giới thiệu và hướng dẫn giải chi tiết đề thi đại học cao đẳng khối a tóan lý hóa

TRAN TIEN Tl/ - PHAN VAN HU&N - HUYNH VAN UT
DE
THI DAI HOC - CAO DANG
Khoi
A
TOAN - LI - HOA
THJ VlfN Tii^HbiNM THUAN
NHA XUAT BAN DAI HQC QUOC GIA HA NQI
F
AH
r.nhasachkhangvje
)T
THANH VIENDWH
I
-
p. Dakao
-
Quan
1
-
Tp.Ho Chi
/
-
39105797
-
39111969
-
391119d
!yahoo.com.vn
nsviet.vn
16 HANG CHUOl

-
HAI BA
TRUNG
- HA
NOI
Dien thoai: Bien
tap
-
Che ban: (04)
39714896
Hanh chi'nh:
(04)
39714899. Tonq bien
tap: (04)
39714897
Fax:
(04)
39714899
Chju
tract!
nhi$m xuat ban:
Giam
doc -
Jong
bien
tap:
TS.
PHAM TH! TRAM
Bien
tap:

BUI
THE
Che
ban &
trinh
bay bia:
TUONG
VY
Doi
tac
lien
ket
xuat
ban:
Nha
sach SAO MAI
3Q
SACH LIEN
KET
GSdl
THIEU
&
HD
GIAI
CHI
TIET
OE THI OH
-
CO
KHOI

A:
TOAN
- LI -
HdA
Ma
so:
1L-306 DH2012.
;
In 2000 cuon,
kho
If
x 24cm.
Tai CTY
TNHH
MTV
in
Dudng
sat Sai Gon.
DC:
136/1A Tran
Phu
-
Q.5
-
TP.
Hp
Chi
Minh.
So xuat
ban:

1264-2012/CXB/07-204/DHQGHN.
Quyet dinh xuat
ban so:
306LK-XH/XB-NXBDHQGHN
In xong
va
nop luu
chieu
quy
I
nam 2013.
BE
THI
TUYEN
SINH
DAI HQG,CAO DANG NAM
2004
-
2005
M6n: TOAN; Khoi A I
Thdi
glan lam bat: 1§C phut jcc +j(-
Cau
I (2,0 diem) (i-x)l (f-
(J
Cho ham so y =
-x'+3x-3
2(x-l)
'[i)^
= CI

:rinib
DRX
q&T *
1) Khao sat ham so (1). ' n. _/)c
2) Tim m de diromg thang y;= m cat do thi ham so (1) tai hai diem
A, B sao cho AB = 1.
Cau
II (2 diem)
- (O
.JUntsb
nga auni hi I - x
gnlrii
gn6lj(3
1) Giai b^t
phucjrirft-iil?
4(^^)x^^^n^nM
log, (y-x)-log^ = l a-
2)tJiaihe phtfcfng
trihh
1
X
Cau III (3 diem)^
14 trong mat phang ydi hi\a do Oj^dio-fiai diem A(0; 2) va
Tim
toa do
trifc
tarn
va toa do
taiii
di/dng tron ngoai tiep

cua tam
giac
OAB. II
2) Trong khong gian
vdri
he toa dp
Oxyz
cho hinh
chop
S.ABCD
c6
day ABCD la hinh'thoi, AC cat iki goc toa do^9i''Biet A(2; 0; 0),
B(0; 1; 0), S(0; 0; 2^2). Goi M la t^ung diem q^anh SC.
a) Tinh goc va khoang
each
giifa'^^'dtfdng thang SA, BM.
b) Gia suf mat phSng (ABJ^)^
^tJ
yCcfqg
th&ng SD tai diem N. Tinh
the tich khoi
chop
SrABM!st:^^^^4
' ' ' '
Cau IV (2 diem)
1) Tinh tiqfer^an I =
1
+
-
VHiT

2) Tim he so cQa trong
khai
Cau V (1 diem)
trivia"
thanh da thufc cQa:
I
-
•
r ijb diiBod
rifirij
gaXi^jrfl
(S
Chq tam
giac
APC khong tuvthoa^man dieu kieh^ yj, +. - <-
cos2A
+
cosB
+ 2 V2
cosC
= 3. Tinh ba goc cusai tan>:giac ABC.
3
Cftu
I.
DAP
AN
1
-x'+3x-3 1 ,
*
Tap x^c dinh: D =

