Z )
2
2

Cau 1:
a) Danh cho ban dpc.
b) Phuong trinh hoanh dp giao diem ctia hai do thj

-4(m-l)>0

m^ -4m + 3• = 4 <=> m = - 1 hoac m = 7 .
m-1

Cau 2: Dieu ki?n: cos5x ^ 0
Khi do phuong trinh cho tuong duong voi 2sin5x.cos3x = 2sin7x.cos5x

Hl/dTNGDANGlAl

A=(m-l)

2.Vm^ -6m + 5.|m-l

iL
ie
uO
nT
hi
Da
iH
oc
01
/

<=>

Cau 9.b: Tim m de phuong trinh: 27" - 32"^^ +15.3" - m = 0 c6 nghi^m - 1 < x < 2

x2

K-4m.3)^2p-6m.5) ^ ^

Taco: f ' ( x ) = : ^ ^ ^ > ^ ^ ^ > 0 = i > f ( x ) la ham so dongbien tren do?in [ l ; e ] ,
^ '
2x
2x
^'

)

OA^ = x^ + ( m - 1 - X j )^ = 2x^ - 2(m - l ) x i + ( m - if

suy ra f ( x ) < f ( e ) = l-\/e < 0 hay Inx-x/x = - l n x +Vx

= 2 g ( x j ) + m ^ - 4 m + 3 = m ^ - 4 m + 3,

Cau 5: Gpi O la giao diem cua A C va B D ^ ( S A C ) n ( S B D ) = S O , chung minh

m-1
Tuong t u OB^ = m^ - 4m + 3, d (O, A B ) = —j=
v2

dupe S O 1 ( A B C D ) , A C = VBA^TBC^ = 2a

OA = OC=a

DatSO = h =>SC = ^|S0^+0C^ =h^+a^ .
203

202
i i


Cty TNHH

Vi SI = 2CI nen IC = ^SC =-Vh^ +
3
3

MTV DWHKhang

Vi?t

Cau 9.a: Dieu kien
Gpi z = a + bi ( a , b e R )

Tarn giac A I C vuong tai I

, - z^+2z + 3
Theo gia thiet, ta c6 z =
z+1

nen A I = V A C ^ - I C ^ = ^VsSa^ - ( o < h < a N / 3 5 )

•\

o (a - bi)(a + bi +1) = (a + b i f + 2(a + bi) + 3

- N / s S a ^ - h ^ V h ^ + a ^ = 2h.a
3

<=> (-2b2 + a + 3 ) + (2ab + 3 b ) i = 0

iL
ie
uO
nT
hi
Da
iH
oc

01
/

2SsAC = A I S C = SO.AC

c^h'* + 2 a 2 h 2 -BSa'' = 0 o ( h 2 + 7 a 2 ) ( h 2 -Sa^) = 0 » h = a>/5

-2b^+a + 3 = 0

2ab + 3b = 0

^ V = isO.S^3CD=a^4^
x2>3

Cau 6: Khong mat tinh tong quat, gia su

[a = -3

"^1b

=o

3
a=—
2
b =±

Voi a = -3,b = 0 ,ta c6 z = Va^Tb^ = 3

Dat

t = yz:

yz <^

_ = _ _ < 3 , k h i do P = 2 t ^ + t 2 - 2 0 t + 72
2
2

3

S

Voi a = — , b = ± — , t a c 6
2
2

Xethamso: f ( t ) = 2t^+ t 2 - 2 0 t + 72 vdi | t | < 3 c6 f (t) = 2(t + 2 ) ( 3 t - 5 )

V4

4

Vay, modun cua so phuc z la 3 hay Vs

I I . PHAN RIENG Thi sinh chi dugc chpn lam mpt trong hai phan (phan A

Ta

B. Theo chUorng trinh nang cao

s/

hoac B )

ro

Cau 7.a: Goi D doi xung voi A qua I thi D ( 5 ; - 7 ) va D nSm tren duong tron

up

A. Theo chUorng trinh chuan

om

/g

( C ) ngoai tiep tam giac A B C : (x - 3)^ + (y + 3)^ = 20.

.c

Gpi J.la trung diem cua H D thi J la trung diem ciia B C nen B C : x - y - 4 = 0

ce

bo

ok

Tpa dp hai diem B , C la nghi^m cua h? phuang trinh: •

"'"(>'•'• 3)
[ x - y - 4 =0

Cau 7.b: Phuong trinh DC qua D va vuong goc (d) la: 3x - 2y + 6 = 0.
Giao diem ciia DC va (d) la: M ( - 4 ; - 3 ) va cung la trung diem DC.
Suy ra tpa dp C(-2;0).
Gpi

C la diem doi xung cua C qua d' thi C ' G A B , phuong trinh C C :

/ I 1^
x - 5 y + 2 - 0 . Giao diem C C va d ' la I
.Suy ra tpa dp C ( 3 ; l ) .
U'2
Phuong trinh A B qua C vuong goc (d) la: 3x - 2y - 7 = 0.

fa

Ma X g < XQ nen hai dinh can tim la B ( - 1 ; - 5 ) va C ( 5 ; l ) .

Cau 8.b: dj di qua M(1;2;0)C6 vecto chi phuong i j j = ( l ; 2 ; - 2 )

w.

Cau 8.a: Goi vecto phap tuyen cua mat ph3ng (P) la n = (a;b;c) ^ 0 .
dj d i q u a N (2; 2; 0)c6 vecto chi phuong Uj = (2;4;-4) = 2iaj d o d o d i / / d 2

ww

Do mat ph^ng (p) chua A , B nen n 1 B A trong do B A = (2; - 1 ; 2 ) , suy ra

Gpi (P) la mat phSng chua dj va d j thi (P) d i qua diem M ( l ; 2 ; 0 ) va c6

2 a - b + 2c = 0 = > b - 2 a + 2c ( l )
Vecto phap tuyen n = M N ; Qj
Hon

= (O; 2; 2), do do c6 phuong trinh y + z - 2 = 0

nCra mat phang (P) t^o voi mat phSng (Oxy) mpt goc ip sao cho
Gpi A = d3 n (p) thi tpa dp A thoa h?:

coscp =

suy ra 5a^ + Sac - 4c^ = 0 (2)

Tu ( l ) va ( 2 ) suy ra a = -2c hoac a = - c
5

X _ y _ z-1
X _ y _ / -Jl
2 ~ 1 ~ 1 <=> 2 " 1 " 1
o(x;y;z) =
V'2'2]
y+z-2=0
z-l+z-2=0

A

^'2'2

.Gpi B = d4 n ( P ) thi tpa dp B thoa hf:
204

205


Tuye'n

chqn

€t Gi&i

thifu

x - 2 _ y _ z-T
2 ~2~ -1
y+z-2=0

dethi

I I Q C - Nguyen

Todn

Phti

Khdttlt,

Ngtii/en

Tii't

Tftu.

Cty

x =y+2

iL
ie
uO
nT
hi
Da
iH
oc
01
/

AM vn r n nniFonff t r i n n

^

4

=

Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong tai A. Dinh

Cau9.b:Dat

B(1;1), duong t h i n g A C c6 phuong trinh: 4x + 3 y - 3 2 = 0, tren tia BC lay

V—2
z
= —

= -

1

Vi?t

A. Theo chiTomg trinh chuan

Vay A chinh la duong th5ng AB di qua B(4;2;0),c6 vecto chi phuong
X—

Khang

j l . PHAN RIENG T h i sinh chi dvegc chgn lam mpt trong hai phan (phan A
hole B)

khong cung phuong voi Uj = ( l ; 2 ; - 2 ) , do do AB cat d j va d j .

phuong trinh

DWH

Cau 6: Cho a, b la cac so' thyc thoa man a^ + b^ = 4a - 3b . Tim gia trj Ion nha't

va nho nha't cua bieu thuc: P = 2a + 3b

tai A , cat d 4 tai B '

Duong thSng AB, d ^ d j ciing chua trong (P), ngoai ra AB = ^'2'~2

^ 2
u =

MTV

SAB = SCB = 90" . Tinh the tich khoi chop S.ABC theo a va goc giua SB voi mat
phing ( A B C ) .

