Luyện giải đề trước kỳ thi đại học tuyển chọn và giới thiệu đề thi toán học phần 2

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

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.

I

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 .

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

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)
A. Theo chUorng trinh chuan

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

m ,
-2 x + — + 1.

Gia sir ham so c6 eye d^i, eye tieu t^ii cae diem (xi;yi),(x2;y2) •
Vi y'(xj) = 0,y'(x2) = 0 nen phuong trinh duong thing (A)qua hai diem
eye dgi, eye tieu la: y =

r2m_2^

+ ^ + 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 ) :

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
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

X =

^ + k27t

6

—

6

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

D
'

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

Voi cosx = sinx <=> x = — +
4

Voi sin2x = — o
2

c

+ k27t

12

57t
X =

/''>•

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

— + kTl

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

,

—
'12 + kn

• |x + ky + z j

/

A-

\

\

\

\

D'

=0

Cau 3: Dieu ki#n: x > 3
Phuong trinh cho tuong duong: 5 ( 2 V x - 3 - N/2X + I ) = 2X - 1 3
Nhan 2 ve voi bieu thiic lien hg-p va dat thua so chung:
(2X-13)(5-2N/X^-N/2X

2N/)r^

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

1

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

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

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

z:>X =

+

4

Tirdo, ta duoc P > m i n P = c ' ' - c + —

4

13
Vgy, phuong trinh cho c6 nghiem la: x = - y , x = 4
Cau 4:1= f x e M x +

1

0

f

0
1

dx
e^+l

l2=

\—

—

Ta c6: P'(c) = 3c^ - 1 va P'(c) = 0 o c =

-. Dat u = e" + 1

du - dx

-Y = ln

+I2

= l + ln

.Ii=xe''

du = e^dx

u-1 1 +e
2

I = Ii

Xet P(c) = c 3 - c + J voi ce[0;3" .

0 ^ +^

fu = X
at ^ •
= Jxe^dx. Dat
dv = e dx

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

e+1
2
e+1

2e
e+1

Cau 5:

.a.a.— = 0 <=> k = — .
4
5

Cau6:a + b + c = 3=>a = 3 - b - c , k h i d ( S P = 3 - b - c + b^+c^
N/2X +1 = 5

+

I ) = 0 c^x = y hoac

-k

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

^
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 =
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.
;|

/

50

25

25

/

I
V

B. Theo chi/orng trinh nang cao

5C(8;0) hoac

C(2;8)
Cdch2.Tu M d\mg M K 1 B C , ( K € A B )

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.

Ijifc^^/

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

Vi ABCD la hinh vuong nen

jCDln
d(C;AB) = CD

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

25

\

=>

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

Ta

25

-;2

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

—=—
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

BK

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

fb = 5a-10
' ^ [ | a - l | = |4a-10

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

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

l ( i ; 3 ; - 2 ) , ban

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

^..32-4t^
1/

^

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

+

r20-4t^'

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

= 25 o t = 2 hoac

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)

z = -3-2t
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

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
( l - 3 i ) z la sothuc <=>b-3a = 0<=>b = 3a

Tim gia tri Ion nhat cua bieu thiic: Q = — - — + —^— + ———
b + 3c a + 3c bc + ca
II. PHAN RIENG Thi sinh chi duhole B)
A. Theo chUcrng trinh chuan

z - 2 + 5i

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

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

=1

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

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

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

O

diem M , bie't duong thang TjTj di qua diem A ( 1 ; - 1 ) .
Cau S.a: Trong khong gian toa do Oxyz, cho diem M(0; - I ; 2), N ( - l ; I ; 3).
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
I. P H A N C H U N G C H O T A T CA C A C T H I SINH

Cau 1: Cho ham so: y = x'' - 3x + 2 c6 do thj la (C).
a) Khao sat su bien thien va ve do thj (C) cua ham so.
b) Tim toa do diem M thuoc duong thang (d) c6 phuong trinh y = -3x + 2
sao cho t u M ke duoc hai tiep tuyen toi do thi (C) va hai tiep tuyen do vuong
goc voi nhau.

Cau9.a: Chosophuc z =

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,

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

MC toi mat cau (S) va diem D ( l ; 2; 5) thupc mat phSng (ABC)

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

Cau 3: Giai h# phuong trinh:

11

1-i

V2x + 3 + 7 4 y + l = 6

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,

Cau 9.b: Giai he phuong trinh:

log2

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

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

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

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

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

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

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

7i_i_:^in2
2

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

Ta CO

(gt)

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 .

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

3 + V4y + l = 6
t ^ O . T a c o : f ( t ) = 2t + ^ ,

BCISC
BCISH

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

f ' ( t ) > 0 vol

V t > 0 nen

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

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

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

14R

t

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

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

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

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

ln2-l '
t J2

__1

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

c

voi

2

dv = — d t
t^

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

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

Xet i{t) = t^+/t

^

du = - d t
t

goc bang nhau nen

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

V2x +

J2

*

2

•

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

1 o
sinx = —
2

jll^dt.Dat.

.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

X

KhangVi?

^

3a

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

sin

M T V m'X'H

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

Cty TNIIII

Tat Thu.

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

Neu B^O, dat m = —, ta c6: cosa =
^

Taco: f'(S) = i - ^

^

=

V2m2+4m + 5

^
^2(m + l ) ' + 3

a.

^_

^

va f'(S) = 0c^S = V6
a nho nha't khi cosa

J_

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

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

Vi I M > 2 nen M nam ngoai (C), do do qua M ke Avtqc 2 tiep tuyeh toi (C).
Gpi J la trung diem I M nen toa dp diem J

m-1

(

X

m + 1^

2

I

+

r

m-l^
^ "
2

2010

1 + Z+ Z^+z")

2x + y + l = 0

( x - l f . ( y + 2 f =4
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

cua (C) va (T).

y_.

.3

Cau 7,b: Tpa dp dinh A la giao diem cua AB va A D nen A ( x ; y ) la nghifm ciia
x+2y-2=0

m +1

^•4)j'

B. Theo chuorng trinh nang cao

2(m + l f + 8

X -•

=- - -1

Suy ra w = 0

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

Tpa dp T,, Tj thoa man h?:

•2

CO

w = z

. Duong tron (T)

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

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.

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)
IH-IM

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

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

--=>M

0;0;-^

Cau 9,b: Phuong trinh thu nhat
+ Vx^ +4 ^y^ + 4 - y = 2 O X + Vx^ +4 = y + ^y^ +4
ologj
Xet ham so: f (t) = t + Vt^+4 , ta c6:

7t^+4 + i

f(t) = l

t +t

•>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

CtyTNHHMTV

DWIi Khang Vi^t

DETHITHUfSCf23
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
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.
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 )

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.

sinx _ ^
cos X

sin x = 0
sin2x = 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 .