R\{l}.
xf2-x)
*
Str bien thien:
y'
=
-^^
f; y'
=
0<»x
= 0, x = 2.
2(x-l)^
ycD = y(2) = ycT = y(0) = |.
Difdng
th^ng x = 1 1^ tiem can dufng.
Difdng
th^ng y = —^x
+
1
la ti^m can xien.
*
Bang bien thien:
-00
0
0 +
0
+00
+00
3
2

2
*
D6 thi:
+00
-oc
2)
Phufdng
trinh
hoanh dp giao diem cua do thi h^m so v<Ji
difcfng
th^ng y = m la:
-x'+3x-3
2(x-l)
-
=
m<»x'+(2m-3)x
+
3-2m = 0
(*)
Gidi
thi$u
&
huang din gi^i chi tiSt thi BH
khtfi
A
PhiTcrng
trinh
(*) c6
hai nghiem phan bi$t khi
va

chi khi:
3
1
A>0<=>4m^-4m-3>0<»m>—
hoac
m<—
(**)
2
• 2
Vdi
dieu
kidn
(**),
during thing
y = m c^t
do
thi ham
so
tai hai
diem
A,
B c6
ho^nh do
Xi,
X2
la
nghiem cua phiTcfng
trinh
(*).
AB

=
1 «
= lC:> X,-X2 = lo(x,+X2)^-4x,X2
=1
1±V5
(2m-3)
-4(3
-2m) =
1
m =
-~-
(thoa man
(**)).
Cfiu
II.
1) Dieu kien:
x > 4.
0
^2(x^-16)
Ta
c6:
^ \-3>-7=^
Vx^
Vx-3
o^2(x'-16)
+
x-3>7-x
o^2(x^-16)>10-2x
* Neu
x > 5

thi
bat
phiTcfng
trinh
dutfc thoa man, vi v6'
trdi
dtfcfng,
ve phai am.
* Neu
4
<
x
< 5 thi
hai
ve
cua bat
phUcfng
trinh
khong am. Binh
phifcfng
hai
ve
ta
dUcfc: 2(x^
-
16) > (10
-
2x)^
o
x^

-
20x
+ 66 < 0 «
10-^/34
<
x
< 10
+V34.
Ket hcfp
vdri
dieu ki^n
4
<
x
< 5
ta c6:
10-V34
<
x
< 5.
Vayx>10-V34.
2)
Dilu
ki§n:
y
>
x
v&
y
>

0.
log,(y-x)-log4-
=
l
o-log4(y-x)-log4^
=
l
«>-log4
y-x_
=
1<::>X
=
3y
y
4
Thg' v^o phifomg
trinh
x^
+
y^
= 25, ta c6:
[
3y
1
l4
J
+ y'=25<=>y = ±4.
So sdnh vdi dieu
ki#n,
ta

dUdc
y
=
4,
suy
ra
x
=
3
(thoa man
y
>
x).
V4y nghi|m ciia h# phifcnig
trinh
\k
(3; 4).
GMi
thi^u
&
hudng
din
giii
chi ti^t
06
thi
QH
khdi
A 5
UA-131ASUPL|;

UA-LU03-OOq9A@3
L6e-696LU6e-Z6^50L6e-t^(
OHdi-
L
ugnD -
09>teQ
d
-
Sui
AQNJIAHNVHliOt
O
o
z
I GdH IVQ
XVHN
)dLj
pQ ma uo
usAni
!L
6L|
IBQ
OBA
qujs
usAna !U
3
'!>|sdgoe!quna
'[lAI3-|/v
Jf
p
:iBq pj

Luaip
UOLJO ai
H 'qoja 'os
JBQ
Buojq
CMUllAIJMO OH-cd
Cau III. -ifb!
irfv
?.v
kfcrf
ifkf
nfirfq
m^M§n-
ififi
(*)
rinhi
•gnXilufi*!
1), Ducfng thing qua jO^
vuSng
.ggc y^i
rBA^j/I;:^^^^^^^
trinh
,
^
Dod^ ^hlng #a%, vii6ft^-g6c
vidri
0A(();
if^^'^
p^xik^
tMiih