» ( x ; y ; z ) = (4;2;0):oB(4;2;0)
« y = 2-2z
2-2z + z - 2 = 0

Duong thSng A B nam trong mat phSng (P) cat

TNHH

diem M sao cho BC.BM = 75. Tim dinh C bie't ban kinh ciia duong tron ngoai

-1

tie'p tam giac A M C bang

1 = 3"

=>-3
Xet ham so f (t) = t^ - gt^ + 15t => -25 < m < 135

.

Cau 8.a: Trong mat phang tpa dp Oxyz, cho duong th3ng ( d ) : ^^=.X_^=£1?
va (P): - X + y + 2z + 5 = 0. Viet phuong trinh duong thiing d ' nSm trong mp

OETHITHlJfSOSl

/g

Cau 1: Cho ham so' y = - — j - c6 do thj la (C)

om

a) Khao sat sy bien thien va ve do thj (C) cua ham so.

.c

ok

ce

Tim toa do diem M thupc duong phan giac goc phan t u thu nha't sac cho

Cau 3: Giai h^ phuong trinh:

3

fa

w.

x = - l + 3t
Cau 8.b: Trong mat phang tpa dp Oxyz, cho duong thang d : y = 2 - 2t va hai
[z = 2 + 2t

ww

3

x-^ H-y-*

3xy ^
= 1
x+ y

+ — ^

7 ^ ( ^ 2 x - y + ^6x + y ) = ^ 3 x - 5 y + 5

'

'dx

Cau 5: Cho hinh chop S.ABC c6 day ABC la tam giac vuong can tai
AB = BC = aN/3, khoang each tir A den mat phSng (SBC)
206

•^ho nha't.

\/x(x\/x+l)

Cau 4: Tim nguyen ham: J = J — ^

B(3;-2), di^n tich tam giac AABC la — va trpng tam G ciia tam giac thupc
(Juong t h i n g A : 3 x - y - 8 = 0. Tim tpa dp diem C.

= >/2+2sin2x

\/tanx + cot2x

Hay tinh gia trj ciia bieu thuc A = z + 2iz .

Cau 7.b: Trong mat phSng tpa dp Oxy, cho AABC. Bie't tpa dp diem A (2;-3) va

bo

6

2 + yj2

1 +i

B.Theo chUorng trinh nang cao

b) GQI A , B la 2 giao diem ciia duong thang A: y = ^ x voi do thj ( C ) .

Cau 2: Giai phuong trinh :

Cau 9.a: Cho so phuc z thoa man d i n g thuc 2z + i.z =

ro

up

I . PHAN CHUNG CHO TAT CA CAC T H I SINK

M A + M B CO gia trj nho nha't.

yjli.

s/

Ta

(P) dong thoi each d mpt khoang bang

bSng

aS

b: Co ba hpp dung 5 vien bi trong do hpp thu nha't c6 1 bi trang, 4 bi
^^n; hpp t h u hai c6 2 bi trSng, 3 bi den; hop thu ba c6 3 bi tring, 2 bi den. Chpn
'^gau nhien mpt hpp roi t u hpp do lay ngau nhien ra 3 bi. Tinh xac suat dupe ca
^biden.
207

Tuyen chgn &• Giai thiju dethi Todn h
Giai cac p h u o n g t r i n h tren ta dupe:

H(/(}NGDANGIAI

x=- +

I. P H A N C H U N G C H O T A T CA C A C THI S I N H

1

4.

6
x-1
x +1

phuong trinh:

1

3

3=^ A

V l + xVx

7'

Is'sJ

ok
bo
w.

sin 2x ?t 0

sin2x

Ji = j|(t2 - i ) d t = ^ t3 - 1 1 + C i =

1(N/I

+

XV^]'

- l(VuW^)+q

ww

rsin2x = lI thoa dieu k i ^ n .
sin2x = 2

Vay, I = -JfN/l + x V ^ f - i f V l + x V ^ l + -Vl + xV^ + C
9V

/

3V

y

3

Suy ra t i i giac H A B C la m p t h i n h v u o n g .
Ta c6: A H // BC c (SBC) => A H //(SBC)

. ^
sin2x

o (Vsin2x -l).(2Vsin2x - >/2) = 0

vl + x v x

V l + xVx

Tuong ty H C 1 B C

1

P h u o n g t r i n h d a cho t u o n g d u o n g v o i : ^2 + >/2JVsin2x =^2+

\/sin2x = 1
^/sin2x =

Dat t = Vl + xVx, t 2 - l = x V x o x ^ = ( t 2 - l ) ^ < » x ^ d x = | t ( t 2 - l ) d t

^
, S H I (ABC) 1
Taco: ^
^
' i=*HAlAB .
S A I A B (gt)J

fa

cosx ^ 0

T a c o : t a n x + cot2x =

\ll + x\[\

"u 5: G p i H la h i n h chieu v u o n g goc cua S tren m p ( A B C ) .

ce

la toa d p can t i m .

2sin^x + cos2x

V l + xVx

Ta
s/
up
7

.c

y = i-9t
^ 2

t a n x + cot2x>0

V l + xVx


ro

Tpa d p M la n g h i ^ m cua h$ x = 3 + 1 6 t = > M

C a u 2: D i e u k i ^ n :

o x = 0, x = - - , x = 5
3

J. j ^ ( ^ ^ ; ^ l x . f ^ i ± ^ x . j - ^ x .

9t

x-y =0

Vay M

C a u 4:

/g

V=
^ 2
K h i d o M la giao d i e m cua A'B va d .

/7

^ 3 x - l + ^5x + l = 2 ^

x = 3 + 16t

P h u o n g t r i n h t h a m so cua A'B la :

2

P h u o n g t r i n h t h u hai, t r o thanh:

om

- 2

"^^3 = 0
2

(ke^)

= x^ + y"' + x^ + y^ - xy + X + y > 0

.A'B = 7(l6;-9).
6

b =2

k7t

X + y - 1 = 0 v i (x + y ) + (x + y ) + x + y - 3xy (x + y ) - 3xy

G Q I A ' ( a ; b ) la d i e m d o i x u n g ciia A qua d nen c6:
.1 = 0

571

; x= — +
12

Bie'n d o i p h u o n g t r i n h t h u nhat: (x + y)"* - (x + y ) - 3xy (x + y ) ^ - 1 = 0

A, B n a m ve c i m g phia d o i v o i d u o n g phan giac d : x - y - 0.

a +2

k7t

iL
ie
uO
nT
hi

Da
iH
oc
01
/

b) Tpa d p A , B la n g h i ^ m ciia

b-

x=—+

C a u 3: D i e u ki?n: x + y > 0

C a u 1:
a) D a n h cho ban dpc.

{a-2).l.

k7r;

=> d[A,(SBC)] = d[H,(SBC)] = aV2
sin2x

Di^g HKISC
BC 1 H C
BCISH

t?i K ( l )

>BC1(SHC)
209


C a u 9.a:

=^ BC ± H K (2) (1) va (2) suy ra H K ± (SBC).

.-

Tir d p d[H,(SBC)] = H K = a%/2

2z + i.z=

=>KC = V H C ^ - H K ^ = yjsa^-la^
tanSCH =

HK

SH

KC

HC

=
a

Dodo:

The tich k h o i chop S.ABC d u p e t i n h b o i :
= i A B . B C . S H = -aS-aS-a^e
6

6

=

2

(dvtt)

v4a-3b 8a2+6ab-9b^
N e u a ^ O , b ^ 0 ^ P = 2a + 3b = ( 2 a + 3 b ) —
r =
r
z

a'^ + b'^

xet h a m so i{t) = ^ ^ ^ ^ ,

voi

1-i

4(^^i)
=_i
i = 2 + 2i
2

2z + i.z = 2 + 2 i o 2 a + 2bi + a i - b i ^ =2 + 2i<=>2a + b + (a + 2 b ) i = 2 + 2i

2 + 2i

s/
up

ro
/g

A. Theo chUorng trinh chuan

om

C a u 7.a: Toa d p d i n h A ( 5 ; 4). G o i E la giao d i e m cua d u o n g t r o n ngoai tiep cua

.c

tarn giac A M C v a i B A t h i ta c6 BA.BE = BM.BC = 75 ( v i M nSm tren tia BC), t i m

=78

Vi G € A : 3 x - y - 8 = 0=>3X(,-yo-8 = 0

(2)

T u ( l ) v a (2) suy ra G ( 1 ; - 5 ) = > C ( - 2 ; - 1 0 ) hoac G ( 2 ; - 2 ) = : > C ( l ; - l )
C a u S . b : ( d ) d i qua M ( 2 ; 3 ; - 3 ) , c6 vecto chi p h u o n g u = ( 4 ; 2 ; l )
Xet d u o n g thang ( d ' ) qua M , (d') n a m t r o n g (P) va ( d ' ) 1 ( d )

u;nr

ce
fa

l ( - l + 3 t ; 2 - 2 t ; 2 + 2 t ) e d , l A + IB = 7l7t2 +13 + 7l7t2 - 6 8 t + 81

Xet d u o n g thSng ( d ' ) qua M , ( d ' ) n a m trong (P) va ( d ' ) 1 ( d )

Xet h a m so f ( t ) = 7 l 7 t ^ T l 3 + 7 l 7 t 2 - 6 8 t + 81

= (3;-9; 6)

a = ± l = > A ( 3 ; 0 ; - 1 ) hoac A ( 1 ; 6 ; - 5 ) .

