luyện thi đại học môn toán

Tuye'n chQtt & Giai thifu dethi Todn UQC - Nguyen Phu Khanh, Nguyen Tai

nen

A H = D H =^

=> ( A B C ) 1

Ke D K 1

va

Thu.

Cty TNHH MTV DWH

£)e t u M CO the ke dugc hai tiep tuyen den ( C ) thi I M > R <=> 2t^ + 4t +1 > 0

BC±(AHD)

(AHD)

A H

=> D K 1

„ , > ^ h o , c , < - : ^ ( . )
(ABC)

Phuong trinh di qua hai tiep diem A , B c6 dang:

=>DAK = 45°, D A H = 45°

( t - l ) ( x - l ) + (t + 3)(y + 2 ) - 9 = 0

=> ADAK vuong can tai K.
:

=>K = H : ^ D H l ( A B C )
Di^n tich tarn giac ABC la:
SABC = •^ABACsin60° =

— ^

a^Vs

a^

AB) =

.

2V2t^+4t + 10
3t + l
Xet f ( t ) =
thoa dieu ki^n (•)
2V2t2+4t + 10
2t + 14

Voi t > — thi f (t) > 0 thi ham so f (t) dong bien tren niia khoang

Ta

hai duang thSng DE va HE va bang DEH

s/

Gpi CF la duong cao xuat phat tir C cua tarn giac deu ABC canh a
1 ™ aVs
, HE = - C F =
2
2
4
DH
= 2 =>DEH = arctan2
nen tan DEH =
HE

,

^(2t2+4t + 10)^

T " =IT

Ke HE 1 AB => DE 1 A B . Vay goc giiia 2 mp (ABD) va (ABC) la goc giua

ro
/g
om
.c

ok

Vay, goc giua hai mp ( D A B ) va ( A B C ) la DEH = arctan2

ce

bo

Cau 6: Xet ham so: f ( t ) = ln(t + 1 ) - Vt+T voi t > - 1

w.

fa

va f ( t ) = 0 o t = 3
^'

ww

f ' ( t ) doi dau t u duong sang am khi qua 3 nen f (3) la gia tri Ion nhat.
Suy ra f (t) < f (3) < 0, do do f(x) + f ( y ) + f ( y ) < 0
hay In(x +1) + In(y +1) + ln(z +1) < Vx + 1 + ^ y + 1 + Vz + 1
I I . PHAN RIENG Thi sinh chi dupe chpn lam mpt trong hai phan (phan A
hole B)

Voi t < - | thi f ' ( t ) = 0 o t = - 7
L^p bang bien thien, suy ra f (t) < ^

hay d ( N , AB) <

^

DMng thiic xay ra khi t = -7 tuc M ( - 7 ; - 6 )
Vay, M ( - 7 ; - 6 ) thi gia trj Ion nhat bSng ^

la diem can tim.

Cau8.a:GQi M e A j = > M ( l + t i ; 2 + 2 t j ; l - t i ) , N e A j

=>N(2-t2;3-t2;-2)

=> M N = (l - 1 2 - tpl - 1 2 - 2 t i ; - 3 + tj).
Aj CO vecto chi phucng

U j = ( l ; 2 ; - l ) , v i la mat phSng ( P ) vuong goc Aj

nen (p) c6 vecto phap tuyen n = U j =(l;2;-l)
Gia thie't dau bai ( P ) cat Aj va A j tai M va N , the nen M N nam trong ( P ) ,
suy ra n . M N = 0 <=> l . ( l - t 2 - t i ) + 2 . ( l - t j - 2 t i ) - l . ( - 3 + t i ) = 0

A. Theo chi/ang trinh chuan

= > t 2 = 2 - 2 t i = > M N = ( t i - l ; - l ; - 3 + t i ) khi do M N = ^ 2 ( t i - 2 ) ^ + 3

Cau 7.a: Duong tron c6 tam l ( l ; - 2 ) , b a n kinh R = 3.

Giai thie't

Vi M e ( d ) nentpa dp M ( t ; t + l ) .

1

up

aS

Taco: f ( t ) = ^ " , ^
^'
2(t + l )

Ta c6: d ( N ;

Taco: f ( t ) =

1
1 aJs
The tich khoi t u dien ABCD la V = - D H . S A B C =

, ^„

3t + l

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ADAH vuong can tai H

nen co CF =

Khang Vift

M N =

VlT o

- 2 ) ^ + 3 = N / T T o tj =0 hoac tj =4

Do do CO 2 diem M(1;2;1) hoac M(5;10;-3)
123


TwygH chgn & Giai thifu dethi Toan hpc - Nguyen Phu Khanh, Nguyht Tat Thu.

^ I t t S . b : T i m d u g c tpa d p d i e m A ( 7 ; 16; 1 4 )

M $ t p h i n g (P) d i q u a M(l; 2; 1) c6 vecto phap t u y e n n = U j = ( l ; 2 ; - l )

G(?i U j , i 3 ^ , r i p Ian l u g t la cac vecto chi p h u o n g ciia d , A va vecto p h a p

p h u o n g t r i n h la: l ( x - l ) + 2 ( y - 2 ) - l ( z - l ) = 0 hay x + 2 y - z - 4 = 0

t^yeh ciia ( P ) . Gia s u u j = (a; b; c) a^ + b^ + c^ > 0

M^i p h a n g (P) d i qua M(5; 10; - 3 ) c6 vecto phap t u y e n n = U j = ( l ; 2 ; - l )

Vi d c ( P )

p h u o n g t r i n h la: l ( x - 5 ) + 2 ( y - 1 0 ) - l ( z + 3) = 0 hay x + 2 y - z - 2 8 = 0

(d':^)=45°

hoac x + 2 y - z - 2 8 = 0

= > a - b + c = 0 <::>b = a + c ( l )
^ 2(a + 2b + c ) ' = 9 ( a 2 + b 2 + c 2 )

i^^^^^^^i

(2)

sVa^+b^+C

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Vay, C O 2 m a t p h a n g ( P ) : x + 2 y - z - 4 = : 0

n e n xx^Ln^

Tir (1) v a (2) suy ra: 14c^ + 30ac = 0 <^ c = 0 hogc 15a + 7c = 0

C a u 9.a: T i m tat ca cac so p h u c z , biet |z - 1 - 2if + zi + z = 11 + 2 i ( l )
G(?i so phijrc z = a + b i (a,b e # ) thoa m a n de bai

x =7+t

V o i c = 0, chpn a = b = 1 , ta t i m d u g c d : y = 1 6 - t

r::>z = a - b i , z - l - 2 i = : a - l + ( b - 2 ) i r : > | z - l - 2 i | = ^ ( a - l f + ( b - 2 ) ^ .
Thay vao (l) ta c6: ( a - i f + ( b - 2 f + ( a + b i ) i + a - b i = l l + 2 i

Cau 9.b: D i e u k i f n: x >

s/

a = b+2

a-b =2

up

a =4

ro

b =2

/g

<::> \c
b = -1

| ( a - l ) ' + ( b - 2 f =9

om

Suy ra c6 2 so p h u c can t i m la z = 1 - i va z = 4 + 2i.

2V5

.c

K h i d o |z| = V i va |z| =

ok

B. Theo chUorng trinh nang cao

bo

C a u 7.b: D u o n g t r o n (C) c6 t a r n ! ( - ; 2), R = 4 va d i e m I thuoc d u o n g thang A

ce

D u o n g t r o n ( C ) c6 t a m J ban k i n h R ' = l va tiep xiic ngoai v o i d u o n g tron

w.

fa

( C ) suy ra q u y tich cua d i e m I la d u o n g tron ( K ) C 6 t a m I ban k i n h R + R ' = 5
=25.

ww

h a y ( K ) : (x + l f + ( y - 2 f

V-2

Ta

= ll

x = 7 + 7t

V o i 15a + 7c = 0, chpn a = 7, c = - 1 5 , b = - 8 , ta tim d u g c d : y = 1 6 - 8 t

< : > ( a - l ) ^ + ( b - 2 f + a - b + i ( a - b ) = l l + 2i
(a-lf+(b-2)'+a-b

z = 14

K h o a n g each ciia I t o i A la I o n nhat k h i I la giao d i e m cua d u o n g t h i n g d d i

z = 14-15t

va x > 0, y > 0

Phuong trinh dau tuong duong:

- y"* = | l + x^y^ j l o g i x - l o g j y
V

5

5

.

Neu X > y t h i p h u o n g t r i n h cho v 6 n g h i ^ m .
N e u x < y t h i p h u o n g t r i n h cho v 6 n g h i ^ m .
Neu X = y t h i p h u o n g t r i n h t h u 2 t r o thanh: x + V2x + 1 = 1 + Vx + 2
O x - l

=V ^ - V 2 x + l =

o(x-l)

,

1 +

, ~^^~])
Vx + 2 + V2x + l

Vx + 2 + 72x + l ,

= O o x= l v i l+-

>/x + 2 + V2x + l

>0

V^y, h? p h u o n g t r i n h da cho c6 n g h i f m (x; y ) = ( l ; l ) .

qua J va v u o n g goc v o i A v o i d u o n g t r o n (K).
d C O p h u o n g t r i n h : 4x - 3y +10 = 0 .
Tpa dQ d i e m I thoa m a n h$:

4x - 3y +10 = 0
| ( x - f l ) 2 + ( y - 2 f =25'

x = 2,y = 6
x = -4,y = -2

Voi l ( 2 ; 6 ) = > ( x - 2 f + ( y - 6 f = 1 , v o i I ( - 4 ; - 2 ) = > ( x + 4 ) ^ + ( y + 2 ^ = 1
Vay, k h o a n g each t u I t o i A I o n nhat bang 5.
125
^'^>A


o£THiTHiirsdi9

3 + 51

*Su 9.a: Tim so phuc z thoa man z + —^
"
z
Theo chi/crng trinh nang cac

I. PHAN C H U N G C H O T A T CA CAC THI SINK
Cau 1: Cho ham so y =

5i = 0

Cau 7.b: Trong mat p h i n g tpa dp Oxy, cho hinh thang vuong ABCD, vuong tai

- 3x^ + (3m - 3)x + 2 c6 do thi la ( C „ ) .

va D. Phuong trinh AD: x - yyjl = 0. Trung diem M eiia BC eo tpa dp M ( l ; 0).
giet BC = CD = 2AB. Tim tpa dp eiia diem A.

b) Tim m de ham so c6 eye d^i, eye tieu eiing voi diem l ( - l ; - l ) t^o thanh

Cau 8.b: Trong khong gian toa dp Oxyz, cho mat phang (?):2x + y + z - 6 = 0

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a) Khao sat su bien thien va ve do thi (C) cua ham so khi m = 1.

tam giae vuong tai I .

va m | t cau (S): x^ + y^ + z^ + 4x + 6y - 2z J-11 = 0 . T u diem M tren (P) dyng

Cau 2: Giai phuong trinh:

tiep tuyen M N den mat cau ( N la tiep diem). Tim M de M N ngan nhat, tinh

sin^ X - Vscos^x - i s i n 2x

Idioang each ngan nhat do.

2 (sin x - cos x) - — s i n 2x

Cau 9.b: Giai phuong trinh sau: ^ l o g ^ (x^ + 2x j - log j (x + 3) = logg - — ^

x 3 + ( 2 - y ) x 2 + ( 2 - 3 y ) x - 5 ( y +l) = 0
Cau 3: Giai h§ phuong trinh:

H(/dNGDiiNGlAl

,

Cau 4

,,

,^ r

fln^x-31nx + 3 .

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

-,
r—dx
J
x
{
l
n
x
2
)
Cau 5: Cho tam di?n Oxyz c6 xOy = yOz = zOx = a . Tren Ox, Oy, Oz lay cac

Caul:

om

trift tieu va doi dau qua moi nghi^m, nghla la phai eo:
A'>0

ok

.c

a^+b^+c^

II. PHAN RIENG Thi sinh chi dugrc chpn lam mpt trong hai phan (phan A

bo

V i l y ' ^ ' ^ ^ = ° nen: jyi'^l) = ( 2 m " 4 ) x i + m + 1
[y'(x2) = 0
y(x2) = ( 2 m - 4 ) x 2 + m + l

fa

w.

Cau 7.a: Trong mat phSng tpa dp Oxy, cho duong tron ( K ) : x^ + y^ = 4 va hai

Khi do: I A = ( x i + 1 ; ( 2 m - 4 ) x i + m + 2), iB = ( x 2 + 1 ; ( 2 m - 4 ) x 2 + m + 2)

A, B) la hai diem thupe ( K ) va doi xung

ww

diem A (0; 2), B ( 0 ; - 2 ) . Gpi C, D (A

voi nhau qua true tung. Biet r i n g giao diem E eua hai duong t h i n g AC, BD
n l m tren duong tron ( K J ) : x^ + y^ + 3x - 4 = 0, hay tim tpa dp eua E .
Cau 8.a: Trong khong gian tpa dp Oxyz, cho hinh thoi ABCD eo dinh B thupc
trye Ox, dinh D thupe mat p h l n g (Oyz) va duong eheo AC nam tren duong

<=>l-(m-l)>0om<2

Voi m < 2 thi do thj ham so luon eo eye dai A(xj;y(xj)), eye tieu B(x2;y(x2))

ce

hole 6)
A. Thee chUorng trinh chuan

b) Taeo: y' = 3x^ - 6 x + 3 m - 3 = 3(x^ - 2 x + m - l )
Ham so'CO eye dai, eye tieu khi y' = 0 eo hai nghi^m phan bi?t Xj, X j , y'

/g

Cau 6: Cho a,b,e>0 thoaman: (a + b - e ) ( b + c - a ) ( e + a - b ) = l .
Chung minh rang:

s/

ro

OABC CO the tich Ion nha't.