Difcfng thing qua A, vtiSiig goc
ydfi
B0|V3;il)
c0 phucf^g
trinh
>/3x + y-2 = 0.
jTi^^y
Giai he hai (trong ba) phucfng
trinh
tren ta Auac tri^ctam H'^s/5;-lj
Difdng
trung trtfc
canh
OA c6phif9rng,trinh jr=r^^
Difcfng trung trUc
canh
OB c6 phtftfng
trinh
>?3x'+'^j-
2_= 0.
Dirdrng
trung triic
canh
AB c6 phtfcfng
trinh
Giai he hai (trong ba) phirang
trinh
treii ta;
jdir^Cft
Jtanv

f^Ucfng. tron
,gg©aj)ti^tam
gi^c
OA^.!^
l^utj^uh^j
yui-uAq
jBd
ifii
3 < x UBVL *
.^,d2a)Tac6:,Q(-^;P;
0),.0(0;
-1;,0),
M(-1; Q; 4).
SA = ^2; 0;
-2\/2J
' ' '''' " ^^-^
=^'9^^
Jsi 9v
isri
^nturiq
BM =
(H;-l;V2l
, -
; ,SOl-(x-v) .sol
.i.x < V nsm Borfi/S =
36
i
§nt)ljriq
Ofiv
"srfT

i
omib fiKiieijl U9ifj iov rinBS o8
;8)
rinhi
^nialurfq
eri sua
rn$iri§n
^BV
Goi
a la
goc
giffa
SA va BM.
Ta,
^6.j,
^
+
=
'[^x
~
r)^x
+
f
cosa
=
cos
(SA,
BM)
SA.BM
SA

.BM
a
=
30".
SA,
BM"
=(-2V2; 0; -2),
AB
=
(-2;
1; O).
SA,JBM]^AB
^:>2^:
.8
n&fi
nV)t
:6u->
Vay
d(SA, BM)
=
SA,
BM
2b)
Ta
c6: MN
//
AB
//
CD, suy
ra

N
la
trung diem cua
SD.
Suy
ra
N
SA=(2;
0;
-2y^-mA^^Qr^-
()/_iKJ|r_!^-^
Koo ,0 < — nia oCI
2
j£ £
•^^"'^A
''s^''='('()^
^\f^^6y^'
' s"^*^^ -^^^'3
"^^^ ^^^^ "^^^
- V,,3^=iFsA,SMl.SB=i^-
o
llr
• V,^^,=-^||_SA,SMJ.SN
3
ny
VS.ABMN
=
Vs.ABM
+
Vg.AMN

^4-^ 7
CauIV.
.0
>
M tsV
1)
Bht t =
=> X
=
+1
=>dx
=
2tdt.
X
= 1
=>
t =
0, X
^^r^
<f^'=^P
Ta
c6: J
=
=
2
0
1
+
t
J

t
It3_lt2+it''-2rn|t
+
li
'1^
= 3
1
^
=
41n2.
3
0
V
-11
t+1
OP
= 2
Jo

+
2-21n2
3
2
•
UA-UJOD-OOL|BA@3JOIS>
•6LU6e-Z.6/lS0L6e-t'69SL
I
u^no
-
og>|PQ -d

-
SuBOH
AHNVHllONH
i|>|ij3esegu-MM/
2
o
mww.nhasachkhan
H
MOT
THANH
VI^NO
Hoans
-
P-
Dakao
-
Quan
1
-
Tp
h
15694-39105797-39111969-3
(store@yahoo,com.vn
achkhangviet.vn
f
ir
y
-2)
[l
+

x^(l-x)J=C°+C;x^(l-x)
+
C,V(l-x)'-h
+ C^x^ (1
-
X
/
+
C^x* (1
-
X
+
C^x'° (1
-
x)'
+
+Q^x'^
(1
- xf +
C,^x'^
(1
-
x)
V
C^x'^
(1
- x/.
Bllc
cua X
trong

3 so
hang
dau nho hdn 8, bac cua x
trong
4 so
hang cuoi
16n
hcfn
8.
Vay
x®
chi c6
trong
cac so
hang thuT thiif nam,
vdi
he
so' tifcfng
iJng
\k:
Q\.(Z\,
Suy ra:
as = 168 + 70 = 238.
Gfiu
V.
Goi
M =
cos2A
+ 2V2
cosB