= ( 3 ; - 9 ; 6)

Suy ra: a = ±1 ^ A ( 3 ; 0 ; - 1 ) hoac A ( 1 ; 6 ; - 5 ) .

C a u S . a : ( d ) d i qua M ( 2 ; 3 ; - 3 ) , c6 v e e t o c h i p h u o n g u = ( 4 ; 2 ; l )

A 6 ( d ' ) = > A ( 2 + a ; 3 - 3 a ; - 3 + 2a), ket h p p gia thiet A M 2 = 1 4

(1)

Gpi A e ( d ' ) => A ( 2 + a ; 3 - 3a;-3 + 2a), ket h p p gia thiet A M ^ = 14

/. /;:\ = ^ C ( 8 ; 0 ) h o a c C ( 2 ; 8 ) .
=(575)'

ww

4x + 3 y - 3 2 = 0
x2 ,
(x-i3f+(y-10f

Gia s u G ( x o ; y o ) , ta c6:

^o-yo-5|=i

v o i d u o n g thang A C . Toa d p cua C la

w.

d u o n g t r o n tarn E , ban k i n h r = 575

•AB:x-y-5 = 0

fAABG _
d ( G ; A B ) = ^ ° " ^ " " ^ _=f ^
^
^
72
AB

bo

ok

d u p e toa d p cua E la E(13; l O ) . Tarn giac A E C v u o n g tai A nen C la giao cua

(5 _ 5
2' 2

V i G la t r p n g tam A A B C = > S ^ A B G = ^ S ^ A B C

Ta

thi P = ^ i i l ^ ,

a^+b^

hoac B)

210

4

(a,b€^).

C a u 7.b: G p i M la t r u n g d i e m A B => M

I I . P H A N R I E N G T h i s i n h c h i d u p e chpn lam mpt t r o n g h a i p h a n ( p h a n A

Gpi

- 2 + 2i

=

B. Theo chUorng trinh nang cao

,

u;np

l + 3i + 3 i 2 + i 3

D o d o A = z + 2i.z = - + - i + 2 i . - - 2 i . - i
3 3
3
3

C a u 6: N e u a = 0 , b = 0 = > P = 0

n g h i ? m cua h f :

J

l-373i + 3(i73f-(i73f
_8
—
'— = -

2a + b = 2
2
2 2 .
^ <=> a = b = - => z = - + - 1
a+2b=2
3

3 3

G o c g i u a SB v 6 i m p ( A B C ) la goc SBH = 45° (do ASBH v u o n g can)

Dat a = t b , ( t ^ O )

1+ i

=

iL
ie
uO
nT
hi
Da
iH
oc
01
/

V =- S - S H
3

^

Dat z = a + b i

g^^HK.HC_ar/2.aV3_^y^
KC

fl-iS]'

T a c o : f ' ( t ) = 17

t

.7l7t^Tl3

suy ra
o

^

t-2

va f ' ( t ) = 0

7 l 7 t 2 - 6 8 t + 81

t V l 7 t 2 - 6 8 t + 8 l ] = (2 - t ) f 7 l 7 t 2 T l 3 '

052t-52 = 0

o t = l
211


Cty TNHHMTVDWHKhang

Cau 9.b: Chpn ngau nhien mot hpp roi tit hpp do lay ngau nhien ra 3 bi. Tinh
xac sua't dupe ca 3 bi den.
Gpi A la bien co lay dupe ca 3 bi den.

Cau 6: Cho cac so x,y,z thoa man x,y,z e [ l ; 4 ] va x > y > 2. j i m gia trj gia trj
y
z
. + —1— + .
nho nha't cua bieu thuc P = —
2x + 3y y + z x + z

A | , A 2 , A3 Ian lupt la cac bien co chpn dupe hop I , I I , I I . Khi do A ^ A j , A3

II, PHAN R I E N G T h i sinh chi dugc chpn lam mpt trong hai phan (phan A
hole B )

la mpt h$ day du xung khac tung doi va P ( A J ) = P ( A 2 ) = P ( A 3 ) = ^

A, Theo chUcrng trinh chuan

Theo cong thiic xac sua't day du, ta co:

Cau 7.a: Trong mat phSng tpa dp Oxy, cho tarn giac A B C vuong tgi A , biet B

iL
ie
uO
nT
hi

Da
iH
oc
01
/

P(A) = P ( A I ) P ( A / A I ) + P(A2)P(A/A2) +

P(A3)P(A/A3)

va C doi xung nhau qua goc tpa dp. Duong phan giac trong ciia goc A B C co

Theo cong thuc xac sua't lya chpn, ta co:
P(A/A,) = ^ . ^ ; P ( A / A , ) . ^ =i ; P ( A / A 3 ) = 0
5

Viet

phuong trinh la: x + 2 y - 5 = 0. Tim tpa dp cac dinh cua tarn giac biet duong
thiing A C di qua diem K (6; 2) .

5

Cau 8.a: Trong mat phSng tpa dp Oxyz, cho diem A(3;0;1), mat phSng ( p ) :

Suyra P ( A ) = 0,1667

x + z - 4 = 0 va duong thSng (d) : • y = 3 . Viet phuong trinh duong thSng A nkm
z=6+ t

Ta

DETHITHlifSdSS

s/

I. P H A N C H U N G C H O T A T CA C A C THI S I N H

ro

up

Cau 1: Cho ham so y = x^ - 3x^ - mx + 2 co do thj (C^)

/g

a) Khao sat su bien thien va ve do thj (CQ) cua ham so.

om

b) Xac djnh m de do thj cua ham so (C^) co cue trj, dong thoi duong thang

.c

di qua hai diem c^c trj cua do thj ham so tao voi hai tryc tpa dp mpt tam giac

1

^^'"^""^eMx

bo

ww

w.

fa

ce

„ , ,
> , sin^x + cos'*x 1/.
^\
Cau 2: Giai phuong tnnh:
—;
= - {tanx + cotx)
sm2x
2^
\fx^
1
Cau 3: Giai phuong trinh: ,
— = , . —7=====
^
V 2 X - 1 - 1 Vx +
3-7^

ok

can.

Cau 4:Tinh tich phan: K = f '

X=:t

^

trong mat ph^ng (P), di qua diem A va khoang each giua A va (d) bang

Cau 9.a: Cho so phuc z co phan ao bang 164 va voi n e R ' thoa
B. Theo chUorng trinh nang cao

—

• = 4i.
z+n

Cau 7.b: Trong mat phMng tpa dp Oxy, cho tam giac ABC can tai C co phuong
trinh canh AB la: x - 2 y = 0, diem l(4;2) la trung diem ciia AB diem M(

4-

thupc canh BC, di^n tich tam giac ABC b^ng 10. Tim tpa dp cac dinh cua tam
giac ABC biet tung dp diem B Ion hon hoac bang 3.
Cau 8.b: Trong khong gian voi h? true tpa dp vuong goc Qxyz, cho hai diem
A ( l ; - 5 ; 2), B(3;-l;-2) va duong thSng (d) co phuong trinh-

^=Zl?.^
4

1

2 •

Tim diem M tren (d) sao cho tich MA.MB nho nhat.