' a +b +c

a) Danh cho ban dpc.

up

diem A, B, C sao cho OA = OB = OC = k > 0. Tim dieu ki?n eua a de t u di^n

Ta

: Tmh tich phan: I =

Tam giae lAB vuong tgi I nen c6: lA.IB = 0
I

o ( x i +l)(x2 + l ) + [ ( 2 m - 4 ) x j + m + 2][(2m-4)x2 + m + 2] = 0
o ( 4 m ^ - 1 6 m + 17)xiX2 +(2m^ - 7 ) ( x i +X2) + m^ + 4 m + 5 = 0 (•)
Theo dinh ly V i - et: xj + X2 = 2, Xj .Xj = m - 1

thing d :

=^

= j . Tim tpa dp cac dinh A, B, C, D ciia hinh thoi ABCP
Khi do (*) tro thanh:

biet di?n tich hinh thoi ABCD b i n g ISyJl
126

(dvdt).
127


Taco: S ^ g c = f ^ ^ - ^ C . s i n e o "

<=> 4m^ - 15m^ + 37m - 2 6 = 0<=>m = l
Do'i chieu dieu ki$n ta c6 m = 1 la gia trj can tim .

Gpi N la trung diem BC,
AN la duofng cao AABC deu.

sinx - cosx = 0 <=> tanx = 1 o x = — + krt, k e Z
4

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Taco: A N = — B C = k V S s i n 2
2

It

sin X + -v/Scosx = 2 o sin x + — = l o x = - + k 2 7 i , k € Z
6
3j
Cau 3: Dieu kien y > -1

^ A G = ^AN = ^ s i n P
3

3
2

Xet AAOG vuong t^i O , ta c6:

Phuong trinh thu nhat tuong duong vai: x^ + 2x^ + 2x - 5 = y ^x^ + 3x + sj

OG2=A02-AG2=k2 l - - s m ^ l 3
2)

o ( x - l ) ( x ^ +3x + 5) = y ( x 2 + 3 x + 5) <^(x^ + 3x + 5)(x - y - 1 ) = 0
x^ + 3x + 5 = 0 ta thay v6 nghi?m.

Dodo,

TH2: X - y - 1 = 0 hay y = x -1 thay vao phuong trinh thu 2, ta dugfc:

V o . A B C f - ^ l - - s i n 2 - (dvtt)

3

2 ^

'

DSt t = sin^ —, do 0 < a < — nen 0 < t < —
2
3
4

s/

Ta

3>/x=3x2-14x + 14 (*)

ro

Khido: V o . A B C = ^ k ^ ,, t^--t^
V
3

/g

3t''-14t2-3t + 14 = 0 < » ( t - 2 ) ( t - l ) ( 3 t 2 + 9 t + 7) = 0 o t = l h o a c t = 2

up

Dat t = \f\i t > 0. Khi do phuang trinh (*) tro thanh:

De V Q ^ B C

om

V a i t = l=>x = l=>y = 0

.c

Vai t = 2=:>x = 4=>y = 3

ok

Vgy, h? cho CO nghi?m la (x;y) = (l;0),(4;3)

voi

bo

Cau4:D|it t - I n x

te

^

ce

fa

ww

]

l

3^
0;-

4j

dat gia trj Ion nhat

: f'(t) = 0 <» t = 2

Do f'(t) doi dau tu (+) sang (-) qua nghi?m ^ nen f (t) d^t gia trj Ion
f

=>AB = 2 k s i n 2
Tuong tif, ta cung c6: BC = CA = AB = 2ksin j

0;\y

AB^ = OA^ + OB^ - 20A • OB • cosAOB = 2k^ (l - cosa) = 4 k 2 s i n ^ j

f

3^

Ta c6: f'(t) = 2t - 4t^, tren khoang

w.

3

voi te

"hat khi va chi khi ham f{t) = t^ -^t^

Cau 5: Tir gia thiet ta c6: k > 0, 0 < a < — . Xet AOAB, ta c6:

ihat tren khoang

3^
"'4

bang — khi = ^12
2

Vi the hi di^n OABC c6 the tich Ian nhat bang — k^ khi va chi khi
36

nen AABC deu.

Trong hinh chop O.ABC ta c6 cac m|it ben deu la tam giac can t^i O day la
am giac deu ABC, nen A.ABC la hinh chop tam giac deu.

sm"^ — = - tuc la a = 2 2
2
Cau6:D|it x = a + b - c , y = b + c - a , z = c + a - b = > x y z = l

Gpi G1^ trpng tam A A B C , the thi OG 1 ( A B C ) , khi do VO.ABC

=-^&ABC-^

3
28

=^AB^

2ksin —
2

'AABC

Cau 2: Phuong trinh cho tuong duong vai (s inx - cosx)^sin x + Vscosx - 2j = 0

THI:

Ja

1

|4m^ - 1 6 m + 1 7 J ( m - l ) + (2m^ - 7J2 + m^ + 4 m + 5 = 0

129


Tuyen

CUQU

&

Gi&i

thifu

dethi

htn

Toan

Dieu can chiing minh tro thanh:

a

!sJ<^uii,'-ii I'hii Kh,ii:h

X+ y+
•

z

^

. \^u\ihi

+

+

Iiil lUu.^

Cau8.a:Goi B(b;0;0), D ( 0 ; d i ; d 2 ) la toa do thoadebai.

+ xy + yz + zx

Goi (Q) la mat phSng di qua B(b;0;0)va vuong goc vai duong thSng AC

x^ + y^ + z'^ + xy + yz + zx _ (x + y + zf

-

fien nhan u ^ ^ = ( 2 ; - l ; l ) lam vecto phap tuyen, nen c6 phuang trinh:

xy + yz + zx
-

2 x - y + z-2b = 0

Vi xyz = 1 nen xy + yz + zx > 3, do do

Goi I la trung diem A C nen I € d => I (-3 + 2 t ; - t ; t )

(x + y + z ) ' ^ xy + yz + zx ^ (x + y + z) _
6

Ta can chung minh:
Dat

"

6

Han nOa, I cung la giao diem cua d va (Q), nen toa do diem I thoa phuang

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6

2

trinh:

x + y + z 5^(x + y + zr__l

2(-3 + 2 t ) - ( - t ) + ( t ) - 2 b = 0 <:>t =

3

b =6

3
1
Xethamso: f ( t ) = t^ — t ^ + - voi t > l
V ;
2
2
Taco: f (t) = S t ' ' - 3 t , voi V t > l thi f ' ( t ) > 0 d o d o f ( t ) dongbie'n voi t > l

Ta
/g

om

.c

Cau 7.a: V i C, D thupc duong tron (K) ma lai doi xiing voi nhau qua true tung

ok
bo
ce

ww

w.

A C : ( b - 2 ) x - a ( y - 2 ) = 0, B D : ( b + 2)x + a{y + 2) = 0

Toa do diem E la nghi^m cua h^:

( b - 2 ) x - a ( y -2) = 0
(b + 2)x + a(y + 2) = 0

^2

YB + YD = 2 y i , tim du(?c:

dj = -6 :

^B ^^D

d j =6

B(6;0;0)
D(0;-6;6)

A ( - 3 + 2 t i ; - t i ; t i ) , C ( - 3 + 2t2;-t2;t2) => A C = (2t2 - 2 t i ; t i - t j ; t j - t j )

Theo de bai ta c6 di^n tich hinh thoi bang 18N/2 , suy ra

o^/6(v-tO^-3V3 = 18N/2 hay t 2 - t i = 2

(l)

A B C D la hinh thoi nen c6: A B = C B => t j +1^ - 6

(2)

i

AC.BD =

IsVi

T u (l) va (2) tadupc t j = 2 : ^ A ( 1 ; - 2 ; 2 ) , t2 = 4 =^C(5;-4;4)
Cau 9.a: Gia su z = x + y i , (x, y e ^ ) suy ra z = x - yi

fa

Plurong trinh duong thang:

Vi E 6 ( K - , ) nenco: 4

s/

ro

hoac B)

up

I I . P H A N R I E N G Thi sinh chi dugic chpn lam mpt trong hai phan (phan A

Taco: a^+h^-^^ ( l ) .

b + 3 b + 3'l

Vi AC nam tren duang thSng d nen goi t j , t2 e ^ sao cho:

Tud6,tadu(?c f ( t ) > f ( l ) = 0

nen toa do 2 diem CO dang la: C(a;b), D ( - a ; b ) (a,b ^O)

2b-3

Vi I la trung diem AC nen I cung la trung diem BD, t u do ta c6 h§:

^ + y + ^^ t > l . K h i d 6 (•) tro thanh t^-|t2+i>() vai t > l

A. Theo chUorng trinh chuan

b+3

Theo gia thiet, ta c6 : z.z + 3 + 5i - 5i.z = 0 <=> x^ + y^ + 3 + 5i - 5xi - 5y = 0
{

2a
x = -4

16 - 6. ^a^ - 4 = 0 O4a2-4b2-6ab+16 = 0 (2)
+_

"fx = 1
x^ +y^ + 3 - 5 y = 0
5x-5=0

<=>

x =i
2

y^-5y + 4 = 0

o

•z = l + i
|y = i "
)
Jx = l
>z = l + 4i
[y = 4^

Vay, CO 2 so'phiic z = l + i , z = l + 4 i .
B. Theo chUtfng trinh ndng cao

T u ( l ) va (2) suy ra 8a^ - 6ab = 0 o 4a = 3b

130

Cau 7.b: GQI H la hinh chieu ciia M len A D ta c6 H

2 72I
3' 3
131


Cty TNHHMIV

Gpi A|\/2a;aj va H la trung diem cua AD suy ra D(;)

I. PHAN CHUNG CHO TAT CA CAC THI SINH

2 yji
=>A
3' 3

Cau 1: Cho ham so y = x"* - 2mx^ - 1 c6 do thj la ( C ^ ) , m la tham so.
2

a) Khao sat su bien thien va ve do thi (C) cua ham so khi m = 1.

2

Cau8.b:Phucmgtrmhm|itcau (S):(x + 2) +(y + 3) + ( z - l )

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2

b) Tim m de ham so c6 ba eye trj tao thanh mot tam giac c6 dp dai c^nh
day gap doi ban kinh duong tron ngogi tiep tam giac do.

=3

Khoang each tit tam mat cau den (P):
2(-2) + (-3) + l - 6 -12
= 2N/6 > R = 3

Cau 2: Giai phuong trinh: tanx + cot2x =

Nen (?) va (S) khong c6 diem chung.

Cau 3: Giai h? phuong trinh:

Ta

cua I len mat phang (?).

s/

Cau 9.b: Dieu ki#n: x > 0

ro

up

?huong trinh cho tuong duong vol

/g

om

.c

ok

+ 3)(x + 2)(x + l ) ] = 3 « (x2+3x)(x2+3x + 2) =3 (*)

bo

o[x(x

+ 2)(x + 3)(x +1)] = log3 3 o [ x ( x + 2)(x + 3)(x +1)] = 3

w.

fa

ce

D|t t = x^ + 3x, phuong trinh (•) tro thanh: t^ + 2t - 3 = 0 o t = -3, t = 1
V6i t = l tuc x^ + 3x = 1 <:> x^ + 3x - 1 = 0 o x = ~^

thoa dieu ki?n.

ww

Vol t = -3 tuc x^ + 3x = -3 o x^ + 3x + 3 = 0 phuong trinh nay v6 nghifm.
Vly, phuong trinh cho c6 nghi?m x =

-3 + M

sin4x

x2-y2+l = 2 ( ^ - V ^ - x )

n (67rx - 371^ jcosx.sin^ x + 47i(l + sin^ x)
Cau 4: Tinh tich phan: I =
'- = _ — 5 ^
^x

Ma IN khong doi nen MN ngan nhat khi IM ngan nhat, tiic la M la hinh chieu

o log3 (x^ + 2x) + logg (x + 3) = log3 3 - logg (x +1)

2(cosx-l)

Vx+l+7y-3+x-y=2

GQi I la tam mat cau, tam giac IMN vuong tai N, ta c6: IN^ + MN^ = IM^

<=>]log3[x(x

rKhangVi$t

DETHITH(jfSd20

DUt AB = x =>BC = CD = 2x =>MH = ^ = - ^ . Vay, AD = - .

Va AD = - s u y ra t = ^
3
3

I )VVI

0

Vl + sin-'x

Cau 5: Cho hinh chop S.ABCD, day la hinh chu nhat c6 AB = 3, BC = 6, mSt
phJing (SAB) vuong goc voi mat phMng day, cac mat phSng (SBC) va (SCD)
cimg tao voi m | t phSng (ABCD) cae goc bang nhau. Biet khoang each giua hai
duong thSng SA va BD bSng S . Tinh the tich khoi chop S.ABCD va cosin goc

giiia hai duong thing SA va BD.
Cau 6: Cho a, b, c la cac so thyc duong thoa man a + b + c = 3.
Chung minh rang :

8\/abc < 9.

11. PHAN RIENG Thi sinh chi duQC chpn lam mpt trong hai phan (phan A
hoac B)
A. Theo chiTorng trinh chuan

Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong can tgi A,
phuong trinh BA: 2x - y - 7 = 0, duong thang AC di qua diem M ( - l ; 1) diem A
nSm tren duong th3ng A: x - 4y + 6 = 0. Tim tpa dp cae dinh cua tam giac ABC
biet rang dinh A eo hoanh dp duong.
Cau 8.a: Trong khong gian voi hq tpa dp Oxyz, cho hai duong thang
x _ y + 1_ z .

132

x - 1 _ y +1_ z-4

133


Cty TNHHMTVjyWH

Tuyen chqn & Gi&i thifu de thi Todn hqc - Nguyen Phu Khdnh , Nguyen Tat Thu.