+ 2V2
cosC
- 3
=
2cos'A-l
+ 2V2.2cos^i^.cos^^^-3.
2 2
Do
sin—>0,
cos^—^<1 nen
M<2cos^
A
+
4V2sin—-4.
2 2 2
Mac khac tarn
gidc
ABC khong
tu nen
cos A
>
0, cos^
A <
cos
A.
/—
A
Suyra:
M<2cosA
+

4V2siny-4
= 2
=
-4sin^-
+
4V2sin 2
2
2
l-2sin^ —
+
4V2siny-4
=
-2
V|y
M
< 0.
>/2sin^-l
<0.
TPfteo
gid
thid't:
M = 0 o
cos
A =
cos
A
B-C
,
cos
=

1
.
A 1
sm—=
-7=
2
4i
A
=
90°
B
=
45°
C
=
45°
8
Gidi
thifu
&
hadng
din
giii
chi
ti«ft
id
thi
E>H
kh5i
A

OE THI TUYEN SINH DAI HQC, CAO DANG NAM 2005 - 2006
Mon:
TOAN; Khoi A
Thdi gian lam bai: 180 phut
C&u
I (2,0 diem)
Goi
(Cm)
la do thi cua hkm so y = mx + — (*) (m la tham s6').
X
1) Khao sat siT bien thien va ve do thi cua h^m so (*) khi m = —.
4
2) Tim m de ham so (*) c6 CLTC tri va khoang
each
tii diem cUc tieu
cua
(Cm)
den tiem can xien cua
(Cm)
bling -1=.
v2
Cau
II (2,0 diem)
1) Giai bat phLfdng
trinh
VSx-l
-Vx-1 > V2x-4.
2) Giai bat phUcfng
trinh
cos^3xcos2x

-
cos^x
= 0.
cau III (3,0 diem)
1) Trong mat ph^ng v(?i he toa do Oxy cho hai dudng th^ng
di:
X - y = 0 v^ d2: 2x + y - 1 = 0.
Tim
toa do cAc dinh hinh vuong
ABCD
biet rkng dinh A
thuoc
di,
dinh
C thuQC
d2
v^ cac dinh B, D thupc true ho^nh.
2) Trong khong gian v6i he toa do
Oxyz
cho dtfdng th^ng d:
^ = -^ = ^ vk mat ph4ng(P):
2x+y-2z
+ 9 = 0.
a) Tim toa do diem I thuQC d sao cho khoang ckch tii I dS'n mSt
ph^ng (?)
bang
2.
b) Tim toa do
giao
diem A cua difcfng thing d va mat phing (P).

Viet phiftftig
trinh
tham so cua di/cfng thing A nkm trong mat phing
(P),
biet A di qua A va vu6ng g6c
vdti
d.
Cau
IV (2,0 dilm)
1) Tinh tich phan I =
sin2x +
sinx
dx.
J
vT+Scosx
2) Tim so nguyen dtfofng n sao cho:
C^„ -2.2CL, +3.2^CL, -4.2^Cl„.,
+ +(2n
+
l).2^''CJ::l
=
2005
(C|;
la s6' to
hcfp
chap
k cua n phan tur).
Gidi lhi§u
& hu<Jn- Jjn
gi&i

chi
ti«ft
dS thi OH
khtfi
A 9
Y.
Cho X, y, z la so dLfOng tho^
iriitri^-H^I^-
= 4.
Chufng minh r^ng:
1
1
-
+
-
1
2x + y + z X + 2y + z x + y + 2z
DAPAN
CauJ.;;;
. Oh
mBii
BUO
rrb ob '
. 1) Ta cq m,= => y = -x+ ^
* Tap xac dinh: D = MAJO}. ,
* Sif
bien
thien:
, 1 1 x'-4 , ^ ^ ^
fmerb