Cau 5: Cho hinh chop S . A B C D co day A B C D la hinh thoi canh b^ng a, SA = SB

9.b: Co hai hop I va I I moi hop ehua 12 bi, trong do hop I gom 8 bi do, 4

= a, SD = ayfl va m|t phSng ( S B D ) vuong goc voi mat p h i n g ( A B C D ) . Tinh

^' frang; hop I I gom 5 bi do, 7 bi trang. Lay ngau nhien tir hop I ba bi roi bo

theo a the tich khol chop S . A B C D va khoang each giua hai duong thJing

^^"g hpp I I ; sau do lay ngau nhien tu hpp I I bon bi. Tinh xac sua't de lay dupe

va S D .

212

^ bi do va mpt bi trSng tu hop I I .


Tuyen chgn & Giai thifti dethi Todn hQC - Nguyen Phu Khdnh , Ngtiyen Tat Thu.

Hl/dNGDANGIAI

"^-x^ + x l n x + l
Cau 4: K = Je^dx = jxe^'dx + Jin xe'^dx + J — d x

I. PHAN CHUNG CHO TAT CA CAC THI SINH

1

Cau 1:
a) Danh cho ban doc.

Kj = |xe''dx = xe"

Vay, K = e^+^ -

Voi m > -3 thi do thj cua ham so c6 cue tri va
r 2m - 2 X . ^2 - rn
y = -^(x-i)-y'+

6- m

ro
/g

6-m o m

Vdi m = 6 thi A = B = O do do so voi dieu ki^n ta nhan m = - -

ce

fa

Vx^-9 - V2x-1 = x - 4

(l)

x+2
.
/
.\
-1 = 0
= 3 = x-4«(x-4)
7x2-9 + V2x-l
.
Vx^-9 + V 2 x - l
x^ - 2 x - 8

o x = 4 hoac N/X^ - 9 + N / 2 X - 1 =x + 2 ( 2 ) . T u ( l ) va (2) suy ra x = 5

214

J

V

;x

g

AO = VAB2-OB2 =

/-2

a- -•

2

4~

a

2

Suy ra the tich khoi chop S.ABD dugc
tinh boi:
'^S.ABD = ^ A . S B D = ^ ^ S B D - ^ O

= isB.SD.AO = l a . a 7 i . i =
6
6
2
• ^ S . A B C D - 2Vs.ABD

3^72

12

^

(dvtt).

( l ) va (2) chung to O H la doan vuong goc chung ciia AC va S D

Doi chieu dieu ki^n, suy ra phuong trinh v6 nghifm.

, .
1

V

Theo Chung minh tren AO 1 ( S B D ) => A O 1 O H (2)

<::> 1 - i s i n ^ 2x = 1 <=> sin 2x = 0
sin2x
2

Cau 3: N / X ^ - 9 - ( x - 3 ) = V 2 x - 1 - 1

J

Trong A S B D dung O H 1 S D tai H ( l ) ^ H la trung diem ciia S D .

ww

l--sin^2x
2
o
sin2x

smx + cosx
cosx sinx J

w.

1 - Uin^2x
_2
Phuong trinh cho tro thanh:
sin2x

bo

ok

Cau 2: Dieu ki?n: sin 2x^0

om

2(m + 3)

9
3
= 6, m = - — , m = - —

.c

<»OA = OB <=>

up

Tarn giac OAB can
m -6

- 'f—dx

-dx-i-+ ll—dx = e'^^^

Matkhac AS = AB = AD=>OS = OB = OD
hay ASBD la tam giac vuong tai S
BD = VSB^+SD^ = Va2+2a2 = aVs

3

Ta

•;0 , B 0;
.2(m + 3)

dx

1{

Do do neu d^ng A O 1 (SBD) thi O e BD.

la duong thMng d qua 2 diem cue tri

s/

6- m

+

"re"

Cau 5: Theo gia thie't ( A B C D ) 1 ( S B D ) theo giao tuyen B D .

Gia su duong thSng d cat 2 true Ox va Oy Ian lugt tai:
A

1

iL
ie
uO
nT
hi
Da
iH
oc
01
/

nghiem, tuc la phai c6: o A' = 9 + 3m > 0 hay m > - 3 .

X +2

- f — d x = e^ '•

1

Ham so c6 cue trj khi y' = 0 c6 2 nghiem phan biet va doi dau qua moi

-2

•e'

K , = fe" Inxdx = eMnx

Taco: y' = 3x^ - 6 x - m

2m

1

1

b) Ham so da cho xac djnh tren M

Suy ra y =

1

Vay d ( A C , B D ) = O H = ^ S B = |
Cau 6: Dat P(z) =

Taco: P'(z) =

2x + 3y

y +z

x + z' z e

X

(x-y)(z2-xy)
(y + z) (x + z)
215


Tuyen chQn & Ci&i thifu dcthi Todn hoc - Nguyen Phu Khdnh , Nguyen lat Ihu.

B. Theo chUcrng trinh nang cao

Neu X = y thi P = -

Cau 7.b: Goi toadp d i e m B ( 2 y B ; y g ) = i > A ( 8 - 2 y B ; 4 - y B )

•J

P(z)>P(y)^

< y^ < x y va P ' ( z ) < 0 =>P(z) la ham nghich bien:

2x + 3y

y+y

x+y

X

y

1

2x + 3y

x+y

2

Phuang trinh duong thang CI: 2x + y - 1 0 = 0
G Q i t o a d p d i e m C ( x c ; 1 0 - 2 x c ) = > C I =S\4-x^\,

X

y

Bai toan tro thanh tim gia tri nho nhat ciia bieu thuc: Q = ^ ^ ^ ^ +
Dat t = - = ^ t € [ l ; 4 ] , k h i d6 Q = Q ( t ) = ^
Q'{t)

-t^ - 6 t - 6

SABC = ^ C I . A B = 10 « | 4 y B + 2 x c - x^y^ -8\ 2

+ -

o x ^ y B - 4yB - 2 x c = - 6 ( l ) hoac x ^ y B - 4yB - 2 x c = - 1 0 ( 2 )
4-Xc =k(2yB-4)

^ + - ^ + ^ voi t 6 [ l ; 4

; 0 => Q ( t ) la ham ngich bien, suy ra Q ( t ) > Q(4)

Vi M

^(2t.3f(t.lf
117

Vay, m i n Q = —

II. PHAN RIENG T h i sinh chi du
Ta
s/
ro

up

C(2b + 5 ; - b ) .

Diem O G B C . Lay doi xiing O qua phan giac ciia goc B ta duoc diem

om

/g

M ( 2 ; 4 ) 6 A B =^ B M = (7 + 2b;4 - b) C K = ( l - 2 b ; 2 + b)

.c

Vi tam giac A A B C vuong tai A nen c6: B M . C K = 0 :=> b = -3 hoac b = 1 .

bo

ok

Cau 8.a: A CO vecto chi phuong u = (a;b;c) ?i 0

w.

t u day ta tim dxxqic b = - a

hoac b =

9^ vi y B > 3

- — + 2xr~ = k

2)

^|yB=-l-72

2xcyB-6yB-5xc+i6 = o

[ x c = - l + V2

' < c y B - 4 y B - 2 x c =-10
2X(-yB - 6yB - 5XQ + 1 6 = 0

xc=2

Vay, tpa dp cac dinh cua tam giac A B C la: A ( 2 ; 1 ) , B ( 6 ; 3 ) , C(2;6)

Cau 8.b: Ta c6 trung diem ciia A B la 1(2;-3; 0)
M A . M B = ( M I + L A ) ( M I + I B ) = ( M I + lAJfMI - L A ) = MI^ - lA^ = MI^ - 9
Suy ra MA.MB nho nhat khi va chi khi MI nho nhat hay M la hinh chieu
vuong goc cua I tren ( d ) .
M 6 d = > M ( - 3 + 4t; 2 + t; - 3 + 2 t ) = > I M = (-5 + 4t; 5 + t; - 3 + 2t)

I M 1 u <=> I M . u = 0 <=> 4 ( - 5 + 4t) + 5 +1 + 2( - 3 + 2t) = 0 o t = I

M)

= * M ( 1 ; 3 ; - l ) , M I = N/38

Theo gia thiet, ta c6
2 = 4 i < ^ _ L t I ^ ^ = 4 i o a + 164i = 4i(a + 164i + n)
z+n
a + 164i + n
a = -656
^
o a + 164ii = - 6 5 6 + 4{a + n ) i <=>-^ .
^
'
[4{a + n) = 164

Tir (2) va (3) ta c6 he:

'
(d) CO vecto chi phuong u = (4; 1; 2)

ww

Hon nua khoang each giua A va (d) bang

fa

ce

Duong thang A nam trong mat phSng (P) c=> u.np = 0 => c = -a

216

B C =i> C M = kMB o

Tu ( l ) va (3) ta c6 h$:

hoac B )
A. Theo chUofng trinh chuan

Cau 9.a: Goi z = a + 164i (a e

G

• 2 x c y B - 6 y B - 5 x c + i 6 = o (3)

khi a = 4,b = c = 1 .