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

-+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
I. PHAN CHUNG CHO TAT CA CAC THI SINH

t = 2 tuc

THI:

C a u 1:

a) D a n h cho ban dpc.
b) P h u o n g t r i n h hoanh d p giao diem:

2x + 3

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

x-2

x+1
x^-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
X

= 2x + m

_

<=> X =

(*)

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

g(2)^0
y
f
^
f

- x+1

5-V37 . „

4

hoac X =

5 + ^y37

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

A

Cdch 2:Dat
^ :

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

Tai hai giao d i e m ke hai tiep tuye'n song song k h i va chi k h i
y ' ( x i ) = y'(x2)<=>Xj + X2 = 4 o m = - 2 .
C a u 2: D i e u ki?n: cos x*0, cos 3 x ^ 0

I

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

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)

5 + N/37

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

THI:

6-m

u=4 ^ > 0 ,

5-N/37

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
Xj ^ X j . Ta CO Xj + X2 =

cos 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 = 0

« 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


Ttiyen

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

IS

X

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

.

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°

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

Tam giac M N l c6 cos 30°

= ^

M N

x-y-2=0
[ y = kx

N I = MN.cos30° = 2aS.— = 3a
2
GQI X la canh cua hinh vuong A B C D
Vi NI^=AN^+AI^<:>9a^= — + —
4
4
Khi do: SABCD =

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

o x = 3aN^

2x + y - 5 = 0
9a^
^

9a' _ 63a^
~~^ =

l./^63a2

~ ^

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

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

c
a; [c b
a)

2

2k

1-k'l-k

y = kx

5

x=

2+k
5k
y=
2+k

Khi do: O A . O B = 10

>B

5k ^
2+k'2+k

OOA'.OB'-lOOo

4 +4k'

25 + 25k^

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

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

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

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

1 x X
= 18a' va S ^ N , = ^ A N . A I • ^ . ^ . ^
2

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

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

2y

1-i

Xet ham so f (t) = t^ + 3t, t €

B. Theo chuorng trinh nang cao

,

(T

I

+

. 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

a + 2b

a + 3b

/

D a t u = x - l , t a d u g c phuong trinh: 2012"
a + 2b

^/a2+b2

Vu^+4-u

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

\

=2

(**)

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

V a ^

a + 2b)

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

r—

) = 29
^

- 2012"

+4 -u

u
+4

-1

In 2012-

ab>0
Vi\/u^+4-u>0

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

DSng thuc xay ra khi
b

5

I

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

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

x= l-t
y = -3 + 2t
z= 3+ t

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

DCTHITHUfSd24
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).
a) Khao sat sv bien thien va ve do thi (C) cua ham so.
b) Tim diem M nSm tren tryc hoanh sao cho tu do c6 the ke dug^c ba tie'p
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:

x3(4y2+l) + 2(x2+l)>/^
=6
x^+ll

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

Khattg Viet

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).
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 = — + — + — + — .
z,,Z2,Z3,Z4

la

Zj

^2

Zg

Z4

B. Theo chi/Ong trinh nang cao

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
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 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:
P=

1

1

2>/aVb^T?7l

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

II. P H A N R I E N G

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

hole B)
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)

cho EF vuong goc voi d ' va EF = 5N/3 .
Cau 9.b: Viet s o phuc sau duoi dang luong giac: z = —^-

'( . n
. .
Snf
sm — i.sm —
I
3
6 )

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

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

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

^1^

Tam giac HBC vuong 6 H

+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

CD = i

"

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

•

D

N C

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^

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

^

+

'—^

2

4

27

1

^rXCOSX

„ sin-' X

71

4

4

U

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

2 "

n

2

+

4

4

1

(dvtt)

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

\^

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

f
'
vsin^ X .

1

X

2

2 sin^ X t

l^r

1

2 ^•' cSin'
m' X
4

1

2

1

4
It
n

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

Suy ra hinh thang can ABCD c6 duong tron npi tiep tam H la trung diem
doan M N vai M , N Ian lugt la trung diem cac canh AB, CD va M N = a
Duong tron do tiep xiic voi BC tai E

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

(a + b + c + 3)^

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

Ll

1 ,
—cotx
1
2; 2

27

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

-dx

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)

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

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)
c e d = > C c;-2 0 - 3 c
A C l BI

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

.AC = c + 2;

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

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

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

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)
[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,

lA = |x - 3|

• AB = X - 3 V2

IK x - 3

doi xung ciia H qua I

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

t = 3 (khong thoa)

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


"wy^w chgtt & Gim thifu dethi

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

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^

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

I

3

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

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

;

OETHITHlJfs625

'

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

3

111 m i > i ^ r m

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

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.

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)

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

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 .
3x + -

+sin

3x--l = -

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

T i m d i e m M thupc A de di?n tich tam giac M B C nho nhat.
C a u 9.b: G i a i h? p h u o n g t r i n h :
9 y - 9 ( x - y + 3).3''-^-243 = 0

C a u 4: T i n h ti'eh phan: I = f — — — d x

^
J7-2eos2x
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

HI/dTNGDANGlAl
I . PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1:

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

a) D a n h cho ban dpc.

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

b) Xet p h u o n g t r i n h hoanh dp giao d i e m :
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+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

Vx^

+ X +

x-l

x-1

<^+x + l

x^+x + 1

Voi t < 1 tuc

x +x+1

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

sinx

^

=

1

1

^2cosx-3

2cosx + 3y

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

sinx

1

1

SA^

AP2

63^

voi b e [ 0 ; 3 "

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

11. PHAN RIENG T h i sinh chi dupe chpn lam mpt trong hai phan (phan A
hoac B)
Cau7.a: I € ( A )

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

2t-3
1
1
^.t = — I n
2t + 3 /
12
2t + 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

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

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

AH^

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

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

Xet g(b) = Vb + V 3 ^

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

-sinx

1 (SMP)

A. Theo chi/orng t r m h chuan

+X+1

Cau 4: Tmh t.ch phan:

=> A H

= SP

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

minP^Vs

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

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

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

• (SAP) ± MP => (SAP)

maxP = — k h i a = —, b = c = —
2
2
4

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

1

(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 =

-

*

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 ^ )
n—7

r s i

•\

^ J-Jx^ +

- 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
B. Theo chUorng trinh nang cao

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
o x ^ =16=>x = 4
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

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
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

- 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


Cty TNHH MTV DWH Khang Vift

Viet phuong trinh duong t h i n g A vuong goc voi d j va cat d j , dg Ian lugt

OETHITH0fs626

tai cac diem A, B sao cho AB = 7 6 .
Cau 9.a: Cho so phiic z thoa man |iz - 3| = |z - 2 - i| va |z + 3i| = |2z + i | .