Viet p h u o n g t r i n h d u o n g t h i n g A cat ca hai d u o n g t h i n g d ^ d j d o n g t h o i

TU
u--. - .
' o AB.ACBC
BC
Theo bai toan, ta co: R = ——
= —r- <=>
4S \ A B C

v u o n g goc v 6 i m a t phang ( P ) : x + 4 y - 2 z + 5 = 0 .

Cau 2: D i e u k i ^ n : x 5^ m - , m e Z .
4

Cau 7.b: T r o n g m a t p h i n g toa d o O x y , cho 2 d u o n g t h i n g Ian l u g t c6 p h u o n g

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„,
u u .
A
sinx

cos2x
2{cosx-l)
P h u o n g t r i n h cho t u o n g d u o n g :
+
=—
'—
cosx sin2x
2sin2xcos2x
1
cos X - 1
<=> cos2x = cos X - 1 o 2 cos^ x = cos x
sin2x
sin2xcos2x

t r i n h la ( d j ) : 2x - 3y - 3 = 0 v a ( d j ) : 5x + 2y - 1 7 = 0 . Viet p h u o n g t r i n h d u o n g
t h i n g d i qua giao d i e m cua ( d j ) , ( d j ) Ian l u g t cat cac tia Ox, O y tai A va B
2

dat gia trj nho nha't.

'AOAB J

Cau 8.b: T r o n g k h o n g gian v o i h§ tga do Oxyz, cho d i e m A(3; - 2 ; - 2 ) va mat
p h i n g ( P ) : x - y - z + l = 0 . Viet p h u o n g t r i n h mat p h i n g (Q) d i qua A, v u o n g
goc v o i mat p h i n g ( P ) biet rang mat p h i n g ( Q ) cat hai t r y c O y , Oz Ian l u g t

ro

up

3

s/

C a u 9 . b : G i a i b a t p h u o n g t r i n h : l o g j Vx^ - 5 x + 6 + log^ V x - 2 > ^ l o g j (x + 3)
3

/g

HI/OFNGDANGIAI

om

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

.c

Caul:

ok

a) D a n h cho ban doc.

X =

cos

X = —

2

Giasu: A(0;-l), B ( - V m ; - m ^ - l ) ,

c(N/m;-m^-l)

Ta CO d i ^ n tich t a m giac S ^ ^ B C = ^ B C . d ( A , B C ) = m^%/m ( d v d t ) .

134

«

2

x =± - + k2n,keZ
3

X e t h a m so: f { t ) = t^ + 2\/t v o i t > 0 , ta c6: f ' ( t ) = 2t + - ^ > 0 i : i . f ( t )
vt

dong

bien t > 0 , k h i do p h u o n g t r i n h : f ( x + l) = f ( y ) o y ^ x + 1
Thay vao p h u o n g t r i n h t h u hai ta dugc: Vx + 1 + 7 x - 2
2 = 5 - x o

=3

fx<5

2 <=>x = 3=> y = 4 (thoa d i e u ki^n)

x 2 - x - 2 = (5-x)'

Vay, h? CO n g h i f m la { x ; y ) = (3;4)

Cau

«(67tx-371^1 COS x.sin^x

4: I = J^^

Tarn giac ABC can tai A nen canh day la BC v o l :

BC = 2 V m , AB = A C = V m + m *

— + krt

P h u o n g t r i n h t h u nha't t u o n g d u o n g : (x +1)^ + 2\/x + 1 = y^ + 2 ^

fa
w.

m o i n g h i e m nen ham so da cho c6 3 cue t r i .

ww

N e u m > 0 t h i y ' = 0 c6 3 n g h i ? m x = 0, x = - V m , x = \/m va d o i da'u qua

X =

D o i chieu dieu kien, n g h i f m p h u o n g t r i n h da cho la: x = ± - + k27:,k e Z

3
x>-l
Cau 3: D i e u kien:
y>3

N e u m < 0 t h i y ' = 0 c6 1 n g h i e m x = 0 va d o i da'u t u ( - ) sang ( + ) nen c6
1 cue t r j ( k h o n g thoa bai toan )

0
1

ce

bo

cos

Ta

tai d i e m phan biet M va N sao cho O M = O N

b) T a c o : y' = x|x^ - m j

= 4m

Vay, m = 1 thoa m a n yeu cau bai toan.

B. Theo chi/tfng trinh nang cao

AB

4m^%/rn

<=> m ' ^ - 2 m N / m + 1 = 0 <=> m = 1 (thoa man)

z - 1 = 5 va 17 z + z = 5 z z
V
>

sao cho

+

Khang Vie

, '
\/l + sin^ x

u = 67IX - 37r^

Dat

dv =

,

dx + j47tVl + sin^ xdx
d u = 6ndx

cosx.sin^ X

sm X

t

3

sin^x

135


Cty TNHH MTV P W H khang

ruye'n chgn & Giai thifu dethi Todn hQC - Nguyen Phu Khdnh, Nguyen Tat Thu.

Khi do I = (67rx - 3n^ j^Vl + sin'

= 471^

Taco: f ( t ) = |

Zau 5: Ha SH 1 A B => S H 1 ( A B C D ) (do (SAB)

1 ( A B C D ) = AB)

Nh$n thay, ( 9 - 8 t ) ^ - t ^ = 9 ( l - t ) ( 9 - 7 t ) > 0

Ke H K 1 CD => t u giac H B C K la hinh chu nhat.
Ta thay B C 1 ( S A B )

Dang thiic xay ra khi a = b = c = 1.

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CD 1 ( S H K ) => SKH = ( ( S C D ) , ( A B C D ) )

JI. PHAN R I E N G Thi sinh chi du-grc chpn lam mpt trong hai phan (phan A

SBH = SKH = > A S H B = ASHK (g - c - g) => HB = HK = BC = 6

holcB)

Do do A la trung diem HB. Ta thay Z7 A B D K la hinh binh hanh

A. Theo chi/orng trinh chuan

=> BD//AK

Cau 7.a: Ggi diem A e A => A(4yo - 6; yg)

BD//(SAK) ma SA e ( S A K )

=> d(BD,SA) = d ( B D , ( S A K ) ) = d ( D , ( S A K ) ) = d ( H , ( S A K ) ) = ^6 = h
1
h^

1
HS^

1
•+

1

HA^

Ta
up

s/

2 . 3 V 5 . 3 V 5

5

V

D|t t = abc, ta

w.

ww

ro

om

+ c^ = (a + b + c)^ - 3{a + b)(b + c)(c + a) < 27 -24abc

3
CO

V
0
a + b + c'

z = 4 + 3k

Gpi giao diem cua A voi dj,d2 Ian lupt la A, B:
A ( m ; - l + 2 m ; m ) , B(l + k ; - 1 - 2k; 4 + 3k)
Mat phSng (P) c6 vecta phap tuyen n = ( l ; 4 ; - 2 )

fa

Vgy p = SAK = arccos-

+ 2 m , d2 : y = - l - 2 k

z=m

.c

2AS.AK

x=l +k

=m

Cau S.a: Phuong trinh tham so ciia d j : y = - l

ok

1

X

bo

45 + 45 - 72

Khi do: ?

« 1 3 y 2 - 4 2 y o + 3 2 = 0 < = > y o = 2 = > X o = 2 hoac yj, = ^ = > X Q = ( l o a i ) .

ce

AS^+AK^-SK^

Cau 6: Ta c6: a^ +

^

Vay,A(2;2), B ( 3 ; - 1 ) , C(5;3) la tpa dp can tim.

/g

giua hai duang thiing B D va S A => p = ( B D , S A ) = ( A K , S A )

=

1

^5(l7y2-42yo+26)

HK^

Ta C O S K = 6V2,SA = A K = 3N/5 .Trong tam giac S A K
cos S A K

6yo-7

cosACB =

- +•

=> Vs.ABCD = ^SH.dt^ABCD = f 6.3.6 = 36 (dvdt).

P la goc

nAC = (yo -1;5 - 4yo)

Tam giac ABC vuong can tai A, nen c6:

i =- i -+- +—=>SH2=36=>SH = 6
6 H S ^ 9 36

Gpi

= > f ' ( t ) < 0 vol mpi t 6 ( 0 ; l )

guy ra f (t) nghich bien voi mpi t e ( 0 ; l ] va f (t) < f ( l ) = 9

S B H = ((SBC) , ( A B C D ) )

Do tarn di^n H . S A K vuong tai H

Vifi

V i A 1 (P) => AB va n = (l;4;-2) ciing phuong <=> AB = tn

I

l +k-m =t

3

- 2 k - 2 m = 4t
4 + 3 k - m = -2t

k =0

o t = -l=^A(2;3;2),B(l;-l;4)
m =2

=>0I Duong t h i n g A d i qua diem A (2; 3; 2) va nhan n = { l ; 4 ; - 2 ) lam vecto chi

Xet ham so: f (t) = ^ 9 - 8 t + 8 ^ voi t € (0;1^

136

.
, ,
, , x-2
y-3
z-2
luong nen A co phuong tnnh
=
= ——
r

137


C a u 9.a: Gia s u z = x + y i (x, y e M^^ => z = x - y i

y

- 1 + yi =5

(X-I)^y2=5

17.2x = 5(x + y i ) ( x - y i )

34

(x-lf
2

X

-X - X

= 5

34
2
=-x-x

nen n g = u , n p

l34x = 5(x2 + y2 J

K h i d o m a t phSng ( Q ) : 2x + y + z - 2 = 0 va ( Q ) cSt O y , O z tai M (O; 2; O) va

5

lNj(0;0;2) (thoa m a n )

X = —

5^N/i79.

6

y =±

= (2;1;1).

6

Vl79

N e u a = - b t h i M N = (0;-a;-a)// u ( 0 ; l ; l ) va n g 1 u

6

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Gia thiet ta c6: •

N e u a = b t h i M N = ( 0 ; - a ; a ) // u ( 0 ; - l ; l ) va n g 1 u

nen n ^ = u , n p

B. Theo chUtfng trinh nang cao

= {0;1;-1).

Khi do mat p h l n g ( Q ) : y - z = 0.

C a u 7.b: Giao d i e m ciia ( d , ) va ( d 2 ) l a M ( 3 ; l ) .

( Q ) c i t O y , Oz tai M ( 0 ; 0 ; 0 ) va N ( 0 ; 0 ; 0 ) (loai).

Cdch 1: S,^oAB = ^ A B . O H v o i H la chan d u o n g cao ha tir O len A B
AB

Vay, ( Q ) : 2x + y + z - 2 = 0 .

N2

J

^
OH

. Vi OH < O M

O H m a x = O M thi

Cau 9.b: D i e u kien: x > 3

n h o nhat.

OH'

P h u o n g t r i n h da cho t u o n g d u o n g :

K h i d o A B nhan O M l a m vec t o phap tuyen. Ta vie't d u o c p h u o n g t r i n h A B

'

^

+

s/
'

a^'
1+ 3-

/g

a

^ . D a t t=J ,

t>0

ok

V^AOAB/

(x - 2) > ^ l o g ^ . i (x + 3)

« ilog3 (x^ - 5 x + 6 ) - i l o g 3 ( x - 2 ) > - i l o g 3 ( x + 3)

« log3 [(x - 2)(x - 3 ) ] > log3 (x - 2) - log3 (x + 3)
Olog3[(x-2)(x-3)]>log3

'x-2^
x+3

«('<-2)(x-3)>^<:>x2-9>l«

x<->/lO
X>^[W

Ket h(?p v o i dieu ki?n, ta dupe ng hi^m cua bat p h u o n g t r i n h da cho la:
x>VlO

w.

fa

ce

bo

a

t^ +1
X e t h a m s o : f ( t ) = 4.—^
— voi t > 0
(3t4.l)

ro

^

om

•AB =

.c

^ AB

;0 , B

up

Theo bai toan, ta t i m du^c:
(3a + h

Ta

Cach 2: P h u o n g t r i n h d u o n g thang d c6 dang: a ( x - 3) + b ( y - l ) = 0 , ( a , b > O)

ilog3 (x2 - 5x + 6) + \\og^-,

ww

2
Gia t r j n h o nhat ciia f ( t ) la - dat dug-c k h i t = 3 hay a = 3b
P h u o n g t r i n h d u o n g thang can t i m la: 3x + y - 1 0 = 0
C a u 8.b: Gia s u n g la m p t vecto phap t u y e n ciia ( Q ) .
Khi do n Q l n p ( l ; - l ; - l )
M a t p h ^ n g ( Q ) cat hai tryc O y va Oz tai M ( 0 ; a ; 0 ) , N ( 0 ; 0 ; b ) phan bi?t sao
cho O M = O N nen a = b o a
138

= b^O hoac a = - b * 0
139


Tuyen chgn & Giai thifu dethi Todn hgc - Nguyen Phu Khdnh, Nxuyht Tai Thu.

Cau 9.^: Tinh modun cua so'phiic z, biet: z = (2 - i)"^ + ( l + i)'* - ^ — i - .

OETHITHUfSdzi

0. Theo chUomg trinh nang cao

Cau 7.b: Trong mSt phSng tpa dp Oxy, cho hinh vuong ABCD eo phuong trinh
duong thing AB: 2x + y - 1 = 0, va C, D Ian lupt thupc 2 dupng thing
d j : 3x - y - 4 = 0, d j : x + y - 6 = 0. Tinh di|n tich hinh vuong.