Cau 7.a: Goi toa do diem B(-2b - 5;b)

AB = V 2 0 | y B - 2

Dien tich tam giac A B C la:

1

iL
ie
uO
nT
hi
Da
iH
oc
01
/

Neu x > y t h i

a = -656
n = 697

Vay, fM A . M B )

' mill

= 29 d a t d u p e khi M ( l ; 3; - 1 )

Cau 9.b: A ; (i = 0 , 1 , 2 , 3 ) la bien co c6 i bi do va ( 3 - i ) bi trSng c6 trong 3 bi
chpn ra t u hpp I. Khi do A ( , , A p A 2 A3

rn?*

day dii, xung khic tung doi

Va ta c6:
217


Tuyen chgn & Gi&i thi?u dethi Todn hgc - Nguyen Phu Khdnh , Nguyen Tat Thu.
4

_

-3

Cty TNHH MTV DWH

| u 6: C h o a, b, c > 0 thoa m a n a + b + c = 3 .

48
220

-12

Clc\2

T i m gia t r j n h o nhat cvia bieu thuc : P = a^ + b^ + c^ + •

-12

at. + be + ca

a^ + b^ + c ^ + a + b + c

56

-12

Khang Vi?t

I I . PHAN R I E N G T h i s i n h c h i duQc c h p n l a m m p t t r o n g h a i p h a n ( p h a n A

220

ho^cB)

Theo cong t h i i c xac suat day d u , ta c6:

f^, Theo chUorng trinh chuan

P(A) = P(Ao)P(A/Ao) + P(Aio)P(A/Ai)

Cau 7.a: T r o n g m a t p h i i n g tpa d p O x y , cho t a m giac A B C v u o n g can tai A . Biet

iL
ie
uO
nT
hi
Da
iH

oc
01
/

+ P(A2)P(A/A2) + P(A3)P(A/A3)

p h u o n g t r i n h canh BC la ( d ) : x + 7 y - 31 = 0 , d i e m N (7; 7 ) thupc d u o n g thSng
Theo cong thuc ti'nh xac suat l u a chpn, ta c6

A C d i e m M ( 2 ; - 3 ) thupc A B va nSm ngoai doan A B . T i m tpa d p cac d i n h cua
^

Cjs

tam giac A B C

1365

Cau 8.a: T r o n g k h o n g gian v o i h? true tpa d p O x y z , cho cae d i e m A ( - 1 ; 1 ; 0 ) ,

Suy ra xac sua't can t i m la P ( A ) = 0,2076

B ( 0 ; 0 ; - 2 ) va C ( l ; l ; l ) . Viet p h u o n g t r i n h m a t phMng ( P ) q u a hai d i e m A va B,
biet khoang each t u C t o i m a t phang

DETHITH(jfs634

Cau9.a: T i m so p h u c z biet (z +1)^2 - V3i) + ( z + 1 ) ( 2 + \ / 3 i ) = 14 va |z| = 2 .

Ta

B. Theo chUtfng trinh nang cao

s/

I. PHAN C H U N G C H O T A T CA C A C T H I S I N H

Cau 7.b: T r o n g m a t ph3ng O x y , hay viet p h u o n g t r i n h h y p e b o l ( H ) d a n g chinh

2 m x + 5 (,(3 j 5 ti^j I J ,

up

Cau 1: Cho h a m so y =

ro

x+m

/g

a) K h a o sat s y bien t h i e n va ve d o thj ( C , ) ciia h a m so.

om

b) V o i m la t h a m so thuc khac 0 va d u o n g thSng d c6 p h u o n g trinh

ok

.c

y = 2 x — . T i m m de d ck ( C ^ ) tai hai d i e m phan b i ^ t A , B c6 hoanh d g X j , Xo

ce

bo

thoa m a n x^ - 9 x i - S x j .

\/2(cosx-sinx)

tanx + cot2x

cotx - 1

fa

1

w.

Cau 2: Giai p h u o n g t r i n h :

ww

Cau 3: Giai p h u o n g t r i n h : V 3 x ^ + 33 + 3N/X = 2x + 7
''x-=^+2x2
KIX,
Cau 4: T i n h tich
ich phan: J = j ,

0 V1 + :

Cau 5: C h o lang t r u tarn giac A B C . A ' B ' C c6 day A B C la t a m giac v u o n g tai A
A B = a, A C -

(P) bang yjs .

, h i n h chieu v u o n g goc cua A ' tren m a t phSng ( A B C ) trun;.

biet rMng ( H ) tiep xiic v o i d u o n g thiing d : x - y - 2 = 0 tai d i e m A c6 hoanh
dp bang 4.
Cau 8.b: T r o n g k h o n g gian v o i h f true tpa d p v u o n g goc O x y z , vie't p h u o n g
trinh d u o n g thSng A d i qua d i e m M ( l ; - 1 ; 0 ) , cat d u o n g thMng

( d ) : ^ = ^ = ^

va tao v o i m a t p h ^ n g (P): 2 x - y - z + 5 = 0 m o t goc 30° .
Cau 9.b: C h o tap h a p A = { l ; 2 ; 3 ; . . . ; 1 8 } . Co bao nhieu each chpn ra 5 so t r o n g
t^p A sao cho h i ^ u ciia hai so bat k i t r o n g 5 so d o k h o n g n h o h o n 2 .

HUdfNGDANGlAl
I. PHAN C H U N G C H O T A T C A C A C T H I S I N H
Cau 1:
a) D a n h cho ban dpc.
b) H a m so da cho xac d j n h va lien tuc tren k h o a n g ( - o o ; - m ) u ( - m ; + o o )
H o a n h d p giao d i e m cua d u o n g t h i n g d va ( C ^ ) la n g h i # m ciia p h u o n g

v o i t r p n g t a m G cua t a m giac A B C va goc gii>a A A ' tao v o i m a t p h a n g ( A B C )
bang 60" . T i n h the tich k h o l lang t r y A B C . A ' B ' C va k h o a n g each t u B ' dci
mat phang ( A ' B C ) .

218

trinh ^ n 2 i ± 5 ^ 2 x - i o 4 x 2 - x - m - 1 0 = 0 ( V x ^ - m )
x+m
2
^
^
Dat g ( x ) = 4x^ - x - m - 1 0
219


Cty TNHHMTVDVVH

Tuye'n chpn & Giai thifu dethi Todn hpc - Nx'nicn I'hu Klu'inh , Nguyen Tai Thu.

De d cat (Cm) tai hai diem phan bi^t A, B khi va chi khi phuong trinh
g(x) = 0 CO hai nghi^m phan biet khac - m tiic phai c6:

2m^ - 5 m

<=>

<

Ap dung Viet cho X j ,

<=>X>A^-11X

Vio
b

Xj+X2=

— :

+ 26N/^-16 = 0

»(N/X-8)(VX-2)(NA(-I) =

^ x ^ +2x^
,
dx . Dat u = x* + 2x, dv =
0^ i.x^+1

1

0

.dx

Cau 4: J = J

a
4
_ c _ m +10

ta c6

33+9Vx = 6x + 21 (2)

Tu (1) va (2) suy ra (x + 17)(>/x +1) - (6x +12) = 6x + 21 -9>/x

161
m > - - 16
m

sVsx^ +

iL
ie
uO
nT
hi
Da
iH
oc
01
/

16m + 161>0

A>0
g(-m)^0'

Taco:

KKaffgVief

at t = Vx^ +1 => t^ = x^ +1 => tdt = xdx .
Dat

Cau 5: Gpi M la trung diem BC. Tu gia thiet ta c6:

Xet dieu kien bai toan xj - 9x, = 8x2 <=> Xj - 9xj = 8

BC = 2a, AG = | A I = y ; A ^ = 60" => A'G = AG.tan60'^ =

- - X i

The tich V cua khoi lang try dugc tinh boi:

Xj - Xj - 2 = 0 o X = - 1 hoac X = 2.