O.S) II JJAD
. y = y• ^0X =-2, X = 2.
4 - x4^ < f - X - - f - xcv xinhi
grrrrtjrif!
,C) (1
ycD
= y(-2)rf -1^?^ =.?^^oSf^fjo3
riniii
-sntyurfq isd
ieiO
(S
Dirdng thang x = 0 la tiem can durng.
^ ^ J • • ;ra9ii>0,8) III
UG3
|y^ng thing
y,n,^;c,)a,tifPI^9$^^^^^
Jfim
-noiT
(I
*
Bang
bien
thien: = - ^
-
t^
,jb Douj^.! |\b gnfh
GOQA
§n6i(y
riniri
rinib

0E;2ob r.oj
DEMS'
;b
+
.Si! 0>ri
i>L«j
odudj
( ,8 f?fixb a«o
.s^^Jj
oebrij
0
rii^ib
gnfirii
gfioljb-oJio
s^xO 6b
BO
JD = G + s
Bfn
n'^)^,
^
jjf° :.>60
gaBorfjf orb aes'b
louriJ
m'sib
ob^BoJ
inxi
{.«
-00,
m
8v

* Do thi:
.(1) gnBiiq jBm .GV b
gn^rii
snt>'ut4
-5
010
.2 gnBd ("5) gneriq
A rnuib osig 6b Boi miT (d
»"ub Buo 03 mfid)
rinhi
gmoliriq j^^eiV
.^\t)v
aog gnouv KV A m^ib A
laid
0,Si) VI USD
—-r
, = I
fiB^iq
rioiJ
ri(T?iT
(I
:qr ^'hm n
gn^olib
asYtrgn oa miT (2
.0x7 nlf [ n Buo JI qBrio q'ori oi 08 B1 P )
2)
Ta
c6:
y' = m -
-Ij-,

y' = 0 c6
nghiem
khi
va
chi khi
m >ft
"^^^
Neu m-y
b
Ijhry-a
©
&-akj^^^ii^A^Vi3x toh
D A
c/
Xet dau y':
.<0
.
1
•a rn§ib gnuiT
-00
i i
- cii n
+00
0
Diem cUc tieu cuajCChi)
1^ M
Tiem can xien (d):
y = mx «•
ttii-'^
='0.

0
m'-2vm
'
:
XluIGUV
din

vm
+
rod
VGV
d(M,
d)=l<^^^^i^4Ti^M^i,ii^w<i>w^i^^ («2
Vaj?
nip 1.
Cau it~™-i=.((q),I)b
,a 4 K + 8- ;i
1)1 ndn
b 9 I :6o
BT
5x-l>p
1) Dieu ki^n:
x-l>0
«>^>l^^i'-fK-
2x-4>q
Taco:
-
>;
V^)^fk:<=g>
^-1

hV^j^4+,>0;"
(rf
-i
^
5x:^§>
2k-4!+«R4I+2^2x^4)(x^
1)
1
ci> -
A
o;
BT
«
X
+
2
>
J(2x-4)(x-l).oy
+ 4x
+
4
>
2x'
-k
\ "^"^
,
f :v n
novuJ
qisdq
twaa-v

oo
§n6riq
JeM
«
x^ -lOx
<p-<^0<)f
^lp.^,,^^^q
^j^^.^
^^^.^^
trinh
da
cho.
^
2) Ta c6:
COS^SXCOS^JT-cos^x
= 0
o
(1 +
cos6x)cosk^'ii
^io^% ¥
^'"-^^
snwijrfq
tc«v
<=>
cos6xcos2x
1
l:'D^4::>
cos8x
+
cos4x

-2 = 0
<=>
2cos^4x
+
cos4x
- 3 = 0 <::>
cos4x
= 1
hoSc
cos4x
=
—''^(iol^^
2
_
,
X 1118(1
+X800£)f
Vay
cos4x
=
lc:>x
=
k-(kG
7)^r ^=^^~^
2
^
I
6o
BT (I
St1

BO VAN HOA -
THONG
TIN
THir
VIEN
QUOC GIA
VIET
NAM
BANC
UNC
CHO CAC
1
T.2: B/>
HL
C&u III.
1) Vi
A €
di nen A(t; t).
Vi
A va C
doi xufng nhau qua
BD va B, D e Ox
nen C(t; -t).
Vi
C e
da
nen 2t - t - 1 = 0 o t =
1. Vay A(l;
1),
C(l;