I. P H A N 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: Cho ham so : y = x'^ - 3x^ + mx +1 c6 do thi la (C^^)

a) Khao sat sy bien thien va ve do thi (C) cua ham so khi m = 0.

x = -t
Cau 8.b: Trong khong gian Oxyz, cho duong thang (d): y = 1 + 2t va mSt cau
[z = - 3 - 2 t

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b) Tim m de ham so c6 cue dai, cxfc tieu. Gpi ( A ) la duong thang di qua hai
diem eye dai, cue tieu. Tim gia trj ion nhat khoang each tir diem I - ; — den
U 4j
duong thSng ( A ) .
cosx + yfz sin

f

I

X

7t

—

(S): x^ + y^ + z^ - 2x - 6y + 4z -11 = 0. Viet phuong trinh mat phing (p) vuong
goc duong thang (d), cat mat cau (S) theo giao tuyen la mpt duong tron c6 ban
kinh r = 4.

\

Cau 9.b: Tim so phue z thoa man ( l - 3i) z la so thuc va z - 2 + 5i = 1.

4j

Cau 3: Giai phuong trinh: sVZx + l + 2x = loVx-3 +13

HMGDANGIAI

ixe" (e^+lj + l
Cau 4: Tinh tich phan sau: I = f
^
—dx.

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

Caul:
a) Danh cho ban dpc.

s/
up

b) Taco y' = 3x^-6x + m .

Ham so c6 eye dai, eye tieu khi phuong trinh y' = 0 c6 hai nghi^m phan

bi?t.Tuclaeanc6: A ' - 9 - 3 m > 0 o m < 3 .

om

/g

ro

e^+l
^
Cau 5: Cho hinh hpp dung ABCD.A'B'C'D' eo day la hinh thoi e^nh a, BAD=a
voi cosa=-, canh ben AA' = 2a. Gpi M la diem thoa man DM = k.DA va N la
4
trung diem cua canh A'B'. Tinh the tich khoi tu dien C'MD'N theo a va tim
kde C ' M I D ' N .

Ta

I

.c

Chia da thiic y cho y ' , ta dupe: y = y'. x _ l _
3 3

ok

Cau 6: Cho 3 so thuc khong am a, b, c thoa man a + b + c = 3 . Tim gia tri nho

nhai cua bieu thuc: P = a + b^ + c^.
II. P H A N R I E N G Thi sinh chi dupe chpn lam mpt trong hai phan (phan A
hoac B)

bo

Vi y'(xj) = 0,y'(x2) = 0 nen phuong trinh duong thing (A)qua hai diem
eye dgi, eye tieu la: y =

r2m_2^

ww

Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong tai A va diem
B(1;1). Phuong trinh duong thSng AC: 4x + 3y - 32 = 0. Tia BC lay M sao cho

(d;): ^ 2

=

=

~^ \ \g (P): x + y - 2 z + 5 = 0. Lap phuong

trinh duong thSng (d) song song voi m^t phSng (P) va cat (d^), ( d j ) Ian lupt
tai A, B sao cho dp dai doan AB nho nhat.
140

+ ^ + 1 hay y = —(2x + l ) - 2 x + l
(

5>/2
BM.BC = 75. Tim C biet ban kinh duong tron ngoai tiep tam giac AMC la — ^ .
Cau 8.a: Trong khong gian Oxyz, cho hai duong thSng ( d j ) :

m ,
-2 x + — + 1.

Gia sir ham so c6 eye d^i, eye tieu t^ii cae diem (xi;yi),(x2;y2) •

ce

fa

w.

A. Theo chUorng trinh chuan

2m

\

Ta thay, duong thang (A) luon di qua diem co'djnh A — ; 2 .

so'goc

|*a duong thing lA la k = | . Ke IH 1 ( A ) ta thay d ( l ; A ) = I H ^ I A = | .
I

Ding thuc xay ra khi I A ± ( A ) o ^ - 2 = - i = - - < » m = l

V^y, max d ( l ; A ) = | k h i m = 1.

141


Cty TNHH MTV DWH Khang Vi$t

Tuye'n chgn b Giai thi?u dethi Todu hgc - Nguyett Phu Khanh , Nguyen Tat Thu.

Cau 2: Phuang trinh cho tuong duong voi phuang trinh:

= id(M,(A'B'C'D')).isABCD

sin3x + 3sinx = 4sin^ x.cosx + 2cosx + s i n x - c o s x
<=> sin 3x + s inx = 2 sin x.sin 2x + cos x - s inx

= —.2a.—.a.a.sina = — , / l
=
3 2
3 V 16
12
• Dat AB = x, A D = y, A A ' = z. Taco

o 2 sin 2x.cos X - 2 sin 2x.s inx = cos x - s inx
<=>(cosx-sinx)(2sin2x-l) = 0 o c o s x = sinx hoac 2 s i n 2 x - l = 0

6

6

+ k27t

— + kTl
12

57t
X =

Khido C ' M 1 D ' N < : > C ' M . D ' N = 0

,

—
'12 + kn

• |x + ky + z j

Cau 3: Dieu ki#n: x > 3

Nhan 2 ve voi bieu thiic lien hg-p va dat thua so chung:
+

I ) = 0 c^x = y hoac

2N/)r^

N/2X +1 = 5

dx

e^+l

l2=

\—

-. Dat u = e" + 1

-Y = ln

+I2

s/

bo

du = e^dx

u-1 1 +e
2

I = Ii

ce

w.

0
1

.Ii=xe''

= l + ln

,
e
, 1 , 2e
= ln
l n - = ln
e+1
2
e+1

2e
e+1

Cau 5:

eMx = (xe''-e'') = 1

fa

du - dx

fu = X
at ^ •
= Jxe^dx. Dat
dv = e dx

.a.a.— = 0 <=> k = — .
4
5

up
ro

.c
ok

—

ww

1

f

0 ^ +^

fk
xy = 0 <=>~a^ - k a ^ + - - 1
2

1

Taco: P'(b) = 2 b - 1 va P'(b) = O o b = i .

om

13
Vgy, phuong trinh cho c6 nghiem la: x = - y , x = 4

0

D'

D^t P(b) = 3 - b - c + b2+c^ voi b e [ U , 3 ' .

/g

4

Cau 4:1= f x e M x +

\

Cau6:a + b + c = 3=>a = 3 - b - c , k h i d ( S P = 3 - b - c + b^+c^
+

,
fx<6
2^/x^ + ^ / 2 ^ = 5<=>27(x-3)(2x + l ) = 18-3x<^• X ^ - 8 8 X + 336 = 0
z:>X =

\

\

Ta

(2X-13)(5-2N/X^-N/2X

A-

\

=0

-k

Phuong trinh cho tuong duong: 5 ( 2 V x - 3 - N/2X + I ) = 2X - 1 3

/

/''>•

- 1
D ' N = D ' A ' + A ' N = - y + ^x.

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X =

^ + k27t

—

D
'

C'M = C'D' + D'D + D M = - x - z - k y

Voi cosx = sinx <=> x = — +
4

Voi sin2x = — o
2

c

Tirdo, ta duoc P > m i n P = c ' ' - c + —
4
Xet P(c) = c 3 - c + J voi ce[0;3" .
Ta c6: P'(c) = 3c^ - 1 va P'(c) = 0 o c =

^
v3

Tir do, ta duoc P > min P(c) = H _ . ^
4 3V3

Vay, (a;b;c) =

5

1 1 1

thi minP =

4

373

tt- PHAN RIENG Thi sinh chi dug-c chpn lam mpt trong hai phan (phan A
ko^c B)
Theo chi/orng trinh chuan
7.a: Gpi I la tarn duong tron ngoai tiep tarn giac A M C
Vi B n i m ngoai duong tron ( l )
nen ta c6: BM.BC = BM.BC ( l )

Taco V^MDN = | d K ( A ' ^ ' C ' ^ ' ) ) S c N D
142

143


Ta c6: P^g^^j^j = B M . B C = B I ^ - R ^

vaiR = ^

AB = ^(a - 5)^ + {-a - 1 ) ^ + (-3)^ = Vza^ - 8 a + 35 = ^2(a - 2 ) ^ + 27 > sVs

(2)
^ '

2

Suy ra, m i n A B = 3V3

A(1;2;2), AB = (-3;-3;-3)

Tir ( l ) va (2) suy ra
V^y, phuong trinh duong t h i n g (d) la:

=

=

.

B I ^ - R ^ =75<»Bl2 =

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Cau 9.a: Ta c6: (2 - i) = 3 - 4i va ( l + i)'' = (2i)^ = -4
Phuong trinh A B :
3 x - 4 y + l = 0 vatimduofc A ( 5 ; 4 )
Al2
GQi l ( x ; y ) ta c6:
Br

•
I

125
4
175
2

, . 3 - 4 1 - 4 - 1 ^ , z =- l - 4 i - 2 i . ^ i = _i9_93.

;|

/

-;2
\

=>

/

I
V

C(8;0) hoac

BK

om

MB

2(b-a) + l.(l0-b-3a) = 0
'
2b+6-b-l
= -J(b-af+(l0-b-3a)^

A B . B K = M B . B C = 75

/g

—=—

ok

3 x - 4 y + l = 0.

ce

bo

V i A la giao diem cua A B va A C nen A(5;4)

' ^ [ | a - l | = |4a-10

bu 8.b: Mat cau ( S ) : x 2 + y 2 + 2 ^ - 2 x - 6 y + 4 z - l l = 0

fa

x = -t

r20-4t^'

w.
+

Duong t h i n g ( d ) : y = l + 2t CO vec to chi phuong u = ( - l ; 2 ; - 2 )

= 25 o t = 2 hoac

z = -3-2t

ww

e A C va AC = 5 o ( t - 5 )

t =8
V^y,C(8;0) hogc C(2;8)
Cau8.a:Dat A ( - l + a;-2 + 2a;a), B(2 + 2b;l + b ; l + b)

l ( i ; 3 ; - 2 ) , ban

lnhR = 5

AB = 5=>BK = 15 =>AK = 10 => A C = V 4 R ^ - A K ^ = 5
^

fb = 5a-10

.c

Phuong trinh duong t h i n g A B qua diem B(1;1) va c6 VTPT (3; - 4 ) :

1/

jCDln
d(C;AB) = CD

ro

(Do t u giac A K C M npi tiep)

^..32-4t^

25

Ta
s/

Vi ABCD la hinh vuong nen

up

G<?i I la trung diem K C => I la tarn duong tron ngogi tiep tam giac A M C

Gpi C

25

Cau 7.b: C , D Ian lupt thupc 2 .duong thSng d j : 3 x - y - 4 = 0, d 2 : x + y - 6 = 0
^
4^ D f b 6
C D = (b-a;10-b-3a)
^
^' ^ ' ^ ' ^ ^
Gpi n = ( 2 ; l ) la mot vecto phap tuyen ciia AB.

Cdch2.Tu M d\mg M K 1 B C , ( K € A B )

A A B C dong d^ng A M B K nen:

25

Ijifc^^/

C(2;8)

CO

25

B. Theo chi/orng trinh nang cao

5-

Phuong trinh duong trung tryc I N cua A C => A C n I N = N

Ta

50

Mat p h i n g (P) vuong goc duong thang (d) nen phuong trinh mat p h i n g (P)
bdang: - x + 2 y - 2 z + D = 0

I Tac6d(l,(P)) =

-1 + 6 + 4 + D

D +9

=>AB = (-a + 2b + 3;-2a + b + 3;-a + b + l )
Theo gia thiet, (P) c i t (S) theo giao tuyen la mpt duong tron c6 ban kinh

Do AB song song voi (P) nen AB 1 np = ( l ; l ; - 2 ) <=> b = a - 4

\ 4 nen ta dupe:

Suy ra: AB = (a - 5; -a - 1 ; -3)
144

IT

145


T-ty

D +9

d(l,(P),) = V R = - r 2

4 ^ o D + 9 = 9<=>

D =0

MTV DWH

Khung Viei

Cau 6: Cho a, b, cla 3 sothuc duong thoa man: (a + c)(b + c) = 4c^.

D = -18
Tim gia tri Ion nhat cua bieu thiic: Q = — - — + —^— + ———
b + 3c a + 3c bc + ca

Vay, CO hai mat phang can tim la: - x + 2y - 2z = 0, - x + 2y - 2z - 1 8 = 0
Cau 9.b: Gia su z = x + y i , khi do ( l - 3i)z = ( l - 3i)(a + hi) = a + 3b + (b - 3a)i

II. PHAN RIENG Thi sinh chi du
( l - 3 i ) z la sothuc <=>b-3a = 0<=>b = 3a

hole B)
A. Theo chUcrng trinh chuan

c^lOa^ - 34a + 29 = 1 <=> 5a^ -17a + 14 = 0 o

=1

Cau 7.a: Trong mat phang tpa do Oxy, cho duong tron ( C ) : ( x - l ) ^ +(y+2)^ =4.

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z - 2 + 5i

- 2 + (5-3a)i| = l « J ( a - 2 f + ( 5 - 3 a f

a = 2=>b = 6

M la diem di dpng tren duong thang d: x - y + 1 = 0. Chung minh rSng t u M ke

7

. 21
a =-=>b = —
5
5

dup-c hai tiep tuyen M T j , MTj toi (C) ( T j , Tj la tiep diem) va tim toa do
diem M , bie't duong thang TjTj di qua diem A ( 1 ; - 1 ) .

7 21.
Vgy, C O 2 sophiic thoa man: z = 2 + 6i,z = - + — i
13

Cau S.a: Trong khong gian toa do Oxyz, cho diem M(0; - I ; 2), N ( - l ; I ; 3).

O

Viet phuong trinh mat phang (R) d i qua M , N va tao voi mat phang (P):
2 x - y - 2 z - 2 = 0 mot goc nho nhat.

DE THI THI/SO 22

Ta

Cau9.a: Chosophuc z =

s/

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

ro

up

Cau 1: Cho ham so: y = x'' - 3x + 2 c6 do thj la (C).