V = S A B C A ' G = ^ AB. AC. A ' G
X j = -1 => X2

Voi

= ^ => m = -5

1

=2

X2

= - - => m = 4

up

s/

Xj

Ta

Voi

ok

.c

^ 1

\^(cosx-sinx)

bo

Phuong trinh cho tro thanh: —
sinx

cos2x

ce

cosx - 1
sinx

+

sin2x

w.

ww

x + 17

^Vx+1

220

v^'+l

V3x2+33 + 2x + 4

BC

^^"3

1 a.3iS ^ aS
3

2a

~

6

Dyng G H 1 A ' l t^i H
BCIGI

(l)

B C I G H (2).

Tu (1) va (2)

GH 1 (A'BC)

Mat khac nhan thay AB' cat mp (A'BC) tai N la trung diem cua AB'.

D61 chieu dieu kien, ta dugc ho nghiem la x = - j + k27t (k e Z)

^"-^^

3"

3

MA

BCIA'G

<=>x = - + k27i hoac x = - - +k27i,(kGZ)
4
•
4
^

Cau 3: Vsx^+33-(2x+4) + 3V^-3 = 0 c^(x-l)

tai I =0 G I / / A K

AB.AC _

=i

Do:

= v2smx o2smx.cosx = >y2smx o c o s x = —

3

AK

fa

cosx

cosx.sm
cosx

om

cosx.sin2x.sinx.(tanx + cot2x) ^ 0

1

va G I I B C

/g

Vay m = -5 hoac m = 4 thoa man yeu cau bai toan.
cot X

Dung A K 1 B C tai K

ro

Ket hop voi dieu kien m > ~ ~ " va m ;i

Cau 2: Dieu kien: •

3

/r layfi

= - a . a V 3 . — ^ = a'^ (dvtt)

7

V3x2+33 + 2x + 4J

= 0 «3V3x2+33=(x + 17)(>/;^ + l)-{6x+12) (l)

TCrdo: d[B', (A'BC)] = d[A, (A'BC)] = 3d[G, (A'BC)] = 3GH
= 3.

A'G.GI
''^

3.A'G.GI
VA'G^+GI^

3.

2a>y3 aS
6 _ 6a

|l2a2

3a2

VsT

lay/El
17

36
221


m i n i

Cau 6: P =

Matkhac: (z + ^)(2-^/3i) + (z + l)(2 + ^y3i) = 14 » 2(z + z) +V3(z - z j i = 10

2(a2+b2+c2)'+5(a2+b2+c2) + 9
2(a2 + b2+c2+3)

<=>4x + 2 N / 3 y = 10<=>y =
fa + b + c)^
> ^ — - — — = 3, v i the neu

thi t e [ 3 ; 9 ) . Xet ham so: f (t) =

2t^ + 5t + 9
2t + 6

The (2) vao ( l ) , tadupc: x^

,.
vai t € [ 3 ; 9 ) .

4 ( t ^ + 6 t + 3l
^ ,
_ ^ > 0 , V t e ( 3 ; 6 ) = > f ( t ) dong bien voi V t € [ 3 ; 9 ) va
(2t + 6 f

5-2x

. 73 ,

= 4 o 7 x 2 - 2 0 x + 13 = 0

3N/3

' Vay, CO hai so phuc can tim la z = 1 + Vsi, z = — + —
7

7
B. Theo chUorng trinh nang cao
Cau 7.b: Do A thupc d : A(4;2)

i

Giasu ( H ) : ^ - ^ = 1 ( * ) ^ A e ( H ) 2 f - ± = 1 ( l )
a
D
a
b

Ta

I I . PHAN RIENG Thi sinh chi dugfc chgn lam mpt trong hai phan (phan A

s/

ho^c B)

up

A. Theo chUorng trinh chuan

(a^ + b^ ^ O ] . (AB;BC) = 45° nen cos45° = ^ ^ [ ^
^
'
V50Va^ +

.c

om

4a = -3b"

/g

33 = 4b

ro

Cau 7.a: Duong thang AB d i qua M nen c6 phuong trinh a ( x - 2 ) + b(y + 3) = 0

bo

(AB) :4x + 3y + l = 0 . ( A C ) : 3 x - 4 y + 7 = 0.

fa

ce

Tir do A ( - 1 ; 1 ) va B(-4;5). Kiem tra MB = 2MA nen M nam ngoai dogn A B

w.

Neu 4a = -3b, chpn a = 3, b = -4 dupe ( A B ) : 3X - 4y - 1 8 = 0,

ww

Ja2+(a-2cf+c2

V

bV-a2(x-2)^=aV
y = x-2

•(b2-a2)x2+4a2x-4a2-aV=0
=>A',=4aU(b2

-a2)(4a2 + a^b^) = 4a^h^ + a^b^ -a^b^

o a V ( 4 + b2-a2) = 0

16b2-4a2=aV

(P) diqua A ( - 1 ; 1 ; 0 ) , B(0;0;-2) nenco: b = a - 2 c
= 73o2a2-16ac+14c^=0 o

=aV

^ Ket hpp voi ( l ) :

Cau 8.a: Gpi n = (a;b;c) it 0 la vecto phap tuyen cua (P)

2a+ c

b V - a V
y = x-2

=>a2=b2+4

(AC) :4x + 3 y - 4 9 = 0 =>A(10; 3), B(10;3) ( k h o n g t h o a )

= S

Mat khac do d tiep xiic voi ( H ) thi h f sau c6 nghi^m bang nhau:

y = x-2

ok

Neu 3a = 4b, chpn a = 4, b = 3 dupe

222

(2)

O X = 1 => y = ^/3 ho|ic x = y =^ y = ^

P > f ( 3 ) = ^ . V a y , minP = |

d(C,{P))

5-2x

iL
ie
uO
nT
hi

Da
iH
oc
01
/

Taco: i'(t)^-^

n y t

Ta c6 |z| = >/x^Ty^ = 2 <=> x^ + y^ = 4 ( l )

a^+b^+c^+S

t = a^ + b^ +

y n n.nunji

^au 9.a: Gpi z = x + y i (x,ye]R)=>z = x - y i

( a 2 + b 2 + c 2 ) ^ + 3 ( a 2 + b ^ + c 2 ) + ab + bc + ca

Ta thay: 9 = (a + b + c)^ > a^ + b^ +

irii V u y

fb'*-Sb^ +16 = 0

b2=4
a'=8

a =c
a = 7c

^

'

8

4

-au 8.b: Gpi N = (d) n A => N(2 + 2t; t ; - 2 +1)
; Ta c6: M N = (1 + 2t; t +1; - 2 +1) va mp (p) c6 VTPT fi = (2; - 1 ; - 1 )

223


Tuyen

chgn & Giai

thteu de thi loan

hoc - Nguyen

l^hu Khfl>t7r7TVgt
Cty TNHH

2 +4 t - t - l + 2 - t

( d ) , ( P ) = 30'' nen: sin30° = cos^MN,n

CaU 5=

7(l + 2t)2+(t + l)2+(t-2)2.7g"2
2t + 3

= j -

iL
ie
uO
nT
hi
Da
iH
oc
01
/

C h u n g m i n h rang:

=" [ J "= "j"

C a u 9.b: Ta can t i m so p h a n t u ciia tap T sau:
l
X e t t a p h(?p H = { ( b i , b 2 , . . v b 5 ) : b i

<...
l < b i <14}

h^=a^-3,

xac

a2+c2

+ 1 > 3 ( a + b + c)^

giac C O 3 canh n a m tren (d j ) , ( d 2 ) true O y
Cau 8.a: T r o n g k h o n g gian v o i h ^ true tpa d p v u o n g goc O x y z , cho mat p h ^ n g

Ta

(P):x + y + z - 3 = 0 va d u o n g thMng A : ^ ^ = ^ = — . L a p p h u o n g t r i n h d u o n g
1
3
1
t h i n g d , n a m t r o n g m a t phSng ( P ) , v u o n g goc v o i d u o n g t h i n g A va each

s/
/g

ro

up

t u . D o d o H i = C^4 = 2002 . Vay |T| = 2002 .