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

Cau 1: Cho ham so y = - - x ^ + x - -

Tim modun ciia so' phuc z .

c6 do thi la ( C ) .

B. Theo chUorng trinh nang cao

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

Cau 7.b: Trong mat p h i n g Oxy, cho duong tron (C): x^ + y^ - 8 x - 2 y = 0 va

b) Goi M la diem thuQC do thj (C) c6 hoanh do x = 2. Tim cac gia trj ciia m

diem A(9; 6). Viet phuong trinh duong thing qua A cat (C) theo mpt day cung

de tiep tuyen voi (C) tai M song song voi duong thSng d: y=(m^

CO

-4^+^^—

Cau 8.b: Trong mat p h i n g Oxyz, cho hai diem M ( l ; 2; 1), N ( - l ; 0; -1). Viet
phuong trinh mp(P) di qua M , N cat Ox, Oy theo thu t u tai A va B (khac O)

Cau 2: Giai phuong trinh:
cos 1 Ox + 2 cos^ 4x + 6 cos 3x.cos X = cos X + 8 cos X. cos'' X.
Cau 3: Giai h$ phuong trinh:

sao cho

x(3x-7y + l) = - 2 y ( y - l )

.

.

,

,

-

2(1 + i)z2 - 4 ( 2 - i ) z - 5 - 3i = 0 . Tinh \z^f + [zj

T

Hl/dfNG DAN GIAI
I. PHAN C H U N G C H O T A T CA CAC THI SINH

Cau 5: Cho hinh lang try dung A B C . A ' B ' C c6 day A ' B ' C la tarn giac vuong

Cau 1:

tai B'. Gpi K la hinh chieu vuong goc ciia diem A ' len duong t h i n g A C . Biet
goc giixa duong t h i n g A ' K voi mat p h i n g ( C A B ' ) b i n g 30" va A ' B ' = a,
A ' C = aVs. Tinh the tich khoi tu dien KA'BC.
4
3
.
., •

Cau 6: Cho a, b, c la cac so thuc duong thoa man - b > a - c > - b . Tim gia tri

a) Danh cho ban dpc.
b) Tiep tuyen tai M c6 phuong trinh: y = -3x + —
3
Tiep tuyen voi (C) tai M song song voi duong t h i n g d:

V:

m ^ - 4 = -3

12(a-b) 12(b-c) 25(c-a)
Ion nhat ciia bieu thuc: P = —^
+—
+ —\—c

a

b

II. PHAN RIENG Thi sinh chi dugc chpn lam mpt trong hai phan (phan A
hoac B)
Cau 7.a: Trong mat phing Oxy, cho hinh binh hanh ABCD c6 B(1;5) va duong
cao A H CO phuong trinh x + 2y - 2 = 0, voi H thuQC BC, duong phan giac trong
cua goc ACB c6 phuong trinh la x - y - 1 = 0. Tim toa do dinh A , C, D.
Cau 8.a: Trong mat p h i n g Oxyz, cho ba duong thang
J

172

9m+ 5

7^

14 o m = - 1
—

3
3
Cau 2: Bien doi phuong trinh ve dang
cosl0x + l + cos8x=cosx+2|4cos''x-3cos3xjcosx

A. Theo chi/ofng trinh chuan

d, :

BN

Cau 9.b: Goi z, va Z2 la hai nghiem phuc ciia phuong trinh:

^x + 2y + 74x + y = 5

'^r In^x + lnx ,
Cau 4: Tinh tich phan: I = J
i(lnx + x + l )
,

do dai 4^/3.

x+1

-1

=

y-3
1

=

z + 3.
1

-,

J
d, :

2

x-1
I

y
z , .
= - = - v a d o :

1

I

X
y
z+2
-==-r = —:;—

- ^ 1 2 1

<=> 2cos9x.cosx + l=cosx + 2cos9x.cosx<=> cosx = l <=> x = k7t ( k e Z )
Vay, phuong trinh c6 1 hp nghiem
Jiu 3: Dieu Ki?n:

11.

cho viet lai:

x + 2y^0
4x + y >0
x(3x-7y + l)=:-2y(y-l)

(1)

7x + 2 y + 7 4 x + y =5

(2)
17.-^


Tuye'u chyn & Giin thiju tic thi Toan h(fc-Nguyen PMi Khanh , Nguyen Tat 77in.
(1) »

X (3x - 7 y + 1 ) = - 2 y (y - 1 ) »

Cty TNHH MTV DWH Khang Vift
4
3
4
V i —b > a - c > —b hay a < —b + c < b + c nen 0 < 4x < z 2 9 x .
5
5
5

Bx^ _ ( 7 y - 1 ) x + ly'^ - 2 y = 0

c > ( 3 x - y + l ) ( x - 2 y ) = 0 « y = 3x + l (S) hoac x = 2y (4)
T h a y (S) vao (2) ta

AXXQC:

24x

V7X + 2 + N/TX + T = 5 d i e u k i ? n : x >

. V49x^+21x + 2 = 11 - 7x

11
^ - y

n-7x>0

o

24z

Q =

=
x+y

17

,
^ / \x
va xet Q ( y ) =

y+z
z+x
24(z-x)(y2-zx)

Ta c6: Q ' ( y ) =

76

50x
+

175x = 119

T h a y (4) vao (2)tadu(?c: ^
In^x + l n x

Cau 4:

(Inx + x + l f

+J9y = 5 o y = l = > x = 2

(i^i„^)3^
1 + lnx-

Dat

t =

Inx

- + l=>dt =

1 + lnx

x+ y

+ 1

dx

48Vx

50z

Vx-Vz

z+ x

( l + Inx)
1

1

3'2
C a u 5: A ' K H = 30° => A ' K = 2 A ' H ,

AA'2

A ' H ^

a^

A ' K ^

Da,t

2y = c + a - b = > b = z + x
2z = a + b - c

c= x+ y

z+ x

50

t+ 1 t^+i

voi

af(t) = i « i ^ . 5 0
t+ 1
t^+i

« t +

Y"Y«t

= i.Tathay, f
/

Su
Suy ra Q > Q ( V ^ ) > £

12(b + c - a )
12(c + a - b )
25(a + b - c )
Cau 6: Ta c6:
6: 0Q = 4 9 - P = —5^
-+ —
- +
c
a
b
2x=b+c-a
a= y+ z