/g

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

om

b) Tim toa do diem M thuoc duong thang (d) c6 phuong trinh y = -3x + 2

.c

sao cho t u M ke duoc hai tiep tuyen toi do thi (C) va hai tiep tuyen do vuong

bo

ok

goc voi nhau.

1+i

Tinh mo dun ciia so'phuc w = z,2010 _^j,2011 ^^2016 ^22021

B. Theo chUtfng trinh nang cao

Cau 7,b: Trong mat phang toa do Oxy, cho hinh thoi ABCD c6 phuong trinh

hai canh AB va A D theo t h u t u la x + 2y - 2 = 0 va 2x + y + 1 = 0. Canh BD
chua diem M ( l ; 2). Tim toa dp cac dinh cua hinh thoi
Cau 8.b: Trong khong gian toa dp Oxyz, cho mat cau (S): (x - 2)^ + (y + 2)^ + (z - 1 f = 1.
Tim toa dp diem M thupc tryc Oz sao cho t u Mice dupe ba tiep tuyen M A , MB,
MC toi mat cau (S) va diem D ( l ; 2; 5) thupc mat phSng (ABC)

ce

Cau 2: Giai phuong trinh: sinxcos2x + cos^ x|tan^ x - 1 j + 2sin"' x = 0

11

1-i

w.

fa

V ^ - 2 - 7 3 ^ =y^-x2+4x-6y +5

V2x + 3 + 7 4 y + l = 6

Cau 9.b: Giai he phuong trinh:

itanx.ln(cosx)
Cau 4: Tinh phan sau: I = J
^dx
cosx
Cau 5: Cho hinh chop S . A B C D c6 day A B C D la hinh binh hanh, canh A B = a,

log2

x +Vx^ +4 + l o g 2 \/y^~+4^-y - 9

xy - 4(x + y) +10 = (x + 2 ) 7 2 y - l

ww

Cau 3: Giai h# phuong trinh:

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

Cau 1:
a) Khao sat su bien thien va ve do thj (C) ciia ham so.

A B C = 60". Hai mat p h i n g ( S A D ) va ( S B C ) la hai tam giac vuong Ian lugt tai
A va C . Dong thai cac mat phMng nay ciing hop voi m^t day mpt goc a. Tinh
the tich khoi chop S . A B C D theo a va a.
146

b) Goi M ( a ; b ) la didm cSn t i m M e (d) => b =-3a + 2
Tiep tuyen cua do thj (C) t?i diem (xo;yo) la y=(3x^-3J(x-Xo)+x^-3xo+^
147


Tuyin chgn & Giai thifu dethi Toi'ui hoc

Nguyen

I'lui Khdnh

, Nguyen

Cty TNIIII

Tat Thu.

D o i cgn: x = 0 = > t = l , x = ^ = : > t =

< » - 3 a + 2 = (3xo - 3 ) ( a - x o ) + x^ - 3xo + 2
2xo - 3axo = 0 o

X Q = 0 V xg =

u = lnt
Khid6:I = - ) l ^ d t =
1

UJ

•

7i_i_:^in2

2

2

D o (SAD) va (SBC) hq>p v o i day

O = A C n B D va song song v o i A D .

s/

= -1

Ta

2sin^ x + s i n x - 1 = 0
c

/g

ro

up

X = — + k27t, X = — + k27c, X = — + k27t
2
6
6

.c

om

D o i chieu d i e u ki?n, p h u c m g t r i n h c6 n g h i ? m x = ^ + k27r, x = ^ + k27i
6
6

ok

C a u 3: D i e u k i f n: x > 2, y < 3

bo
f ' ( t ) > 0 vol

V t > 0 nen

ww

w.

t ^ O . T a c o : f ( t ) = 2t + ^ ,

fa

ce

3 + V4y + l = 6

BCISC

(gt)

BCISH
BC1(SHC)=>BC1HC
va i

DA I S A

(gt)

DA1SH

D A 1 (SHA)

B
D A 1 H A . Suy ra A , H , C thSng hang. V ^ y H s O .

S A B C D = 2S^ABC - A B . B C . s i n 6 0 ° =

Va A O = i A C = i A B . s i n 6 0 ° =
^- ^
2
2

f ^ > / x - 2 J = £ ( 3 - y ) o x - 2 = 3- y hay x = 5 - y , thay vao p h u o n g t r i n h t h u
2 ta duQ-c: ^ 1 3 - 2 y + yfiy + l = 6

-

—
4

Vs.ABCD-3^ABCD

C a u 6: D a t x = - , y = C

D § t a = 7 l 3 - 2 y , b = 74y + l
b = 6-a
{a = 3 , „
o
<=>^
hoac
3a2-12a + 9 = 0
b = 3

Ta CO

Ta t i n h d u g c BC = A B c o s 6 0 ° = - d o do
2

h a m so' f ( t ) d o n g bien t r e n n u a khoang [0;+oo)

14R

t

Suy ra H thupc d u o n g t h i n g d i qua

s i n x ^ l - 2 s i n ^ xj + 2sin^ x - 1 + 2sin^ x = 0 o

a+b = 6
'6 he:
Ta CO
h?: ••( ,
,
3a^+b^ = 2 7

ln2-l '
t J2

__1

d(H,AD) = d(H,BC).

s i n x c o s 2 x + cos^ x t a n ^ x - l j + 2 s i n ^ x = 0

voi

2

dv = — d t
t^

goc bang nhau nen

C a u 2: D i e u ki?n c o s x * 0

Xet i{t) = t^+/t

^

du = - d t
t

Cau 5: G p i H la h i n h chieu cua S x u o n g mat p h 5 n g ( A B C D ) .

V | y CO hai d i e m thoa m a n de bai la: M

V2x +

J2

*

.I = - - l n t
t

H a i tiep t u y e n nay v u o n g goc v o i nhau o k j k j = - l o a ^ = ^ o a = ± ^ ^ ^
81

1 o
sinx = —
2

jll^dt.Dat.

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k i = f ( 0 ) = - 3 , k 2 = f ' 3 i l = 27a^ - 3
4

X

^

3a

Co hai tiep t u y e n d i qua M v o i h? so goc la

sin

KhangVi?

Cau 4: Dat: t = cos x = > d t = - s i n x d x

Tiep t u y e n d i qua M ( a ; b )

o

M T V m'X'H

a-Vs

=>SO = A O . t a n a = - ^ t a n a .
4
(dvtt).

thi x , y > 0 .

C

T u gia thiet, ta c6: (x + l ) ( y + l ) = 4 o x

+ y + xy = 3

a = l
b = 5

Khido, Q = _ ^
y+3

+- y - + 3 L
x+3
x+y
149


Tuyi'n chgn b Giai thifu

Todn UQC - Nguyen

dethi

Phii Khdnh , Nguyen

Tilt

Thu.

Do tinh doi xurigcua x, y ta dat S = x + y, P = xy =>S + P = 3

Vi diem N thuoc ( P ) nen thay tpa dp N vac ( P ) ta dupe: A = 28 + C

=i> P = 3 - S > 0 hay S < 3 . De ton tai S, P ta c6:

Gpi a la goc tao boi hai mat phang ( P ) va ( Q ) , ta c6:

S2-4P>0

<oS2-4(3-S)>OoS>2

B

cosa =

S ^ - 2 ( 3 - S ) + 3S ^ 3 - S ^ S ^ 3 3
3S + ( 3 - S ) + 9
S
2 S 2

Tom lai, voi 2 < S < 3, luon c6 Q:

V 5 B 2 + 4 B C + 2C2

Neu B = 0 thi a = 90" .

Dat f{S) = | + | - | voi 2 < S < 3

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Neu B^O, dat m = —, ta c6: cosa =
^

Taco: f'(S) = i - ^

va f'(S) = 0c^S = V6

a nho nha't khi cosa

maxf(S) = f(2) = l khi (S;P) = ( 2 ; l ) , suy ra maxQ = l khj a = b = c = l

J_

^

=

V2m2+4m + 5

^
^2(m + l ) ' + 3

a.

^_

^

<=>m = - l o B = - C .

Vay mat phang (R): x + y - z + 3 = 0

2
I I . PHAN RIENG Thi sinh chi dupe chgn lam mgt trong hai phan (phan A
hoac B )
A. Theo chi/crng trinh chuan
Cau 7.a: Duong tron (C) c6 tam l ( l ; - 2 ) ban kinh r = 2, M nam tren d nen

Cau9.a: Ta c6 : ^
1+

Ta
s/

Ta

ro

. Duong tron (T)

/g

Gpi J la trung diem I M nen toa dp diem J

X

2

I

+

r

m-l^
^ "
2

2(m + l f + 8

bo

ce

2x + y + l = 0

m

x+2y-2

_lY
2

= -1

=1

=i2"l«(l + i + i ^ + i " )

.i2010(i + i _ i _ i )

^

0

= -1
3.

4 5'
['3'3

5
Phuong trinh duong phan giac goc A la

fa
y_.

1 + Z+ Z^+z")

x+2y-2=0

w.

ww

m +1

X -•

.3

Cau 7,b: Tpa dp dinh A la giao diem cua AB va A D nen A ( x ; y ) la nghifm ciia

( x - l f . ( y + 2 f =4
Tpa dp T,, Tj thoa man h?:

^•4)j'

B. Theo chuorng trinh nang cao

2 ,
T u M ke dupe 2 tiep tuyeh MTi, MT2 den ( C ) , nen Tj, Tj la hai giao diem
cua (C) va (T).

2010

.c

m + 1^

=- - -1

Suy ra w = 0

ok

(

om

IM
duong kfnh I M c6 tam J ban kinh tj := — c6 phuong trinh (T) la:

•2

CO

w = z

up

Vi I M > 2 nen M nam ngoai (C), do do qua M ke Avtqc 2 tiep tuyeh toi (C).
m-1

i _

Suy r a z = ( - i ) " = . - i i U - i 4 . 2 ^ 3 _ r

M ( m ; m + 1) =:>IM = ^ ( m - l f + ( m + 3 f = ^ 2 ( m + l f + 8
m +1

- ill^

i

2x+v+l,

2(m + l f + 8

"(di):x-y +3 = 0
(d,):3x + 3 y - l = : 0 '

Truong hop ( d j ) : x - y + 3 = 0.

)

= > ( T I T 2 ) : ( m - l ) x + (m + 3)y + m + 3 = 0

Duong thSng ( B D ) di qua M va vuong goc voi ( d , ) nen ( B D ) : X + y - 3 = 0

Vi A G ( T I T2) nenco: m - l - m - 3 + m + 3 = 0 o m = 1=>M(1;2)

Suyra B - A B n B D = > B ( 4 ; - l ) , D = A D n BD => D ( - 4 ; 7 ) .

Cau S.a: Mat p h i n g (P) di qua M nen c6 phuong trinh dang :
A(x -0)

150

+ B ( y + 1)

+ C(z-2)

= 0 voi

Gpi I = BD n { d ^ )

A^+B^+C^^^O.

1(0;3). Vi C doi xung voi A qua I nen C

4 13
3'3j

Truong hop ( d 2 ) : 3x + 3y - 1 = 0 . Ban doc lam tuong tu.

LI

151


Twygti chpn & Gi6i thi^ dithi Todii hoc Nguyen Phu Khatth , hi^micn Tat Thu.

CtyTNHHMTV

Cau 8.b: Mat cau (S) c6 tarn l(2;-2;l), ban kinh R = 1, M(0;0;m) e O z . M | t
phMng ( A B C ) c6 vecto phap tuyen n = IM = (-2;2;m - l ) ; m^t phing (ABC) di

qua D(1;2;5) nen c6 phuang trinh:
2 ( x - l ) - 2 ( y - 2 ) - ( m - l ) ( z - 5 ) = 0 hay 2 x - 2 y - ( m - l ) z + 5 m - 3 = 0
X = 2-2t
Duong thing IM: y = - 2 + 2t
z = l + (m-l)t
Gpi H la giao diem ciia ( A B C ) voi IM thi to? dp cua H la nghi?m cua h?;
X = 2-2t
X = 2-2t
y = - 2+ 2t
y = - 2 + 2t
z = l + (m-l)t
z = l + (m-l)t
4m + 6
2x-2y-(m-l)z +5m-3 =0 t =m^-2m +9
Do MA la tiep tuyen ciia (S) nen tam giac MAI vuong tai A va AHIIM, cho
n e n t a c o I A 2 = I H I M O I H I M = 1 (do H n i m tren tia I M ) , IH=(-2t;2t;(m-l)t)
0;0;-^

bo

ok

.c

Cau 9,b: Phuong trinh thu nhat
+ Vx^ +4 ^y^ + 4 - y = 2 O X + Vx^ +4 = y + ^y^ +4
ologj
t +t

fa

7t^+4 + i

f(t) = l

ce

Xet ham so: f (t) = t + Vt^+4 , ta c6:

ww

w.

•>0
f(t) dongbiehtren R nen f (x) = f (y) o x = y
Phuong trinh thu hai tro thanh: x^ - 8x +10 = (x + 2) V2x-1 (*)
Dat u = V2x-1 voi u > 0, thay vao phuong trinh (*), Igp bi?t so
t2+4

A = 25(x + 2)^ => u = ^^-^ hoac u = -^^-^ (khong thoa)
3
2
Voi u = ^ ^ ta du(?c x + 2 = 3V2x-l c6 hai nghifm x = 1, x = 13, ta tim
3

duoc (x;y) = (l;l),(l3;13)
152

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

2x + 3

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Cau 1: Cho ham so y = — c6 do thi la (C)
a) Khao sat su bien thien va ve do thi (C) cua ham so'.
b) Tim m de duong thang (d): y = 2x + m cat do thj (C) tgi hai diem phan
bi?t sao cho tiep tuyen ciia (C) t^ii hai diem do song song voi nhau.
Cau 2: Giai phuong trinh: sin^ X sin^ 3x = tan 2x (sin X + sin 3x)
cosx cos3x
Cau 3: Giai phuong trinh: 2^x^ + 2J = sVx^+l.