.c

om

OETHrTHilfSOSS

ok

I . PHAN C H U N G C H O T A T C A C A C T H I S I N H

z - 3i - 1 = 1
Z

fa

w.

ww

x2+y2+iL = 9
xy"' + 4y^ = x^ + xy

(an z la so'phuc)

*

Cau 7.b: T r o n g m a t p h i n g Oxy, cho hypebol (H): — - ^ = 1 v a d i e m M ( 2 ; l ) .

2
3
p h u o n g t r i n h d u o n g t h i n g ( d ) d i qua M , bie't r i n g d u o n g t h i n g d o e i t

C a u 2: Giai p h u o n g t r i n h 2 sin^ x + Vs. sin 2x + 1 = 2 (cos x + yfs.sin x

C a u 3: Giai h ^ p h u o n g t r i n h :

+i-1 : m -z

B. Theo chaofng trinh nang cao

tao v a i hai d i e m eye dai, eye tieu cua do thi J

h a m so (C^^) m p t tarn giac c6 dien tich nho nhat.

8
d u o n g t h a n g A m p t khoang bang - j = .
V66
Cau 9.a: T i m t h a m so thue m de he p h u o n g t r i n h p h u c c6 n g h i ^ m d u y nhat

{C^)

ce

bo

y = 2x^-3(2m + l)x^+6m(m + l)x + l codothjla

a) Khao sat s u bien thien va ve do thj (C(,) ciia h a m so.

224

-+1

va ( d 2 ) •• 4x + 3y - 1 2 = 0 . T i m toa d p tam va ban k i n h d u o n g t r o n npi tiep tam

b^=a^-4

M a t khac m o i b p (bi,b2,...,b5) t r o n g H la m p t to h o p chap 5 ciia 14 phan

Inxdx

U^+b^

Cau 7.a: T r o n g m a t p h ^ n g toa d p Oxy, cho hai d u o n g thang (dj) : 4 x - 3 y - 1 2 = 0

(bj,b2,...,b3)

De thay k h i d o f la m p t song anh, suy ra |T| = |H .

C a u 4: T i n h tich p h a n : K = | 3 x - -

+1

hoac B)
A. Theo chifoTng trinh chuan

>2

Xet anh xa f cho t i r a n g u n g m o i bo (apa2,...,a5) v o i bo

b) Xac d j n h m de M ( 2 m 3 ; m )

U^+b^

I I , P H A N R I E N G T h i s i n h c h i d u g c chgn l a m m p t t r o n g h a i p h a n (phan A

T = (ai,a2,...,a5):ai
Caul:Chohamso

theo a.

Cau 6: C h o a,b,c la cac so'duong thoa m a n a^ + b^ + c'^ = 3 .

X — l y + l z

=a^-2,

Viet

phang ( P ) va ( Q ) v u o n g goc v o i nhau, c6 giao tuyen la

Xinh ban k i n h m a t cau ngoai tiep A B C D va d A , ( B C D )

x-1 _ y+ 1 _ z
V6i t = 0, phuong trinh ^ ' ^
^
2

d i n h n h t r s a u : b j =a^, b j = 3 2 - 1 ,

Khang

trong ( Q ) '^V d i e m D sao cho A C , B D ciing v u o n g goc v a i A va A C = B D = A B .

o lOt^ - 18t = 0 o t = 0 hoac t = |

9

DWIi

d u o n g thSng A . Tren A lay hai d i e m A va B v a i A B = a. T r o n g ( ? ) lay d i e m C ,

6 r +2t + 6

Vai t = - , p h u a n g trinh ^ •

MTV

tai h a i d i e m A , B ma M la t r u n g diem eiia A B .
tha

8.b: T r o n g k h o n g gian v a i he true tpa d p v u o n g goc O x y z , cho cac d u o n g
'^S C O p h u o n g t r i n h
A, • i i _ y _ z - 3
.
x - 2 _ y - l _ z
x+2 y +1 z - 1

1 •_
, A , .
- ——, A^3q ••
•
—
2
1
-3
-3
Vie
' ^ t p h u o n g t r i n h d u o n g t h i n g A d i qua d i e m A ( 4 ; - 3 ; 2 ) cat Aj,A2 va

VUo

'"^g g o c v o i d u o n g t h i n g A3 .
225


a^z^

Vx^ -2x + 4.1og2 y = X

a^ +

C a u 9,b: Giai h? phuong trinh J y 2 - 2 y + 4.1og2 x = y (x, y e i?)
X < 4,

+2^

4

= !

_

z2

Z

y<4

=9
=> z"* + 2z^ - 3z^ - 8z - 4 = 0

1
+ _

z

hay (z^ -4)(z + lf =0=> (x;y) = (2;l),(-2;-l),(2;-2),(-2;2)
HUOTNGDANGIAI

I. PHAN CHUNG CHO TAT CA CAC THI SINH
1:

a) Danh cho ban dQC.
b) Ham so da cho xac djnh tren R

B(m + l;2m3+ 3m^)

=^K = K i + K 2 = 3 .

Suy ra AB = yfl va phuong trinh duong thang AB:

Ta
V2

s/

bo

a
a =—
2

C a u 3 : D $ t z = - , ta dupe h§:
y
226

1

w.

fa

o 3sin^ x + 2>/3.sinxcosx + cos^ x = 2|cosx + Vs.sinx j
COSX

ww

(cosx + \/3.sinxj = 2(eosx + Vs.sinxj

+V3.sinx = 0

—
+ krt
6

X =

7t

x = - + k27t
3

X

3

=1

2
a
' =2

cosx + V3.sinx = 2

IS

a^ = 2ya
2a ^ =2aa + 2pa

ce

2 sin^ x + N/S.sin 2x + sin^ x + cos^ x = 2(cos x + Vs.sin x j

cos

3 ll3e^-l
'2
4

+ y 2 + z^ - 2ax - 2Py - 2yz = 0

B, C, D G S

ok

C a u 2: Phuong trinh cho viet lai:

2

1

^

a2=2pa

.c

khim = 0

(S):

dat dupe

/g

min d(M; AB) =

om

Ta c6: d(M, AB) - ^ E l t i => d(M; AB) > 4V2
V2

L2

'I

-l|xdx =

Phuong trinh mat cau

ro

toi AB nho nhat.

1
^ sm
. x =l
—eosx

+—

=>K, = — I n x
2

A(0;0;0), B(0;a;0), C(0;0;a), D(a;a;0)

up

Do do, tam giac MAB c6 di?n tich nho nhat khi va chi khi khoang each tir M

= -

dx

C a u 5: Chpn h? trye tpa dp Axyz sao Cho:

+ y - 2 m ' ' - 3m^ - m - 1 = 0

tan X

u = lnx=>du =

Tinh K j bang each doi bien t = in x

Tga dp eae diem eye dgi, eye tieu eiia do thi la:

\/3.sinx = -cosx

3 x — Inxdx =3jxlnxdx-2Jlnx.-dx

dv = xdx => V =

=> Vm G R, ham so luon c6 eye dai, eye tieu.