^ 5Qz _ 48t

V^ + V ^

o48t'*-100t^-104t2-loot+ 48 = 0 0 1 2

2a

a^^/l5

48N/X

(t + l ) ^ ( t ^ + l f

Dvrng d u o n g cao B I ciia tarn giac A B C t h i B I 1 ( C A ' K ) n e n B I la d u o n g cao

Vay, the tich k h o i t u d i ? n B . A ' C K la :

,

va Q ( V ^ ) =

T a c o : f . ( t ) . 4 8 t ^ - l O O t ^ - 1 0 4 t ^ - l O O t + 48

Sa^

A ' K = Ir/lS A A - = a N / l 5 , A C ' = A ' C = 2 a V 5 O A - = a V 5 , O K =
2

cua k h o i chop B . A ' C K va B I =

>0

247
—->24+
::^Q>64
y + z
50x
50 •>40
z+ x

De thay k h i y -> -oo t h i

l n x ( l + lnx)

va Q ' ( y ) = 0 o y = ±Vzx

f
(x + y ) ( y + z)
24x

24z
50x
=
+ ——
x+y
y+z
z+x

^

^1^
v2y

= 56, f

t

v3y

-25

t + l l - 5 0 = Gi

= 57

\

-

= 5 6 hay 4 9 - P > 5 6 o P < - 7 .

I

D a n g t h u c xay ra k h i a = 2c, 3b = 5c

^ A

Vay, m a x ? = - 7 k h i a = 2c,3b = 5c

P H A N R I E N G T h i s i n h chi dugrc chpn lam mpt trong hai phan (phan A
Hoac B)
Theo chUtfng trinh chuan
Cau 7.a: B C d i qua B ( 1 ; 5 ) va v u o n g goc A H nen B C : - 2x + y - 3 = 0
Toa d p C la n g h i ^ m cua h | :

-2x + y - 3 = 0
x-y-l=0

>C(-4;-5)
^
175


Tuyen

chgn b Giai thifu

dethi

Todn hgc - Nguyen

Phu Khdnh

, Nguyen

Ggi A' la diem doi xung B qua duong phan giac (d): x - y - 1 = 0, BA n (d) = K.
Duong thSng KB di qua B va vuong goc (d) nen KB c6 phuong trinh x + y - 6 = 0
Toa dp diem K la nghi^m cua h?:

x+y-6=0
[x-y-l =0

>K

7 5

>A'{6;0)

2'2

Phuang trinh AC: x - 2y - 6 = 0, A = CA' n A H => A(4; -1)
Trung diem l(0;-3) cua AC, dong thoi I la trung diem BD nen D(-1;-11) .
C a u 8.a: dj c6 vecto chi phuong la u i = ( - l ; l ; l ) . Voi
=:> A ( l + a;a;a), B(b;2b;b - 2)
Ta c6:

Aedj/Bedj

l + a - b + 2b-a + b - 2 - a = 0

A B = N/6 ^

[(b-l-af+(2b-af+(b-2-a)^=6

b= ]

hoac

ri.|iz-3| = | z - 2 - i | ^{y

K/.,),/x Vift

=2»b = l
b
Matkhac A M = v / s B N » A M ^ = 3BN^ « ( a - l ) ^ + 4 + 1 = 9 <=>
=l «

a=3
a= -l

x + 3y - 4z - 3 = 0

Voi a = - 1 :r> - = 0 (khong thoa)
c

b = -2

C a u 9.a: z = x + y i (x, y e

nvvil

1 2
1
-+-+-=1
x y z
• (P): - + ^ + - = 1, vi (P) di qua M, N nen ta c6: a b c
1 1
1
a c

Voi a = 3=^c = - | =>(P): f+ ( - |

a = -5

MIV

Caul8.b: Gia sir (P) cat Ox, Oy, Oz Ian lupt tai A(a;0;0), B{0;b;0), C(0;0;c)
Cau

AB = (b - 1 - a;2b - a;b - 2 - a ) .

AB.u[ = 0

fa = l

~CfyTNlni

Tat Thu.

=> iz - 3 = - y - 3 + xi va z - 2 - i = (x - 2) + (y -1)
+ 3f +

+(y-if

={x-2f

x = - 2 y - 1 (l)

V^y, matphSng c a n t i m ( P ) : j^ + 3 y - 4 z - 3

= 0

Cau 9,b: Co A' = 4(2 - i f + 2 ( l + i)(5 + 3i) = 16.

Vay phuang trinh c6 hai nghifm phuc:

z + 3i| = |2z + i| o |x + (y + 3)i| = l2x + (2y + l ) i

, 3

5.

1 1 .

2

< » x 2 + ( y + 3 f = 4 x 2 + ( 2 y + l f c ^ 3 x 2 + 3 y 2 - 2 y - 8 = 0 (2)

+

Z2

=9

Tir ( l ) va (2) suy ra 3(-2y -1)^ + 3y2 - 2 y - 8 = 0

DETHITHUfSd27

,

1
<=> 15y^ + lOy - 5 = 0 o y = - 1 hoac y = 3

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

Vol y = - l = > x = l = > z = l - i
1
5
Voiy = - ^ x = - - ^ z

5 1.
= --.-i=:>

C a u 1: Cho ham so y = x"* - 2mx^ + m c6 do thj (C^)

V26

•s

a) Khao sat su bien thien va ve do thj (Cj) cua ham so'.
B.Theo chuorng trinh nang cao

C a u 7.b: Toa dp tarn duong tron la l(4;l);ban kinh R =

^

Goi A la duong th5ng qua A va cat duong tron tai M, N phuang trinh cua A
CO dang la: y = k(x - 9) + 6.

+1

C O ba diem eye trj t^o thanh mpt tam giac c6 ban kinh vong tron npi tiep Ian
Hon 1.
C a u 2: Giai phuong trinh: sin^ x=cos^ x+cos^ 3x

GQI H la trung diem M N , ta c6: I H =
|4k-l-9k + 6

b) Tim tat ca cac gia trj thuc cua m de do thj ham so (C^ ) y = x'* - 2mx^ + m

R2-

MN

N2

= Vl7-12=>/5=d(l;A)
•au 3: Giai h? phuong trinh:

•k = 2 : ^ y = 2 x - 1 2
1

1

21

J9x+y
V
x

W

]

2x

\

y
>

y -9
= 18
1x2
)


Tuyen

cht?n

& GiaithJQU

aethi

,

Toan

HQC - Nguyen

Phu RhAnh

, Ni;,nf,'„

Tnf

IHU.