Ta

e22 + lnx(2 + ln2x)
:
Tinh
tich
phan:
1=
f

-i—
klx
Cau 4
X .In X
Cau 5: Cho hinh chop S.ABCD c6 day ABCD la hinh vuong, SA vuong goc voi
day. Gpi M , N Ian lugt la trung diem ciia SB va AD. Tinh the tich ciia khoi chop
M.NBCD biet duong thang M N tao voi mat day mpt goc 30° va MN = 2a>/3 .
/ b c^
Cau 6: Cho a,b,c e ri;3]. Chung minh rang: 5 — + — + — c b a
yb c a)
II. PHAN R I E N G Thi sinh chi dvtqic chpn lam mpt trong hai phan (phan A
hole B)
A. Theo chUtfng trinh chuan
Cau 7.a: Trong mat phSng voi h^ tpa do Oxy, cho hai duong thang dj :x-y-2=0,
P2 : 2x + y - 5 = 0. Viet phuong trinh duong thSng A di qua goc tpa dp O cat
r d j , dj Ian lupt tai A, B sao cho OA.OB = 10.
Cau 8.a: Trong mat phang tpa dp Oxy, cho hinh chu nhgt ABCD c6 M(4;6) la
trung diem ciia AB. Giao diem I ciia hai duong cheo nam tren duong thSng (d)
CO phuang trinh 3x - 5y + 6 = 0, diem N(6; 2) thupc canh CD. Hay viet phuang
trinh cgnh CD biet tung dp diem I Ion hon 4 .i W s i '
.\
Cau 9.a: Tim modun ciia so phuc z biet: z = 1 + i . ( l . 2 i )

s/

ro

--=>M

om

^m^ - 2m + 9Jt = l o 4 m + 6 = l o m =

DETHITHUfSCf23

up

= 1 o (-2t).(-2) + 2.(2t) + (m - l).(m - 1 ) t = 1

/g

IH-IM

DWIi Khang Vi^t

153


Tuye'n chgn & Giai

thifu

dethi

ToAn hgc - Nguyen

Phu Khdnh

, S ^ i n i c n TalThu.

B. Theo chi/orng trinh nang cao

sinx = 0

C a u 7.b: T r o n g mat phSng v o i h ^ tpa d p Oxy, cho ba d i e m A ( - l ; - 1 ) , B(0; 2),

sinSx

C(0; 1). Viet 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 A sao cho t o n g k h o a n g each

_ cos 3x

t u B va C t o i A la Ion nha't.

sin2x = 0

<=>

sinx = 0
cos X = 0

<=> X = k7t

p h u o n g t r i n h da cho t u o n g d u o n g v o i :

va

2 (x + l ) + ( x 2 - x + l ) = 5 ^ ( x + l ) ( x 2 - x + l ) (*)

mat ph5ng (P); 2x + y - 2z + 9 = 0. Gpi A la giao d i e m cua d v o i (P). Viet p h u o n g

t r i n h d u o n g thSng A n a m trong (P) biet A d i qua A va v u o n g goc v o l d .

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Qjch i . C h i a ca 2 ve p h u o n g t r i n h (•) cho x^ - x + 1 , ta d u p e :

2y(4y=•2+3x2) = x4(x2+3)

x +1

C a u 9.b: Giai he p h u o n g t r i n h :

,x

2012'* ( ^ 2 y - 2 x + 5 - x + 1 ) = 4024

I. PHAN CHUNG CHO TAT CA CAC THI SINH
a) D a n h cho ban dpc.

x^-x + 1

X

Ta

= 2x + m

s/

< » 2 x 2 + { m - 6 ) x - ( 2 m +3)-0,x;^2

x-2

x+1

= 4 o 4 x ^ - 5 x + 3 = 0 v 6 n g h i ^ m v o i m p i xeM .

TH2: t = r tuc - J i l i _ ^ i « x 2 - 5 x - 3 = 0

(*)

>0
7
<=>(m-6) + 8 ( 2 m + 3 ) > 0
phan bi§t va khac 2 <=> • ^ ,
g(2)^0
y
f
^
f

^ :

ok

m ^ + 4 m + 60 > 0 (luon diing).

<=> X =

Cdch 2:Dat

.c

om

/g

A

_

bo

ce
fa

C a u 2: D i e u ki?n: cos x*0, cos 3 x ^ 0

ww

y ' ( x i ) = y'(x2)<=>Xj + X2 = 4 o m = - 2 .

w.

Tai hai giao d i e m ke hai tiep tuye'n song song k h i va chi k h i

5-V37 . „

hoac X =

5 + ^y37
5-N/37

5 + N/37

2

2

v = V x ^ - x + l > — , k h i d o p h u o n g t r i n h (•) t r o

u = 2v <=> Vx + l = 2Vx^ - x + 1 , binh phuong 2 v e r o i riit gpn ta dupe:

THI:
I

u=4 ^ > 0 ,

4x^ - 5x + 3 = 0 , phuong trinh nay v 6 nghi^m voi mpi xeR .
TH2: V = 2u o Vx^ - x + 1 = 2Vx + l , binh p h u o n g 2 ve roi riit gpn ta dupe:
X^-5X-3.0

P h u o n g t r i n h cho t u o n g d u o n g v o i p h u o n g t r i n h :
tan X sin X + tan 3x sin 3x = tan 2x(sin x + sin 3x)

4

2(u^ + v ^ ) = 5 u v « > ( u - 2 v ) ( 2 u - v ) = 0c:>u = 2v hoac v = 2u

Voi V
m €€#M tthhii du(
d u o n g thSng cat d o thj h a m so tai hai d i e m c6 hoanh d p
Vm
6-m

- x+1

V^y, n g h i ^ m p h u o n g t r i n h cho la: x =

ro

( d ) cat ( C ) tai 2 d i e m p h a n bi|t k h i va chi k h i p h u o n g t r i n h (*) c6 hai n g h i ^ m

up

b) P h u o n g t r i n h hoanh d p giao diem:

t = 2 tuc

THI:

C a u 1:
2x + 3

-+1 . 5 . / - . ^ ( . * ) . D a t t = j - . i i ± i - . 0
- x+ 1
lx^-x + 1
(x^ - x + 1

Khi d o p h u o n g t r i n h (• •) t r o thanh: 2 t ^ - 5 t + 2 = 0 <=>t = 2 hole t = ^

Hl/dNG DAN GIAI

Xj ^ X j . Ta CO Xj + X2 =

cos X

sin x = 0

QiU 3: D i e u kien: x > - 1 .

x-l_y+3_z-3
C a u 8.b: Trong mat phang toa dp Oxyz, cho d u o n g thang d : ^ ^ ~ ^ ^ ~ Y "

o

sinx _ ^

« x =^ h o S c x

2

|Vay, nghi^m phuong trinh la: x =

=

^

.

2
x-^ ' ^ ^

<=> (tan X - tan 2 x ) s i n x + (tan 3x - tan 2 x ) s i n 3x = 0
sin(-x)

smx
-sin3x = 0
-smx + cos3xcos2x
cos X cos 2 X

I 4 : l = 21i+l2 v 6 i l , = j i ^ ^ x ,
' x'^.ln'^x
e

•

l 2 = ' f ^ x
i x^

^

J " ' ^ V d x . D a t t = x l n x = ^ d t = (lnx + l ) d x
„ X .In X
154

155


choii

i'-f Ci&i

Ihicii

ile thi

Toiiii

- Ni;iii/<'" f " '

hoc

2e^

1

Tat

IhU.

CtyTNHHMTV

D^t t =

thi t e

X

__1

X

f(a)>f(t) = 10t-t2+ — - t^ t

,tudaytac6:

.V3

Xethamso f ( t ) = 1 0 t - t 2 + A _ i vai t e

X

X

Cau 5: Gpi I la trung diem cua A B ta c6 M I // SA.
=>MI1(ABCD)=^MI1(NBCD).

IS

.

va £'(t) = 2^^

1^

1

t^ t
L i p bang bien thien suy ra f (t) > f ( l ) = 12
11. P H A N RIENG Thi sinh chi dvtgc chpn lam mpt trong hai phan (phan A
ho?c B)
A. Theo chUcrng trinh chuan

e2

Khi do: I , = — I n x

^

Cau 7.a: Do A qua O, nen c6 phuang trinh dang: x = 0 ho^c y = kx
Neu phuang trinh A: x = 0, khi do A = A n d i : x - y - 2 = 0=> A ( 0 ; - 2 )

V^NBCD =|-MI-SNBCD

A n d j :2x + y - 5 = 0=>B(0;5)=>OA.OB = 10 (thoaman)

Va goc giira M N voi mp day chinh

Neu phuang trinh A: y = kx

la goc M M = 30°

Ta

Do A = A n d j nen tpa dp cua A la nghi^m ciia h$ phuang trinh:
= ^

s/

M N

up

Tam giac M N l c6 cos 30°

=>SfjBCD = S A B C Q - S ^ j N =18a

~~^ =

~ ^

VN.MBCD = 3-MI-SNBCD = 3 - ^ ^ ^ - " ^ =

fa b c^ fa c b
Cau 6: f(a) = 5 —+ —+ - — - + —+ lb
c
a; [c b
a)

2k

1-k'l-k

y = kx

5

x=

2+k
5k
y=
2+k

>B

5k ^
2+k'2+k

OOA'.OB'-lOOo

4 +4k'

25 + 25k^

( 1 - k ) ' - (2 + k ) '

ww

l./^63a2

2x + y - 5 = 0

Khi do: O A . O B = 10

Mat khac: sin30° =
o M I = MN.sin30° = 2 a 7 3 . i = aVs
^
MN
2
^Ayfic

2

Do B = A n d j nen tpa dp ciia B la nghifm cua h^ phuang trinh:

om

/g

9a' _ 63a^

fa

2

9a^
^

ce

1 x X
= 18a' va S ^ N , = ^ A N . A I • ^ . ^ . ^

ok

.c

o x = 3aN^

2
x=i-k
2k
y =
1-k

w.

Khi do: SABCD =

[ y = kx

bo

Vi NI^=AN^+AI^<:>9a^= — + —
4
4

x-y-2=0

ro

N I = MN.cos30° = 2aS.— = 3a
2
GQI X la canh cua hinh vuong A B C D

k'+lj

=(k' + k-2)

63a3N/3

o

k'+l = k2+k-2
k'+l =- k ' - k +2

= 100

k=3
k =-l,k = ^

~[2~
Luang trinh cua duong thMng A la y = 3x, y = - x , y = j x
•(a):

a'bc

^u8.a:Gpi P(xp;yp) d o i x u n g v a i M ( 4 ; 6 ) qua I nen

V i a , b , c e [ l ; 3 ] nen 5 c ^ 5 > 3 > b va f'(a) = O o x = Vbc
11 thupc (d) nen
156

DVVH Khattg Vift

T u bang bien thien suy ra f (a) > f(Sc] = 10 /- - - + 5 - - 2, v
/
v b b
c
vc

1
du = —dx

dv = — d x
e2

, Nf^iiyctt

1
e

u = Inx
I,/ji^x.Dat

Khunh

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Ttiyen

4 + Xp =2xj
6 + yp=2yi

_ f f c Z p ] + 6 = 0 o 3xp - 5yp - 6 = 0

(l)
157


Tuye'tt chpn & Gi&i thiC'u dethi Toan hqc - Nguyen fnu A / i . i i m

vynyen x » > ^ r . » .

Lai CO P M I P N <=>'PMPN = 0<::>(xp-4)(xp-6) + ( y p - 6 ) ( y p - 2 ) = 0
,
T u (l) va (2), suy ra: 34yp^ - 162yp +180 = 0 « yp = 3 ho|c

1 + 3i + 3i^ + i ^

^

=

(2)

vol vecto u ' = (l;0;l) => A.: y = - l
z=4+t

30

Cau 9.b: Neu x = 0, t u phuong trmh thii nhat suy ra y = 0. Khi do khong thoa
phuong trinh thu hai.

'

Neu X 5* 0, chia ca 2 ve phuong trinh dau cho x^, ta dugc:

= l i ± ^ . Itill^llllil
=-14 + 2i
•2
1-i

Xet ham so f (t) = t^ + 3t, t €

B. Theo chuorng trinh nang cao

I

+

a + 2b)

om

) = 29
^

bo

w.

Cau 8.b: A = d n (P), tpa dg cua A la nghi^m cua h$

z= 3+ t

+4 -u

(**)

u
+4

-1

In 2012-

I

va

^
Vu2+4

Suy ra ham so' g ( u ) dong bien tren R. Mat khac g(0) = 2 nen u = 0 la

•Wghi^m duy nhat ciia (* *) . T u do x = 1 va y

ww

y = -3 + 2t

Vi\/u^+4-u>0

ce

fa

5

x= l-t

=2

Xet ham so g ( u ) = 2012" f V u ^ T I - u = 2 tren

- 2012"

2=i>a = 2,b = 5=>A: 2x + 5y + 7 = 0

b

Vu^+4-u

Taco: g'(u) = 2012" l n 2 0 1 2 f V u ^ - u + 2012"

.c

. 5^ )(a^ +

ab>0
DSng thuc xay ra khi

\

ok

d , - = l = ( 2 | a | . 5|b|) < - j ^ i l '
^ «'+b^^

r—

Ta

V a ^

up

^/a2+b2

s/

a + 2b

ro

(T

/

D a t u = x - l , t a d u g c phuong trinh: 2012"

/g

,

a + 2b

a + 3b

. De thay f (t) la ham so dong bien tren

Thay vao phuong trinh thu hai, ta dugc: 2012 x - l J ( x - l ) ^ 4 - ( x - l ) = 2

CO phuong trinh: a(x +1) + b(y +1) = 0
d(CA) =

M

^ '

Do do t u (*) ta dugc — = x hay 2y = x*^.