-1).
Trung diem
cua
AC 1^ 1(1;
0).
f IB = lA =
1
Vi
I la
tam ciia hinh vuong nen
ID
= IA =
1
[BeOx
DeOx
B(b;
0)
D(d;
0)^
b-1
=1
d-1
=1
b
=
0; b = 2
d
=
0;
d

= 2
Suy
ra B(0; 0) vk
D(2;
0)
hofic
B(2; 0)
v^ D(0;
0).
Vay bon dinh
cua
hinh vuong la: A(l;
1), B(0; 0),
C(l;
-1),
D(2;
0)
hoac:
A(l;
1), B(2; 0),
C(l;
-1),
D(0;
0).
x = l-t
2a) PhUcfng trinh tham
so cua d:
y
=
-3

+ 2t
z = 3 +
t
Ta
c6: led nen
1(1
- t; -3 + 2t; 3 +
t),
d(l,(P))
=
-2t + 2
d(l,(P))
=
2<»|l-t|
=
3o
t
=
4
t
=
-2'
V^y
c6 2
diem
I:
Ii(-3;
5; 7), U3; -7; 1).
b) Vi
A e d n§n

A(l
- t; -3 + 2t; 3 +
t).
Ta
c6 A e (P) <:> 2(1 - t) + (-3 + 2t) - 2(3 + t) + 9 = 0 t = 1.
vay A(0; -1;
4).
Mat ph&ng
(P) c6
vecttf
phdp tuyg'n
n = (2;
1;
-2).
Dufirng
thing
d c6
vector
chi phtfcfng
u =
(-1;
2; 1).
Vi
Ac(P) yk
Aid nfen
A c6
vectcf
chi phirctog =[n, u]=
(5; 0; 5)
'x

= t
V$y phucmg trinh tham
s6' cua A 1^:
y
= -l
z = 4+t.
Cftu
IV.
1)
Ta c6 I = J
(2cosx
+ l)sinx
,
^
•,
=—dx.
V1+3C0SX
12
QMI
thi«u
&
hudng din
gi&i
chi
ti«t
i6
thi
DH
khtfi
A

Dat
t = Vl + 3cosx
cosx = -
dt
= -
3
3sinx
2>/l
+ 3cosx
dx.
71
x
= 0=>t =
2,
x = -=>t = l.
2
I
=
-+1
2^
3.
dt
= -
9
•(2t'+l)dt
fie
)
2
2
fl6

f2
,^
-
—+t
—+
2
—
- +
1
I
3 J
,^9
l3
;
l3
;
34
27
_ 2
~ 9
2)
Ta c6:
(1
+x)^-^
= CL,+CL,x + CL.x^+CL,x^+ +
C^::!x^-'
VxeR.
Dao h^m hai
ve ta c6:
(2n

+
1X1
+
x)^-^
= CL, +2CL,x + 3CL,x^
+ +(2n
+
l)Ct;x^"
VxeR.
Thay
x =
-2,
ta c6:
CL, -2.2CL, +3.2^CL, -4.2-^CL,
+
+
(2n
+
l)2^"C^::;
=
2n
+
l.
Theo
gia
thiet
ta c6: 2n + 1=
2005
=> n =
1002.

Cfiu
V.
\2
Y6i
a,
b>0tac6:
4ab<(a
+ b)'<=>-^<-^»—
a
+
b
4ab a + b 4
a
bj
Da'u
xay ra
khi chi khi
a = b.
Ap dung
ket qua
tr&n
ta c6:
1
2x
+ y + z 4
Tilcfng
tif:
x+2y+z
4
' <l

x
+ y + 2z 4
1
1
1
2x
y + z^
M
1 ^
4
1
1
2y
x + z^
M
1 ^
2x
4
1
1
• + -
(I
1 1
—+•
4L2y
4
—+ -
Vsiy
2z
x + y

1
4
1
1
• +
2z
4
1
1
1
—+ -
X y
{x
2y 2z)
1
1 1
—
+ — + •
y
2z 2x^
1
1 1
- + — + •
{z.
2x 2y;
(1)
(2)
(3)
1
1