VxV''+3xe''+e''+K

Cau 4: Tinh tich phan: I = I
0

dx
xe^+l

Cau 5: Cho tu di^n deu SABC. Gpi (P) la mat phSng di qua duong cao SO ciia
tu di^n; mat ph3ng (P) cat cac mat phSng (SBC), (SCA) va (SAB) Ian lupt theo

r

n y I J V H H MIV UVVH Khang VI

HlTtifNGDANGlAl

I. PHAN CHUNG CHO TAT CA CAC THI SINH
Cau 1:
a) Danh cho ban dpc.
b) m > 0 thi do thj ham so da cho eo 3 cue trj

cac giao tuyen SM, SN, SP. Cac giao tuyen nay Ian lupt tao voi mat phing

A ( 0 ; m ) , B(-Vii^;m-m2),c(>A^;m-m2) =^S.^^^ =

(ABC) cac goc a, p, y. Chung minh: tan^ a + tan^ (3 + tan^ y = 12.

p = vm^ +m + V m

Cau 6: Cho cac so thuc duong a, b, c doi mot khac nhau thoa man 2a < c va
Si

\

L^i'^o r = ^ > l « V m - % m 2 >m^+ m

c

ab + be = 2c^ . Tim gia tri Ion nhat cua bieu thuc: P =
+ •;
+
.
•
a-b b-c c-a
II. PHAN RIENG Thi sinh chi dug^c chpn lam mpt trong hai phan (phan A

-m-l>Q

Cau2: Phuong trinh dupe bien doi duoi dang:

hoac B)

".^V;^,

Il£2i^=ll£2£l^ + cos2 3x

o 2 eos^ 3x + (cos 4x + cOS2x) o 2cos^ 3x + 2 cos 3x.cos X=0

A. Theo chUorng trinh chuan

o (cos 3x + cOS5x) cos 3x=0 o 2 cos2x.cos x.cos 3x=0

Cau 7.a: Trong mat phang tpa do Oxy, cho parabol (P): y = x^ + 2x - 3 . Xet

eos2x = 0

hinh binh hanh ABCD A { - 1 ; - 4 ) , B(2;5) thuoc (P) va tam I cua hinh binh

cos X

hanh thupc cung AB cua (P) sao cho tam giac lAB c6 dif n tich Ion nhat. Hay

o

2x=r-+k7t

cos2x = 0

x=-+k4
2

2

cos3x = 0

3x= —+kn
2
Vay, phuong trinh c6 2 hp nghifm

Cau 8.a: Trong mat phing tpa do Oxyz, cho 3 diem A(l; 2; -1), B(2; 1; I ) ; C(0; I ; 2)
i = ^^-^ = ^-i^. Hay lap

0

cos3x = 0

xac djnh toa do hai diem C, D.

va duong thSng (d) c6 phuong trinh la: (d):

=

Cau 3: Nhan thay.

phuong trinh duong thMng ( A ) di qua true tam cua tam giac ABC, nam trong

Dat:u =

^

2x

9x + ^
,

v

.

Q

= 9xy +

x=-+k6
3

18x2 y2
+ Z_ + 2 .

^

mat phang (ABC) va vuong goc voi duong thang (d).
Cau 9.a: Mpt hpp chiia 5 bi xanh, 7 bi do va 8 bi vang. Lay ngau nhien 8 vien
bi. Tinh xac suat de lay dupe 8 vien bi c6 du ca 3 mau .

Cau 4: I = Ij

I^, trong do: I, = /(xe" + l)dx va
0

B. Theo chUcrng trinh nang cao
Cau 7.b: Trong mat phSng tpa dp Oxy, tam giac ABC can tai A, c6 dinh B va C

Tinhli=

thupc duong thing di: x + y + 1 = 0. Duong cao di qua dinh B la d2: x - 2y - 2 = 0,

,

,

X—

1

=

V—2 z
= - va hai

diemA(l;l;0), B(2;1;1). Viet phuong trinh duong thSng A di qua A, A i d
sao cho khoang each tir B den duong thang A la Ion nha't.
Cau 9.b: Tim so phuc z thoa man z^/2^+l=|z|(2 + 6iz).
178

I= X

ffxe'<+l)dx:Dat|""''
^
• ldv

dv = ie''dx

1 (x + l)e''
Tinh I2 = p
J
dx: Dat t = xe" +:
o xe^+l

diem M(2;1) thupc duong cao di qua dinh C.
Viet phuong trinh cac canh ben cua tam giac ABC.
Cau 8.b: Trong mat phang Oxyz, cho duong thang d :

= f^^^il^x
0 '^e" +1

Cau 5: Gia su- S.ABC la tu dien deu canh a.
Khi do A H =

2

AO =

3

SO =

h
V

9

Dieu phai c h u n g minh: tan^ a + tan^ p + tan^ y = 12
^

SO^

S O ^ SO^

, ,

kez


Cty TNHH MTV DWH

6a^
Vay,thay v i chung minh ( l ) ,
OM^

ON^ ' OP^

SO^

OM^

ON^

Goi goc BMP = m
Xet AMHOco

OM.-en—^

6sinm
H M = OH.cotm = aVScosm
6 sinm
6sinm

2

Cau 6: Gia thiet: 2a < c nen c6 - < - .
c 2
H

Hon nua, theogia thiet ta cungc6: - . — + — = 2<=> — = — - 1
c c c
c
b

a N / I c o s m - 3asinm
=

Vi - < - nen - > - . Dat t = - thi 0 < t < c 2
c 3
c
3

B

6 sinm
> o x * xTx, i r r . a V S c o s m a a\/3cosm+ 3asinm

va BM = H M + HB =
+- =
.
6sinm
2
6sinm

P . - C _

,

a>/3cosm-3asinm 23 N
NM
O
6 sinm.-

>/3cosm-3sinm NO
<=>

.