Cau 7.b: Gia su A di qua diem A va c6 vecto phap tuyen la n = (a;b) ;t 0, nen

Gpi d = d(B,A) + d(C,A) =

+ 3 . ^ = x3+3x
X

= ^(-14)'+22==10V2

a + 3b
d(B,A) = - ^ = = ,
Va^+b

>3

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1-i

2y

A{0;-1;4)

^•

Vay, hf phuong trinh c6 nghi^m duy nhat (x; y) =

2x + y - 2 z + 9 = G
Taco: V T C P c u a d l a : u ^ = ( - l ; 2 ; l ) , V T P T cua (P) la: Hp = ( 2 ; l ; - 2 )
Vi

158

Aid
Ac(P)'

; nen VTCP cua A la: u ^ =

Ud;np

=(-5;0;-5) cung phuong
159


Cty TNHH MTV DWH

di qua A(-3; 0; 2) va cit (A) t^i B sao cho m | t cau tam B tie'p xiic voi hai m a
phSng (Oxz) va (P).

DCTHITHUfSd24

Cau 9.a: Goi

bon nghi^m ciia phuang trinh z'*-2:'-2zi^+6z-4s0
1 1 1 1
tren tap so phuc tinh tong: S = — + — + — + — .

I. PH/VN C H U N G C H O T A T CA C A C T H I S I N H

Cau 1: Cho ham so y = -x^ + 2x^ -1 c6 do thi la (C).

z,,Z2,Z3,Z4

la

Zj

a) Khao sat sv bien thien va ve do thi (C) cua ham so.

^2

Zg

Z4

B. Theo chi/Ong trinh nang cao

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b) Tim diem M nSm tren tryc hoanh sao cho tu do c6 the ke dug^c ba tie'p

Cau 7.b: Trong mat phSng voi h? tpa d p Oxy, cho hinh thang can ABCD c6 dien
tich bang 18, day Ion CD nam tren duong thang c6 phuong trinh: x - y + 2 = 0
Biet hai duong cheo AC, BD vuong goc voi nhau va cat nhau tai diem l(3;l). Hay
V i e t phuong trinh duong thJing BC biet diem C c6 hoanh d p am.
Cau 8.b: Trong khong gian voi h? tpa d p Oxyz, cho mat phJing (P): x + 2y-z + 5 = 0

tuyen den do thi (C).
2
^ .2
2cos2x + 4
Cau 2: Giai phuang trmh: tan^ x + 9cot^ x +
= 14
sin2x
Cau 3: Giai h^ phuang trinh:

Khattg Viet

x3(4y2+l) + 2(x2+l)>/^

=6
x^+ll

2 + 2^4y^ + l l = x + Vx^+l

va duong thSng d : ^

= Z l l = .£Z^. Goi d ' la hinh chieu vuong goc ciia d len

mat phang (P) v a E la giao diem cua d va (P). Tim tpa d p F thupc (P) s a o
2.(x + 2sinx-3)cosx
Cau 4: Tinh tich phan: 1^^
^—
dx
,
sin"^ X

Cau 9.b: Viet s o phuc sau duoi dang luong giac: z = —^-

'( . n
. .
Snf
sm — i.sm —
I
3
6 )

2>/aVb^T?7l

(a + l)(b + l)(c-Hl)

II. P H A N R I E N G

ok

1

bo

1

Thi sinh chi dugc chpn lam mpt trong hai phan (phan A

ce

P=

.c

om

/g

ro

up

Cau 5: Cho khoi chop S.ABCD c6 day la hinh thang can, day Ion AB b^ng bon
Ian day nho CD, chieu cao cua day bang a. Bon duong cao cua bon mat ben
ung voi dinh S c6 dp dai bSng nhau va bang b. Tinh the tich cua khoi chop

theo a, b.
Cau 6: Cho a,b,c la cac so thvrc duong. Tim gia tri Ion nhat cua bieu thu-c:

s/

Ta

cho EF vuong goc voi d ' va EF = 5N/3 .

w.

fa

hole B)

ww

A. Theo chUorng trinh chuan

Cau 7.a: Trong mat phSng voi h? tpa dp Oxy, cho duong tron (C)
x2 + yz - 2x + 4y - 20 = 0 va duong thang (d): 3x + 4y - 20 = 0 . Chung minh d
tie'p xiic voi (C). Tam giac ABC c6 dinh A thupc (C), cac dinh B va C thupc d
trung diem canh AB thupc (C). Tim tpa dp cac dinh A, B, C biet tryc tam cu.^
tam giac ABC trung voi tam cua duong tron (C) va diem B c6 hoanh dp duong
Cau 8.a: Trong mat phSng tpa dp Oxyz, cho mat phing (P): 2x - y + 2x + 6 = 0
va duong thSng (A) :
160

=

=

. Viet phuang trinh duong t h k g (d)

Hl/dNG DAN GIAI
I. P H A N C H U N G C H O T A T CA C A C T H I S I N H

Cau 1:
a) Danh cho b^n dpc.
b) M(m;0)6Oxvadu6ngth5ngdquaM: y = k(x-a)
Gia su d tie'p xiic voi (C) tai diem c6 hoanh dp XQ khi h?:
-x^+2x^-1 = k ( x „ - m )
,

k =-4x3+4X0

(1)
(2)

CO nghiem x^ .
°

Thay ( 2 ) vao ( l ) , tu do ta c6: x^ - 1 hoac 3x^ - 4mxo + 1 = 0 c6 nghi?m Xg
Qua M ke dupe 3 tie'p tuyen den d o thj (C) thi phai ton tai 3 gia tri k phan
bift. De y: x = 0, x = ± 1 thi k = 0 nen c6 1 tie'p tuyen.
^ , 2
7
2cos2x
4
Cau 2: tan'^ x + 9cot x +

+ —-— = 14
sin2x
sin2x
tan^ x + 9cot'^ x + cotx - tan x + 2(tan x + cot x) = 14
<=>(tanx + 3cotx)^ + tanx + 3cotx-20 = 0

161


Tuye'n chqn b Giai thifu

dethi

Todn HQC - Nguyen

PM Khdnh , Nguyen

Cau 3: Dieu ki#n: x > 0
Nh^n thay, (O; y) khong la nghi?m ciia

Tat

Thu.

thi

H M = H N = HE = - la ban kinh
2
duong tron va SE = SM = SN = b

phuong trinh

Xet X > 0, phuong trinh thu 2 tro thanh:
.SH = | V 4 b 2 - a 2

b>^
2)

I

2y + 2 y V 4 y 2 + l = ^ + ^ J - ^ + l (*)

D a t C N = x thi BM = 4x,CE = x, BE = 4x
•>0 V e

vr
do do ham so dong bien nen f (2y) = f

^1^

Tam giac HBC vuong 6 H

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Xet ham so f(t) = t + tN/t^Tl . Ta c6: ('{t) = l + ^J^Tl+~
+1

nen H E 2 = E B . E C C ^ — = 4 x 2 < » x = - ,
4
4

o2y = X

^

Thay vao phuong trinh (*): x^ + x + 2(x^ + l)>/x = 6
Ve trai cua phuong trinh la ham dong bien tren (0;+oo) nen c6 nghi^m duy
nhat X = 1 va he phuong trinh c6 nghi?m

V^y,

'4^

1
2

U

Ta

-dx

2 ^•' cSin'
m' X

2 sin^ X t

4

Ll

1 ,
—cotx
1
2; 2

2

4

(dvtt)

'—^

+

2

(a + b + c + l f
>A
L

4

2 "

1

1

w.

ww

Vay I = I i + l 2 = 2 N / 2 - 3 .

doan M N vai M , N Ian lugt la trung diem cac canh AB, CD va M N = a

27

27

Xet ham so': f (t) =

• (..2f

1
Ta c6: f ' ( t ) = — - +

81

= ^ hay

J

d^t

vol t > 1

va f'(t) = 0, t > 1 <=> t = 4

T u day, ta c6 f ( t ) < f | j

C a u 5: Goi H la chan duong cao cua chop thi H phai each deu cac c^nh cua day
Suy ra hinh thang can ABCD c6 duong tron npi tiep tam H la trung diem

(a + b + c + 3)^

Dat t = a + b + c +1 nen c6 t > 1. Liic nay, P <

(2sinx-3)cosx
2f2sinx-3 . / . \ / r 7
^
dx= I
d(smx) = 2 V 2 - sin^x
^
2
sm x

va trong truong hg-p nay ta chung minh dug-c H nSm trong day.

27

Luc do, bieu thuc da cho tro thanh: P < •
a + b+c+ 1

fa

It
n

Duong tron do tiep xiic voi BC tai E

+

up
/g

1

.c

4

l^r

ok

4

2

om

1

f
'
vsin^ X .

2 u2
2 .
a'' + b'^ + c^ + 1 >

bo

71

X

N C

27

ce

„ sin-' X

1

D

ys.ABCD=l-^-l^l^^

ro

4
n

^rXCOSX

•

s/

^

4

AB = 2a, suy ra S^BCD = ^

C a u 6: A p dyng bat d5ng thuc trung binh cpng - trung binh nhan.

2.(x + 2 s i n x - 3 ) c o s x ,
^ xcosx_,
2 (2sinx-3)cosx
C a u 4: 1=
dx= f
^dx+
^
dx
sm X
„ sm X

7t

\^

CD = i

"

o

P ^ 1 . Do do gia trj Ion nhat cua P =

8

8

dugc khi a = b = c = 1.

11. PHAN RIENG T h i siifh chi dugic chpn l a m mpt trong h a i phan (phan A
hoac B)

A. Theo chi/orng trinh chuan
C a u 7.a: Duong tron ( C ) c6 tam l ( l ; - 2 ) va ban kinh R = 5
163


Tuye'n chgn &• Gi&i thi$u dethi Todn hgc - Nguyen Phu Khdnh , Nguyen Tat Thu.
3-8-20|

IH = d(l,CD) = 2V2 =^ CI = 4 = V2t2-4t + 10

hoac t = - l = > C ( - l ; l )

= 1 = 5 = R .Suy ra d tiep xiic voi (C)
Gpi H la tiep diem ciia (C) va d. Toa dp H la nghi?m ciia h? phuang trinh

H ( a ; a + 2 ) G ( d ) , IH = (a - 3;a +1), IH 1 CD « a - 3 + a +1 = 0 ^ a = 1

H(l;3) ^ D(3;5) =^ CD = 4^2
(IC): y = 1, A(x;l) 6 IC (x > 3)

SAU^T^=-

<»36 = (x-3|%/2+4N/2J'271+x-3^'

o 3 6 = ( | x - 3 | + 4)^ < ^ | x - 3 | = 2 o x = l (khong thoa ) ho|ic x = 5 =>A(5;l)
AB//d:x-y-4 =0
- ^ B(3;-1) = AB n DI BC: x + 2y - 1 = 0
DI: x = 3
CauB.b: E e ( d ) = * E ( - 3 + 2t;-l + t;3 + t)
E e ( P ) : x + 2 y - z + 5 = 0=>t = l=>E(-l;0;4)
L a y d i e m M ( - 3 ; - l ; 3 ) € ( d ) , ta c6: EF = ME.nj^ = (-1;1;1)

/g

ro

up

s/

Ta

= 1 0 o ( b - 4 f + r2o-3b -2 y = 100

om

4 4 - 3 c , BI=(-11;2)
.AC = c + 2;

ok

AC.BI = 0 o - l l ( c + 2) + 2 ^ i j ^ = 0 o c = 0 C ( 0 ; 5 )

bo

A C l BI

AIAB vuong can
(AB + CD).(IH + IK)

.c

c e d = > C c;-2 0 - 3 c

• AB = X - 3 V2

IK x - 3

x^ = - 2 A(-2;-6)
.yA=2yi-yH

YA =-6
Goi M la trung diem canh AB. Do HA la duong kinh nen HM 1 AM
Tam giac HAB c6 HM vua la trung tuye'n vua la duong cao nen AHAB can
20-3b'l
tai H =^ HB = HA = 2R = 10, B e d B b;-

doi xung ciia H qua I

b = -4
.(b-4)^ + 12-3b^ = 1 0 0 o b ^ - 8 b - 4 8 = 0 o
b = 12
Do X B > 0 ^ B(l2;-4)

lA = |x - 3|

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3x + 4 y - 2 0 = 0
x = 4 .H(4;2)
+ y ^ - 2 x + 4 y - 2 0 = 0 [y = 2
Do I la true tam AABC va IH 1 BC =i> A e IH . Ket hop A e (C) => la diem

r20-3b
-2
HB = 1 0 o ( b - 4 ) ^

t = 3 (khong thoa)

w.

fa

ce

Cau 8.a: B e (A) B(t -1;6 - t;2t - 5), mat cau tam B tiep xuc voi hai mat phSng
(Oxz) va (P) <:^d(B,(p)) = d(B,(0)) «|l2t-12| = 3 | 6 - t

ww

Cau9.a: z ' * - z ^ - 2 z 2 + 6 z - 4 = 0 «(z-l)(z + 2 ) ( z 2 - 2 z + 2) = 0 (l)

[x = -1 -1
Phuang trinh tham so duong thing EF: • y = t => F (-1 -1; t; 4 +1)
z =4+ t
Theo bai toan, ta c6: EF = 5>/3 =:> EF^ = 45 <=> t^ = 25 <=> t = -5 hoac t = 5
Vay, CO hai diem F thoa man la : (-6;6;10) hoac (4;-5;-l)
Cau 9.b:
2011

Khong mat tinh tong quat ta goi 4 nghi?m cua (l) la
Z j = l , Z 2 = - 2 , Z 3 = l + i, Z 4 = l - i .