1 1
- + —+ -
Vx
y z
1.
2x
+
y4-z
x + 2y + z x + y + 2z 4
Ta thay trong
cAc ba't
ding thuTc
(1), (2), (3)
thi dfiu
"=" xay ra
khi
vk
chi khi
x = y = z.
Vay ding thufc
xay ra
khi
va chi
khi
x = y = z = —.
Gifli thi^u
&
hadng
d£.n
gilii

chi
tifit
06 thi BH
khfil
A
13
DE THI
TUYEN SINH
DAI HQC NAM
2006
-
2007
Mon:
TOAN;
Khoi
A
Thdi gian
lam bai: 180
phut^
-'^
I.
PHAN CHUNG
CHO TAT CA CAC THI
SINH
Cau 1(2
diem)
/
1.
Khao
sat sir bien thien va ve do thi cila ham Sa: ? 0

y
= 2x^ - 9x^ + 12x - 4. , - ,
2. Tim m de phtfjng trinhsau c6 6 nghiem phan! bidir 1 ]
2x'
-9x'+12
= m.
Cau II (2
diem)
1.
Giai
phiTofng
trinh:
M 3 xv '"'-x!;;;p
T
2|[cos''x
+ sin*
x)-sMxcOsx
'
V2-2sinx
~ ' -sT(S
2.
Giai
he phiTOng
trinh:
Cau 111(2
diem)
- e _ v
Trong khong gian vdi he tea do
Qxyz^
cKo/fo'^

lap "pnijfcfng
ABCD.A'B'C'D'
vdi A(0; 0; 0), B(l; of ()), EKO; irO), A'(6; 0;
D.Goi
M va
N
Ian luat la trung dili^xua AB va (3D. ' - nSi :6'j KI i^idi sig odriT
1.
Tinh khoang
each
giiJa
hai diTdng thfing AC va MN. .V us3
: J2. Viet phaorng
trinh^
matiphSng, chufa, A'C va tao vdri mat phang
•i
j' • : / ^
•;r,^'^-
I " \'-
•
ni; ;0:> 'J < d .n loV
Oxy mot goc a biet
cosa
=
-
Cau IV (2
diem)
(I
i
l^ih

tich phSn: I =
, sin2x
„ -s^s" x + 4sin^ X
2i
Cho hai so thtfc x ^ 0, y ^ 0 thay doi va thoa nxan dieukien:
^
V
•
j r - . + y)xy
:F^X?
+-y
i
- ~~ + ~ =i
I
xi x£ 7,8 i;.^ /./t
yrfjii
t|+.xj •/£
xfvC
+ x
Tim gia tri Idn nhfi't cua bieu thCifc; A = —r = .
r
1^
s£ + v + x
II.
PHAN
Ti;
CHQf^
i; '
7/7/ s/V?A7
chQn

cau V.a
ho$c
cau V.b
Cau V.a.
Theo
chiidng
trinh
THPT
khong ph4n b^ (2
(^iemjr
-'^^^
(iijiliiTrong
mat phing vofi, he 193
4<t.Oxjc,,^p^0<^^4tf^|^g;tih|^^;
.p
di:
X + y + 3 = 0,
d2:
X - y - 4 = 0, dg: X - 2y ^i),,, ^ ^'
-^jj
Tim toa do diein M nkm tren dtrcfng thSng ds sao cho khoang
each
tiT
M
den diicfnfT thang di bang hai lari^khoang
each
t^ M deh di^crng
ih^il^^
d2.
i

as 3
2. Tim he so cua so
hang
chdfa
x trong khai
tri^n
nhl thijfc Niutotn
cua f^ + xO , bie'trkng
C^„,,+C^„„+
-11h
ngiiy§n
dxiOng,
. . vH
imrQ
Oft
iH'i-d
orio
fib dnrtl '^'•fix-d'^ BHO
{i' ^f;pn
••'-P.
Cl la so to htfp
chapk
cua'n
phan
tuf). ,
roEri
Cau V.b.
Theo
chifomg
trinh

THPT
phan
ban thi
diem
(2
diem)
1.
Giai phirong
trinh:
3.8" + 4.12" - 18" - 2.27" = 0. ,
2. Cho hinh tru cp cac daj^^la hai hin^h ti^on tarn 0^ ya Q', b^h kinh
day
bang
chieu
cao va
bkng
a. Tren diiofng trpn,day tam O lay
diem
A,
tren diidng tron day tam O' lay
diem
B sao cho AB = 2a^ Tinh the tich
cua khoi tuf
dien
OO'AB.
^ 't
DAP AN
Caul. I
ill
1. * Tap xac dinh: D = M. ;