,

NM

= 1 <=>

o

,

Xet f ( t ) = l
^'

N/Scosm-3sinm

Ap dung djnh h' Menelauyt cho AOHM c6
,

aN/3cosm + 3asinm 23 PM
PO
6 sinm.2

73cosm-3sinm

PO

PM

2 sinm

PM

PO

<=> •

,

V3cosm + 3sinm

2 sinm

P M - O P _ V3cosm + 3 s i n m - 2 s i n m
OP
OM
OP
C:>OP =

180

"

2sinm

c

2t2-t-r l-t"2(l-t)''

— + /
, voi t G
2t + l 6 ( 1 - t )

2t + l^^(Tri)

0;1
4.1

, d o d 6 f ( t ) dong bie'n tren

gjj^

• t l -

27
Vay, gia trj Ion nhat cua P = ~
PHAN RIENG T h i sinh chi duoc chpn lam mpt trong hai phan (phan A

H k o a c B)

Theo chi/orng trlnh chuan
^ K a u 7.a: I thupc cung AB cua (p) sao cho di?n tich lAB ion nha't <=> I xa AB

N/3cosm + sinm
2 sinm
2 sinm

—

3 hay maxP = —
27 , dang
• thiVc
,
Do do, gia tri ion nha't cua ham so dat tai t = —
4
5
xay ra khi ab + be = 2c^
8a = 3b = 4c, ch^ng han ta chpn (a,b,c) = (3,8,6)
2a =c

2 sinm

BM
H PM
PO
BH A
AO

+

2sinm

=

NO

^

Ta c6: f ' ( t ) > 0 voi V t e 0 ; -

=

2sinm
NM
NO
N M + N O \/3cosm-3sinm+ 2sinm

+-

^-^
c c

A p dung djnh h' Menelauyt cho AOHM c6
CM
H N
NM
O
H AO

m

Vay, ( 2 ) dung. Do do c6 dieu phai chiing minh.

sinm

a

^2sm^

a^

(3).

aVScosm

(3cosm - V S s i n m f

IS^sin^m + cos^mj

Mr

.CM = H M - H C =

Op2;

(3cosm + >/3sinm)^

ta chung minh (2).

Khg^w

H K h a t , tuc I la tiep diem cua tiep tuye'n (d)//AB cua (P).
Phuong trinh duong thang A B : y = 3x - 11=> ( d ) : y = 3x+ c

iV3

r/3cosm + sinm 6sinm

3cosm+ ^/3sinm

(d) tiep xuc (P) tai diem I

I

2'~4

•C - 1 ; -

2 )

, D

181


Cty TNHH MTV DWH Khang Vift

Zau 5: Ta c6 AC = BD = 2a. Gpi SO la duong cao va SO = h
Trpn h$ true tga dp Oxyz sao cho:
O(0;0;0), A(a;0;0), B(0;a;0), C(-a;0;0), D(0;-a;0), S(0;0;h)

Cau 6: Khong mat tinh tong quat ta c6 the gia su: x > y > z
Xethamso: f ( x ) =

1

—H

1

Xac djnh tarn va ban kinh m|it cau ngoai tiep hinh chop.
Do hinh chop S.ABCD la t i i giac deu
(S):x^+y^+z^-2zoZ + d = 0

h2-2znh + d = 0

Xetf(y) =

1 1 (l + y . z ) ^
1 + —+
-zj
y

1

'">. Chung minh dup-c: f ( y ) < f ( l ) = 2 + i l ( 2 + z ) '

h^-a^
2h

Mat khac:
SA.SB
cos a = SA.SB

Xet f ( z ) =
a^+h^

1 -cosa

a
•R=
r,Ol
2^cosa(l-cosa)

=

a(2cosa-l)
2^cosa(l-cosa)

Xac dinh tam va ban kinh mgt cau npi tiep hinh chop.
Ta c6: J e OS

> 0, do x^ ^ y z ,

1 1
1 + - + - (i+y+z)^-

d = -a^
Zn =•

x^yz

suyra f ( x ) dongbiehva f ( x ) < f ( l ) = .

'a2+d = 0

A,S€(S):

(x + y + z) .

xyz + ^2x^ - y z j ( y + z)

Taco: f ' ( x ) = (x + y + z)

nen I € OS =:i> l(0;0;Zo)

1

+ —

j ( 0 ; 0 ; r ) , OJ = r

^S.ABCD - g S T p / V g A B C D

I I . 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: ( C ) : (x - 3 ) ^ + (y +1)^ = 4 c6 tam l ( 3 ; - l ) va ban kinh R = 2
Gia su duong thSng qua P c6 vec to phap tuyen
n(a;b)=>d:a(x-l) + b(y-3) = 0

2a\

De d la tiep tuye'n ciia (C) thi khoang each t u tam I den d bang ban kinh:

SxQ = 4 S ^ A B =4.isA.SB.sina = 2(a^ + h ^ ) s i n a
=>STP = S A B C D +SxQ =2(a^ + h ^ ) s i n a + 2a^

a./cosa(l-cosa)

\2

(2 + z)'^ ,chu'ng minh dupe f (z) < 27

aJcosa(l-cosa)

1 + sina-cosa
1 + sina-cosa
Tim a de mat cau ngoai tiep va npi tiep triing nhau.
a^cosa(l - cosa)
I = J o O I = OJo1 + sina-cosa
<::> (sina - cos a)(sin a - cosa +1)

a(2cosa-l)
2^cosa(l-cosa)

<» a = 45°
o sma = cosa
sina-cosa + l > 0

3a - b - a - 3b

a^+b^
•b(4a-3b) = 0:

-2<»

2a-4b

777b^

= 2 o(a-2b)^-a2+b2o4ab-3b2=0

b = 0=>d: a ( x - l ) = 0 3 > d : x - l = 0
b = - i a : ^ d : a ( x - l ) + - a ( y - 3 ) = 0=>d:3x + 4 y - 6 = 0
3
3

T a c 6 : P I = 2V5, PE = PF = V P I ^ - = V 2 0 - 4 = 4 .
Tam giac lEP dong dang voi IHF suy ra: — - — = — = ^

= Vs

I H EH IE
2
IF
2
EP
4
r2
8
= > I H - - ^ = -==,EH = - ^ = - L ^ P H = P I - I H = 2 7 5 - - ^ = - ^
V5
V5
V5 V5
N/S
V5

Vay I = J o a = 45°
242

243


fuyen chVay duang tron c6 tarn C ( 5 ; l ) va c6 ban kinh

Cau 8.a: (S) c6 tarn l ( l ; 2; 3) va ban kinh R = 9
Giasu(P) coVTPT n = (A;B;C), (A^ +

+

> 0)

(P)//BC nen n l B C = (-!;!;4) =^n.BC = 0 o A = B + 4 C ^ n = (B + 4 C ; B ; q
(P) d i q u a A ( 1 3 ; - l ; 0)=^ phuong trinh (P):