X

j

Kj = jx inxdx. Dat

Tac6:y^ = 6x^ -6(2m + l)x + 6m(m + l ) v a y ' = 0<=>x = m, x = m + l

A(m;2m^+3m^+l),

K=

iL
ie
uO
nT
hi
Da
iH
oc
01
/

Cau

Cau4:

(keZ)

n(BCD)=

cau

LI

6:

D|t

2

BC,BD =a2(0;l;l)z:>(BCD):y + z - a = 0=>drA,(BCD)] = - ^
X =

a 2 + b ^ y = b 2 + c 2 , z = a 2 + c 2 thi x,y,z> 0 va x + y + z = 6

Tudo 6 = x + y + z>3^xyz => ^ ^ 1
^xyz 2
227


t-ty ii\n.n mi v uvvn

Tuym chQtt & Gi&i thifu dethi Todn hgc - Nguyen Phu Khdnh , Nguyen Tat Thu.

i^.nang vtft

Goi ( Q ) la mat phang chiia d va song song voi A. Khi do ta chon
Taco - + 1 = - + - + 1 > 3 3 / A . T u a n g t v - + 1 > 3 3
X X X
Vx^
y
Suy ra

X

4+ d
ra C O d(A;d) = d(A;(P)) = d ( M ; ( p ) ) = - ^

>273
>27
2„2,2
^^y

z

= -2(4;l;7) suy ra ( Q ) C6 dang 4x + y + 7z + d = 0.

nQ =

Ke't hop voi gia thiet ta dupe:
+1

3(a + b + c)^ = sfa^ + b^ +

a2+c2

• + 1 >27

4+ d

8

N/66

N/66

iL
ie
uO
nT
hi
Da
iH
oc
01
/

+1

Hay

+ 2(ab + be + ca)

THI:

<3 a 2 + b 2 + c 2 + 2 ( a 2 + b 2 + c 2 y = 27

Neu d = 4

Chpn diem N

Nen

+ 1 >3(a + b + c)^ .

+1

+1
U^+b^

c^+b^

'

( Q ) : 4x + y + 7z + 4 = 0.

1 13

2

I I . PHAN RIENG T h i sinh chi dupe chpn lam mgt trong hai phan (phan A

TH2; Ne'u d =-12

Ta

hoac B)

4x + 3 y - 1 2 = 0

A(3;0)eOx

up

4x-3y-12 =0

ro

A:

s/

A. Theo chUtfng trinh chuan

Cau 7.a: Gpi A la giao ciia d j , d 2

/g

Vi BC thupc Oy cho nen gpi B la giao cua dj voi Oy: cho x = 0 suy ra y = - 4

.c

om

B ( 0 ; - 4 ) va C la giao cua d j voi Oy :C(0;4 ) . Chung to B, C doi xung nhau

ok

qua Ox, mat khac A nSm tren Ox v i vay tam giac ABC la tam giac can dinh A.

bo

Do do tam I duong tron npi tiep tam giac thupc Ox suy ra l(a;0)

fa

ce

. u
u-^
•' .
l A AC 5
l A + IO 5 + 4
OA
Theo tinh chat phan giac trong : — =
= - =>
:— =
<=> •
lO
lO
lO
AO 4

, =IBCOA

=

3

i.5.3 = 15
2

=2 i t ^ i l ^ ^ a i i l l i ! )

dlA

( Q ) :4x + y + 7 z - 1 2 = 0 .

Chpn diem N ( l ; 1; 1) e ( P )

(Q) = d

x -1 _ y-1
Suy ra phuong trinh: 'A- ^
^

z -1
-1

Cau9.a: Gpi z = x + yi ( x , y e R )
Theo gia thiet, ta c6
x - l +(y-3)i| = l

( x - l ) 2 + ( y - 3 f =1

x - l + ( y + l)i = | m - x - y i

(x-l)2+(y + l f

=(m-x)2+y2

(x-l)^(y-3)^=l

(•) la h? phuong trinh tpa dp giao diem ciia duong tron
^ r=

15

=

5

5

( C ) : ( x - l ) ^ + ( y - 3 ) ^ =1 va duong thSng ( A ) : 2(m - l ) x + 2y + 2 - m^ = 0

phuong trinh c6 nghi^m duy nhat

A coVTCP u ^ = ( l ; 3 ; - l ) , M ( l ; 0 ; 0 ) e A .
Uj I n p

-1

Duong tron (C) c6 tam l ( l ; 3 ) va ban kinh R = 1

Cau 8,a: Ta c6 (P) c6 VTPT np = ( l ; l ; l ) ,

de(P)

-1

2 ( m - l ) x + 2y + 2 - m ^ = o '

ww

9

9
4

w.

40A^43^4^j
9

€(P)n(Q) = d

1
13
x+V
Suy ra phuong trinh: d :
3_-^
3 _z + l

Dang thuc xay ra khi x = y = z = 1.

Do

4 + d = 8<=>d = 4 hoac d = -12

m^ - 2 m - 6
= (-4;2;2)

d(l,(A)) = R

( A ) tiep xuc voi (C)

= 10 (m-l)'-7=^4(m-l)'+4

4 { m - l ) ^ +4
229

228


Tuyttt chQtt & Giai thi$u dethiToiiii

N^-ui/i-ii Phi'i KluUih /Nguyen

hoc

lat lltu.

Dlt t = (m-l)^(t>0),tac6:

t - 7 | = V4t + 4 <=>(t-7)^ =4t + 4«>t^ -18t + 45 = 0<» t = 3, t = 15

-—, —

"l'"2j

* t = 3 o ( m - l ) ^ =3<=>m = l ± 7 3

X

/

V

Cau 9.b:

x = 2 + at
y =^ 1 + bt
d cat ( H ) tai 2 diem A, B thi A, B c6 tQa dp :
x^ y =^ 1

Datf(t) =

s/

Ta

= 1..3{2.atf-2(2.btf=6

ro

om

/g

^3a2-2b2 / O

A'-4(3a-bf-4(3a2-2b2)>0

.c

Dieu ki?n:

up

o ( 3 a 2 - 2 b 2 ) t 2 + 4 ( 3 a - b ) t + 4 = 0 (l)

fa

V2b2-3a3

w.

2b^-3a

ce

bo

ok

Khido A(2 + ati;l + bti),vatoad9cua B : B(2 + at2;l +btj), suy ra neu M
la trung diem cua AB thi: 4+a (tj + tj) = 4 o tj + tj = 0
-^t2=±

.

ww

4(b-3a)
Ap dyng vi et cho (l) tj + tj = —
^ = 0 <» b = 3a
3a - 2b

x-2

=>d:

=

a

x-2

y - 1
o

b

y - 1
=

a

.

=>d:

,

e

n

"~r~

iL
ie
uO
nT
hi
Da
iH
oc
01
/

fx = 2 +at
Cau 7.b: Giai su d c6 vec to chi phuong u=(a;b),qua M(2;l) =>d:|^_^^^^

oi^-il^

= 3(1,-2,1)

1 ~ -2

B. Theo chucmg trinh nang cao

,

5_

Duong thing A di qua diem A \ c6 VTCP ( l , - 2 , l ) c6 phuong trinh la:
x-3_y+3_z-2

* t = 15 <=> (m -1)^ = 15 o m = 1 ± N/TS
Vay, gia trj can tim la m = 1 + \/3 hay m = 1 ± ViS
-

Ui,U2

Ket hgp vai A vuong goc vai A3 nen ta c6:

phuong trinh tuong duong vai:

\/t^-2t + 4

log2(4-y) =
log2 ( 4 - x ) =

2x + 4

Vy^ - 2 y + 4

, t € ( - o o ; 4 ) ta c6:

4-t
>0,Vt<4.
-2t + 4 Vt^ - 2 t + 4
Suy ra f (t) la ham so dong bien tren (-«;4)

Neu x>y=r>f(x)>f(y)=>log2y >log2X=>y>x v6 ly.

Neu X < y => f (x) < f (y) ^ logj y < logj x y < x . Vay x = y
Thay x = y vao phuong trinh thu nhat ta duoc g(x) = f (x) - log2 x = 0.
Matkhactaco g(2) = 0 va f'(t) =

"^"^

L_
( t 2 - 2 t 4 . 4 ) > A 2 l ^ t.ln2

Suy ra g(x) = g(2) => x = 2 =>x = y = 2.
Thu lai ta thay (x;y) = (2;2) thoa man he.
Vay he phuong trinh c6 nghi^m duy nha't (x; y) = (2; 2).

3 x - y - 5 = 0.

3a

= 0 suy ra di

Aj,A2
U i ; u 11^
2 .MN
Cau 8.b: Aj di qua diemM(0,0,3) va c6 VTCP
= (2,1,-3);
qua dong
diem
phSng suy ra A nam trong m|it phang chua Aj, Aj
N (2,1,0) va CO VTCP u^ = ( l , 2, - 3 ) . Ta c6
230

231