2011

cos

Thay vabieu thuc S = \z?\ \ \ zl- - l4 + \ —
zl + 4— ^ = 7 (^^i)2 4
B. Theo chUtfng trlnh nang cao
Cau7.b: AICD can tai 1(3; l ) , C(t;t + 2 ) e ( d ) voi t < 0 , IC = \ / 2 t ^ - 4 t + 10,

X

sinl^-isin^
3
6

2V2

-i2on

1

2

2011

2

^ 201l7t~ + i s m ' 201 l7t^
3 ;

N2012

cos V

6y

+ isin V 6 ,

= cos

7t

cos V 3 , + i s m
"3j

+ isin

12011

f 1006711
165


cos
Z = COS

= 2

3016

4i

(

2Q\\n^

'

100671^

.

3 J

COS

Todn h
+ isin

+ isin

f

10057tl

I

3

J

r

I

+ isin

(

I

PhU Khanh

, Nguyen

lai

111 m i > i ^ r m

cua h i n h thang A B C D de C I = 2BI, t a m giac A C B c6 d i $ n tich bang 12, d i e m I

100671^
3

CO hoanh d p d u o n g va d i e m A c6 hoanh d p a m .

,

Cau 8.a: Trong m | t phSng tpa d p O x y z , cho A ( 1 ; 2 ; 2 ) , B ( - 3 ; - 2 ; 2 ) , C ( - 2 ; - 2 ; 1 ) .
Viet p h u o n g t r i n h m a t cau d i qua A , B, C tiep xuc v o i m a t p h i i n g ( O x y ) .

= 2^^^>/2(cos7t + isin7t).

I

3

Cau 9.a: C h o so p h u e z thoa m a n |z| - 2z = - 3 + 6 i .

;

T i n h gia t r j bieu t h u c z + IzP

OETHITHlJfs625

B. Theo chiTtfng trinh nang cao

Cau 7.b: T r o n g m a t p h a n g v o i h§ tpa d p v u o n g goc O x y , cho d u o n g t r o n (C)

I . PHAN CHUNG CHO TAT CA CAC THI SINH
C a u 1: Cho h a m so y =

x 2 + y 2 = 4 v a d u o n g thSng ( d ) :x + y + 4 = 0 . T i m d i e m A thupc ( d ) sao cho

tir A ve dupe 2 tiep tuyen tiep xiie (C) tai M , N thoa m a n di#n tich tam giac A M N

'

bMng 3%/3.

a) K h a o sat su bien thien va ve d o t h j ( C ) ciia h a m so.

Cau 8.b: T r o n g k h o n g gian Oxyz, cho ba d i e m A ( l ; 2; 3), B(3; 0; - 1 ) , C ( l ; - 2 ; 0)

b) T i m m de d u o n g th5ng A : y = x - m ck ( C ) tai h a i d i e m p h a n b i ^ t A , B

va hai d u o n g thang A , : i i z 2 ^ y + 2 ^ z - 3
^
^
2
-1
1 '

Ta

sao cho k h o a n g each tCr A d e n true hoanh gap hai Ian khoang each t u B den

3x--l = -

up

+sin

2

£zi=yzi = .LLl
_ i

2

1

Gpi A la d u o n g t h i n g d i qua A v u o n g goc v o i A j cat A j .
T i m d i e m M thupc A de di?n tich tam giac M B C nho nhat.

/g

ro

3x + -

s/

true t u n g .
C a u 2: Giai p h u o n g t r i n h : sin

i\iiHiig T T ^ I

cheo n a m tren d u o n g t h i i n g A : 2x + y - 3 = 0. Xac d j n h tpa d p cac d i n h con la

201l7iY
3 J.

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"wy^w chgtt & Gim thifu dethi

C a u 9.b: G i a i h? p h u o n g t r i n h :

om

C a u 3: Giai bat p h u o n g t r i n h : 3 V x ^ - 1 < 2x^ + 3x + 1 .

.c

9 y - 9 ( x - y + 3).3''-^-243 = 0

bo

ok

C a u 4: T i n h ti'eh phan: I = f — — — d x
^
J7-2eos2x

ce

C a u 5: C h o h i n h chop S.ABCD c6 d a y la h i n h v u o n g eanh a. S A 1 ( A B C D ) . SC

w.

fa

h p p v o i m a t phSng ( A B C D ) goc 60" . G p i M , N , P Ian lu(?t la t r u n g d i e m ciia

I . PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1:
a) D a n h cho ban dpc.
b) Xet p h u o n g t r i n h hoanh dp giao d i e m :

ww

cac eanh B C C D , A D . T i n h goc giOa S M va N P . T i n h k h o a n g each t u d i e m A

HI/dTNGDANGlAl

x +1

den m a t ph5ng ( S M P ) , the tich h i n h chop S.ABCD .
C a u 6: C h o cac so t h u c k h o n g a m a, b, c thoa m a n a + b + c = 3 . T i m gia t r j Ion
nha't va gia t r j nho nhat cua bieu thiie P = a + \/b + \/c .
I I . PHAN R I E N G T h i sinh chi dugc chpn lam mgt trong hai phan (phan A
hoac B )
A. Theo chUtfng trinh chuan
C a u 7.a: T r o n g m a t p h J n g v o i h? tga d p v u o n g goc O x y , cho h i n h thang

-2x + 2

= X

- m <=> 2x^ - ( 2 m + l ) x + 2 m + l = 0

1
7
m < - - hoac m > - t h i A eat ( C ) tai hai d i e m p h a n bi?t A , B c6 hoanh d p

khacl
Taco: A ( X , ; X , - m ) , B ( X 2 ; X 2 - m ) = > d ( A ; O x ) = X, - m , d ( B ; O y ) =

Theo bai toan, ta c6: X j - m = 2. X j , theo V i - et:

A B C D ( A B // C D ) . Bie't hai d i n h B ( 3 ; 3 ) va C ( 5 ; - 3 ) . Giao d i e m I ciia h a i d u o n g
166

167


2m+1
' "
2
^
2m
+
l
1
<=> m =
m . -

T u do ta dugc

Xj.Xj

• 4

Cau 2: cos^ 3 x - I ^

2

2
sin^ 6x = 0 o x =

Cau 6: v^b +

^
6

Chia hai vecho x^ + x +1, ta dugc : 3

x-1

<^+x + l

x^+x + 1

up
ro
/g

x +x+1

< l < : = > x - l < x ^ + x + l<=>x^>-2 (luon dung)

om

Voi t < 1 tuc

.c

V a i t > 2 tuc | - A - l - > 2 c : > x - l > 4 ( x ^ + x + l)<=>4x2+3x + 5<0 (v6nghi#m)
+X+1

ok

(X

1

1

^2cosx-3

2cosx + 3y

l"r( 1
K h i d o : I = - 1.
6^121-3

Cau 5: Goi K la trung diem cua AB thi M K // NP
Xet tam giac S K M c6:

Cau7.a: I € ( A )

1

SA^

AP2

63^

voi b e [ 0 ; 3 "

khi a = b = 0,c = 3 hoac a = c = 0; b = 3

l(t;3-2t),t > 0

I C = 2 I B o l 5 t 2 + 1 0 t - 2 5 = 0 c ^ t = - - (khong thoa t > 0 )
3
hoac t = 1 => I ( l ; l ) . Phuong trinh duong thang I C : x + y - 2 = 0

ce

.Dat t = cosx=>dt = - s i n x d x

2t-3
1
1
^.t = — I n
2t + 3 /

12
2t + 3

1

A. Theo chi/orng t r m h chuan

fa

=

AH^

=> d [ A , ( S M P ) ] = A H .

11. PHAN RIENG T h i sinh chi dupe chpn lam mpt trong hai phan (phan A
hoac B)

sinx
= -(2cosx-3)(2cosx.3)

w.

-sinx

^

sinx

ww

Cau 4: Tmh t.ch phan:

sinx

minP^Vs

bo

Vay, bat phuong trinh da cho c6 nghi?m x > 1 .

1 (SMP)

< ^ 2 ( b + c) = ^ 2 ( 3 - a )

Xet g(b) = Vb + V 3 ^

3 t < t ^ + 2 < = > t < l hoac t > 2
x-l

=> A H

= SP

bien voi mpi a e [ 0 ; 3 ] va g(a)>g(o) = \/b+ %/c = N/b + V 3 - b

voi t > 0, ta dug-c bat phuong trinh:

1

1 SP

1 (SMP)

Xet g(a) = a + V b + v/c voi a 6 [ 0 ; 3 ] . Ta c6: g'(a) = l > 0 , suy ra g(a) dong

-+2

Ta

+ X +

• (SAP) ± MP => (SAP)

maxP = — k h i a = —, b = c = —
2
2
4

s/

Vx^

x-l

(SM,NP) = SMK = arccos

K h i d o P
3^/x3-l<2x2+3x + l o 3 ^ / ^ . ^ / x 2 + x + l < ( x - l ) + 2(x2+x + l )

r~x-i

AP1 MP

Trong tam giac vuong S A D c6

Cau 3: Dieu kien: x > 1

Dat t =

S A ± M P

=^

Trong ( S A P ) ke A H

= - » cos^ 6x = 1
2

ol--sm

Taco:

15

1

sin^ 9 - 1

, 1 - 2

8

2

4 + sm

c:.l-2cos^ 3 x - -

Nen cosSMK =

-

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2m+ 1
,
2
'

*

1
SABC = ^ A C . d (B; A C ) =:> A C = 672 .

0

Vi A 6 l C = > A ( a ; 2 - a ) nen c6: ( a - 5 ) ^ = 36

a = 11 hoac a = - l

=>A(-1;3)

Phuong trinh duong thang C D : y + 3 = 0
(SM,NP) = (SM,KM)

I Tpa dp D la nghi^m cua hf:

x-y=0
,
V
^
^D(-3;-3)
y+3=0
y ' f

^au 8.a: Gpi I ( a ; b ; c ) , R Ian luot la tam va ban kinh mat cau can tim.

168

169


Ta CO phuong trinh m | t phMng (Oxy) la z = 0. Mat cau Hep xiic vai mat
phang (Oxy) c:>d(l,(Oxy)) = R hay R = c .
a = -l
IA = IB
a=l
b=0
Theo bai toan, ta c6 h?: IA = IC
b = 2 hoSc c = 3
IA = d(l,(Oxy))
c=5
Cau 9.a: Gia sir z = x + yi (x, y e ^ )
r s i

•\

^ J-Jx^ +

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n—7

GpiN la giao diem ciia mat phSng (P) va duang thiing Aj, suy ra N la
x= l-f
.
y = l+2t'
,
;hi?mcuah^:
=> t'=-1 => N(2;-l;-2)
2x-y+z-3=0
Phuong trinh duang thiing A di qua A(1;2;3) va nh^n A N = (l;-3;-5) lam
x=1+m
V T C P la : y = 2 - 3 m
z=3-5m

- 2 x = -3

Tu gia thiet, ta c6: ^x^ + y^ - 2(x - yi) = -3 + 6i<=> | ^
2y-b
+
9
.
(
2
x
3
f
,
x

>
|
^
^
^
4
^
3
j
Vx^+9 = 2x - 3
y =3
y =3
Vay, z + z 2 + z = 5 + 25 + 125 = 155

Vi MeA=i>M(l + m ; 2 - 3 m ; 3 - 5 m ) = > M B = (-m + 2;3m-2;5m-4),
MC = (-m;3m-4;5m-3)
M B ^ M C ^ = 1225m'* - 3850m^ + 4739m2 - 2696m + 600
( M B . M C ) ^ = (35m2

s/
up
ro
/g
om

.c

Cau7.b:Diem A e d ^ A(a;-4-a).Dat MAN = 2a, O A = x > 0
OM 2
AM

. „ 47x^-4
Taco: sma = T ^ = Tr^, cosa = - — =>sin2a =
OA
OA OA
-.Voi SAMN=3>/3 o 4 ( x 2 - 4 '=27x''
x2-4
SAMN

Ta

B. Theo chUorng trinh nang cao

ok

o x ^ =16=>x = 4

ww

w.

fa

ce

bo

VoiOA = 4 o a^ + (4 + a)^ = 4 o a = -4 hoac a = 0
Vay, toa dp diem A can tim A (-4; O) hoac A(0;-4)
x = 2 + 2t
x=l-f'

Cau 8.b: Aj: ] y = -2 - 1 va : y = l + 2 f
z = - l + t'
z =3+t
Gpi mat phSng (P) di qua A(l; 2; 3) va vuong goc vol duang thSng Aj
nen nhan ^ = (2;-l;l) lam VTPT, suy ra phuang trinh m | t phSng (P) la:
2x-y+z-3=0

170

- 55m + 2o)^ = 1225m* - 3850m^ + 4425m2 - 2240m + 400

SMBC = ^ ^ M B ^ . M C ^ - J M B . M C ) ^

= lV314m2-456m+ 200

Xethamso: f(m)=314m^-456m+ 200.
Taco: f(m) = 628m-456 va f'(m) = 0 o m = —
^ '
^ ^
157
Dua vao bang bien thien ta tha'y de min f (m) khi m = 114
157
Cau 9.b: Dieu ki?n: x > l
y >0
Phuong trinh dau tuong duong voi l o g ^ ^ (x -1) = l o g y o y = x - 1
I' Thay vao phuong trinh thu 2 tro thanh: 9^ - 36.3^' - 243 = 0
y = 2=>X = 3
3y =9
3^=27 y = 3 = > X = 4

171