M A = 7(2 + 2 t ) 2 + ( l + 2t ) 2 + t 2 = s/gt^ + ] 2 t + 5 = sj{3t +

(B + 4C)x + By + C z - 1 2 B - 5 2 C - 0
B + 4C + 2B + 3 C - 1 2 B - 5 2 C
,
,
,
=^
V(B + 4 C ) 2 + B 2 + C 2

(P) tiep xiic (S) <^ d[I,(P)] = R «

O B 2 - 2 B C - 8 C = 0 O ( B + 2 C ) ( B - 4 C ) = 0C:>B + 2C = 0 ho^c B - 4 C = 0
Voi B + 2C = 0 chgn B = 2 , C - - 1 ,

Voi B - 4 C = 0 chon B = 4,C = 1 ,

1
Zj

Z2

Z3

1

1
1
—
+
—
+
—
Zl
Z2
Z3

Z—
i

MB = V 4 t ^ + ( l + 2 t ) 2 + ( - t + 5)2 = V 9 t 2 - 6 t + 26 = ^/(3t-1)^+25
Trong mp(Oxy) xet cac vecto u =(3t + 2; 1), v =:(-3t + l ; 5)
Ta c6:

u + V

•3S,

|u| = V(3t + 2 ) 2 + 1 , V =v'(-3t + l)2 + 25

Ta luon c6 bat dSng thuc dung:

Vay, ( M A + MB)^i„ =3^5

1
1

+—
Z2 + —
Z3

d?t duoc khi M

x^ - 4x + 3 > 0
^au 9.b: Dieu ki^n:

x^ - 4 x + 35^1

Z1Z2 + Z 2 Z 3 + Z 3 Z 1

x-3>0

ZIZ2Z3

x-3^1

ZiZ2+Z2Z3 + Z3Zi|_

u + V

D5ng thuc chi xay ra khi u va v cung huong <=> - ^ i i ^ = 1 <=> t = - i
^
^
-3t + l 5
2

= 0

Cau 9.a: Ta c6 Z j Z j = Z2Z2 = Z3Z3 = 1 nen
Z1+Z2+Z3

2f+\

3>/5 < V(3t + 2f+\ yji^t +1)2 + 25 hay M A + MB>3>/5

ta duoc phuong trinh (P);-2x + 2 y - z + 28 = 0

ta dugc phuong trinh (P): 8x + 4y + z-100

CauS.b: M 6 ( d ) n r > M ( l + 2t; 3 + 2t;-t)

X >3

T H I : Neu x > 4 thi log4 Vx^ - 4 x + 3 > log41 = 0 va log4(x - 3 ) > log41 = 0.

Z1Z2 + Z 2 Z 3 + Z 3 Z 1

IZ1Z2Z3

Do do bat phuang trinh tuong duang: log4(x - 3) < log4 -/x^ - 4 x + 3

B. Theo chUorng trinh nang cao
x=t
[x = 7 - 2 m
, C thuQC d' cho nen C: \
Cau 7.b: B thu^c d suy ra B : •!
[y = - 5 - t

iy = m
Theo tinh chat trgng tarn: => XQ =

( t - 2 m + 9)

^
m-t-2
^
= 2,yG =
^
= ^

< » x - 3 < V x 2 - 4 x + 3<=>Vx-3< >/x^

(dung V x > 4 )

TH2: Neu 2 + N/2 < x <4 thi log4 Vx^ - 4 x + 3 > log4 1 = 0
va Iog4 (x - 3) < iog4 1 = 0 . Suy ra bat phuang trinh v6 nghi^m.
TH3: Neu 3< x <2 + x/2 thi log4 \lx^-4x + 3 < log^ 1 = 0

^

[ t - 2 m = -3

va l o g 4 ( x - 3 ) < l o g 4 l = 0.

[t = - l = > B ( - l ; - 4 )

Duong thMng ( B G ) qua G(2; 0) c6 vec to chi phuong u = (3; 4 ) , cho nen ( B G )
—

244

= ^ « > 4 x - 3 y - 8 = 0=>d(C;BG) =

Do do bat phuang trinh tuong duong: log4(x - 3) < log4 Vx^ - 4 x + 3
O X - 3 < V X 2 - 4 X + 3 O N / X - 3 < V X ^ ( d i i n g Vx€(2; 2 + V 2 ) )

20-15-8
5

245


t

DETHITHl]fS638
I. PHAN CHUNG CHO TAT CA CAC THI SINH

Cau 9.a: Tim so'phuc z c6 modun bang 1, dong thoi so'phuc w =
modun Ion nhat.
B. Theo chUorng trinh nang cao

+ 2z - 1 c

Cau 7.b: Trong mat p h i n g Oxy, cho ABC npi tiep trong duong tron tam I v

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

A(3; 3). Diem . M ( 3 ; - l ) nam tren duong tron I va thupc cung BC khong chu

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

diem A. Gpi D, E Ian lupt la hinh chieu ciia diem M len cac duong thang BC, AC

b) Tim tpa dp cac diem tren duong t h i n g y = -4 ma t u do c6 the ke den do

Tim tpa dp cac dinh B, C biet r i n g tryc tam tam giac ABC la diem H(3;1)

thj (C) diing hai tiep tuyen.

duang thang DE c6 phuong trinh la x + 2y - 3 = 0 va hoanh dp ciia B nho hem 2

Cau 2: Giai phuong trinh:

Cau 8.b: Trong khong gian voi h ^ tpa dp Oxyz, cho diem A (3;-2;-2) va ma

sin^x + sin^

X

Cau 3: Giai phuong trinh:

+ sin^
log2

— + X

= 273

sin

I

7t
X + —

.cosx

s

p h i n g ( P ) : x - y - z + l = 0. Viet phuong trinh mat p h i n g ( Q ) d i qua A, vuong

goc voi mat p h i n g (?) biet rang mat p h i n g ( Q ) cat hai tryc Oy, Oz Ian lupt

6;

tai diem phan bi^t M va N sao cho O M = O N

x + logg (2x +1) = 2

Cau 9.b: Cho so'phuc z thoa man z = 1 .
cot X - tan x
Cau 4: Tinh tich phan : I = j I sin 2 X cos

Tim gia trj Ion nhat ciia M = z'' - z + 2 .

^x

HMGDANGIAI

8

Cau 5: Cho hinh hop chu nhat ABCD.A'B'C'D' c6 AB = 2, A D = 4, A A ' = 6.

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

Gpi I , J la trung diem AB, C' D ' .

a) Danh cho ban dpc.

Gpi M , N thoa A M = m A D , BN = mBB' (O < m < l ) .

b) Gpi A la diem n i m tren duong t h i n g y = -4 nen A ( a ; - 4 ) .

Tinh khoang each t u A deh(BDA'). Xac djnh ban kinh r cua duong tron

Duong thang A qua A voi hf so'goc k c6 phuong trinh y = k ( x - a) - 4

giao cua mat cau (S) ngoai tiep A B D A ' va ( B D A ' ) .

Duong thang A tiep xiic voi do thj (C) khi va chi khi h? phuong trinh sau
CO nghif m:

Cau 6: Cho 0 < c < b < a < l . Tim gia tr| Ion nhat cua bieu thuc:
P = a2(b-c) + b2(c-b) + c2(l-c)

-x3 + 3 x - 2 = k ( x - a ) - 4

x3-3x-2 = 3(x2-l)(x-a)

I I . PHAN RIENG Thi sinh chi dug^c chpn lam mpt trong hai phan (phan A
hole B)

-3x^ + 3 = k

A. Theo chUcrng trinh chuan

(x + l)r2x2 -(3a + 2)x + 3a + 2 = 0

Cau 7.a: Trong mat phang Oxy, cho hai duong tron: ( C j ) : x ^ + y ^ =13

va

[-3x2 ^3^1^

( C j ) : (x - 6)^ + y^ = 25 cat nhau tai A ( 2 ; 3) .Viet phuong trinh duong t h i n g di

~3x2+3 = k
(1)
(2)

qua A va cat (Cj),(C2) theo hai day cung eo dp dai bang nhau.

I

x= l
Phuong trinh (1) tuong duong voi: ^ g ( , ) ^ 2 x 2 _ ( 3 , , 2 ) x ^ 3 a . 2 = 0

Cau 8.a: Trong khong gian voi h? tpa dp Oxyz, cho A ( 3 ; 5 ; 4 ) , B ( 3 ; l ; 4 ) . H a y

1

Qua A ke dupe hai Hep tuyeh deh (C) khi va chi khi (2) c6 2 gia trj k

tim tpa dp diem C thupc m5t p h i n g ( P ) : x - y - z - l = 0 sao cho tam giac ABC

Wiac nhau, khi do ( l ) c6 diing 2 nghi^m phan bi^t

can tgi C va c6 di?n tich b i n g 2>/T7 .
246

1

Xi,X2,

dong thoi

thoa

= - 3 x i +3, k j = -3x2 + 3 CO 2 gia trj k khac nhau
I

247


MI/CM

-

chou & Gi&i titieu ae

pan noc - nguyen i-nu r
t/ii l

rvgwygn

Truang hap 1:
g(x) phai thoa man c6 mpt nghi^m bang - 1 va nghi§m khac - 1
fg(-^) = o

6a + 6 = 0

hay

Voi x > 2 , t a c 6 : log2 x > log2 2 = 1, log^ (2x + ] ) > logg(2.2 +1) = 1
Voi 0 < X < 2, ta c6: logj x < logj 2 = 1, logg (2x +1) < l o g , (2.2 +1) = 1
cos^ x - s i n ^ X

• a = - 1 kiem tra (2) thay thoa.
Cau 4: I = J -

-

cot x - tan X

" sin 2 X cos 2 x - ^

Truang hap 2:

8

smx.cosx

^ x =

4

g(x) phai thoa man c6 mpt nghifm kep khac - 1
\2
(3a+2) -8(3a + 2) = 0
hay

3a+ 2

4
cot2x
1
<^X = 2^/^t-™*2x
= 2V^fdx
J sin2x(cos2x
CI
+ sin2x)I
1 + cot2x sin^2x

r3(3a+ 2 ) ( a - 2 ) = 0
3a + 2 ^ -2

^-1

Dat t = cot2x

v3
In
1 -cos2x + 1 - cos
[ 3

X
J

t

i

^ — d x =>--di
2x
2

l

't

.

—\— dx.

=

sin 2x

o

Doi can: x = — =>t = l , x = — =:>t = 0
8
4
n

71

—

'

T^"'-

Vay, cac diem can tim la A ( - 1 ; - 4 ) , A (2;-4) hoac A

Cau 2: Ta c6 sin^ x + sin

dt =
sin

a = — hoac a = 2, kiem tra (2) thay thoa.
3

I = 2V2

'

+ sin^ - + X = 2V3sin X + — . C O S X - 3
.
6j

t

^

1+t

= Vij_i^t = V2' 1 - -

2

i l +t

i

t +1

= V 2 ( t - l n | t + l|)^ = 7 2 ( l - l n 2 )
2x +1 + cos

3

sm X + cos X C O S X - -

Cau 5: Chpn he true tpa dp Axyz sao cho:
A ( 0 ; 0 ; 0 ) , B(2;0;0), C(2;4;0), D(0;4;0),

2K

3 - c o s 2 x - 2 c o s - cos2x
^
—
• = 3sinxcosx + VScos^ x - 2
2
<» 3 = 3sin2x + yf^ilcos^ x - 1 ] o >/3sin2x + cos2x = N/3
. n
. , >/3
^ 1 N/3
<» s m 2 x . — + cos2x.- = — « sin 2x + - - sm —
3
2
2
2
6j
2x + i i = i + k 2 7 i
6 3
2x + —= 7t — + k27i
6
3

<=>

x = — + krc
12
(keZ)
X=—+

4

kn

Cau 3: Dieu kien: x > 0
Nhan thay x = 2 la nghiem cua phuong trinh cho v i log2 2 + log5(2.2 + l) = 2

A ' ( 0 ; 0 ; 6 ) , B'(2;0;6), C'(2;4;6), D'(0;4;6)
^ l ( l ; 0 ; 0 ) , J(l;4;6), M(0;4m;0), N(2;0;6m)
1. Tinh khoang each t u A den ( B D A ' ) .
( B D A ' ) : - + ^ + - = l o 6 x + 3y + 2 z - 1 2 = 0
^
^ 2 4 6
^
=>d[A,(BDA')] = y
2. Xac djnh tam K va ban kinh R cua
mat cau (S) ngoai tie'p ABDA'.
(S): x^ + y^ + z^ - 2ax - 2by - 2cz = 0

(A6(S))

Ta chung minh x = 2 la nghif m duy nhat.
That vay, ham so y = logj x,y = logj (2x +1) deu c6 cac co so Ion hon 1 nen
cac ham so do dong bien.
248

4-4a = 0
B, D, A ' e ( S ) :

dx

^ sin2x cos2x.cos - + s i n 2 x . s i n 4
4j
8

[a = l

16-8b = 0 =>ib = 2
36-12c = 0

c=3

dt