* Sir
bien
thien: y' = 6(x^ - 3x + 2), y' = 0 « x = 1, x = 2.
Bang
bien
thien:
-00'
r-2-
+C0
+00
-00
ycD = yd) = 1, ycT = y(2) = o.
y
/
I
C)
-4
.11 utO
GiSlhiitrS'hijfii'ng'dan
gi^.l chr
tiet
de thi DH
khoi'A
V5
O OHD 0N(
NVfl
WVM laiA VI9 30(10 N3IA
IlHl
Nil
ONOHJ. - yOH MYA 00

•V,,,'-
2. Phi/ang
trinh
da cho tLforng
dLfOng
vdi:
2|xf-9|xf+12|x|-4 = m-4.
So
nghiem
cua
phiTcfng
trinh
da cho b^ng so giao diem cua do thi
h^m
so y = 2|x|'' -9|x|^
+12|x|-4
v6i
dudng
th^ng
y = m - 4.
Ham
so y = 2|xf -9|x|^
+12|x|-4
\k ham chSn, nen do thi nh^n Oy
l^m
true
doi
xufng.
TCf
do thi cua h^m so da cho suy ra do thi ham so:

y
= 2|xf-9x^+12|x|-4. .
Ay
TCf
do thi suy ra
phifcmg
trinh
da cho c6 6
nghiem
phan
biet khi va
chi
khi:
0<m-4<l o 4<m<5.
CauII.
>/2
1.
Dieu
ki?n:
sin x 9^
(1)
Phiftfng
trinh
da cho
tifchig
difdng
vdi:
2
(sin^
x + cos* x) - sin x cos x = 0

o2
1—sm 2x
4
-—sin2x
= 0
2
16
<»3sin^2x
+ sin2x-4 = 0<=>sin2x = l<=>x = ^ +
kJi
(keZ).
Do
dieu
kien
(1) nen: x =-^ +
2m7i
(meZ).
Gidi
thi$u
& hi/dng din giii chi tidt d€ thi OH khdi A
2.
Dieu kien:
x >
-1,
y > xy > 0.
Dat
t
=
Txy (t > 0).
Til

phiiang
trinh thuf
nhat
cua he suy
ra:
x + y = 3 + t
Binh
phUofng hai
ve
cua
phiiofng
trinh thuf
hai
ta
diTcfc:
x
+ y +
2 + 27xy
+ x + y + l
=16
(2)
Thay
xy =
t^,
x + y = 3 + t
vao
(2) ta
difdc:
3
+

t
+ 2 + 2Vt'+3
+ t
+ l
=16
»2Vt'+t
+
4=ll-t
0<t<ll
0<t<ll
4(t- +1
+ 4)
=
(11 -1)'
^
[31-
+
26t -105
= 0
«t
=
3.
Vdi
t = 3
ta CO
X + y =
6,
xy = 9.
Suy
ra

nghiem
cua he
la
(x;
y) =
(3;
3).
Cau
III.
1. Goi
(P) la
mat
phlng chufa
A'C
va
song song
vdi MN. Khi do:
d(A'C,
MN)
=
d(M,
(P)).
Taco:
C(l;
1;
0),
M
^1
-;0;0
,

N
A'C
=
(1;
1;-1), MN
=
(0; 1;
O)
/
1
-1
-1
1
1
1
\
A'CMN A'CMN
\
1
0
0
0
0
1
/
=
(1;
0;1).
Mat phang
(P)

di qua
diem
A'(0; 0; 1), c6
vectcf phap tuyen n=(l; 0;
1),
CO
phucfng
trinh
la:
l.(x - 0) + 0.(y - 0) + l.(z - l) = 0<i>x + z- l = 0
1
+
0-1
1
Wyd(A'C.MN)
=
d(M,(P))
=
^j,^-^^.
2.
Goi mat
phang
can
tim
la (Q):
ax
+
by
+ cz + d = 0
(a'^

+ +
c^
> 0)
Vi
(Q) di
qua A'(0;
0; 1) va C(l; 1; 0)
nen:
c
+
d
=
0
o
c
= -d =
a
+
b.
3VD
OHD
ONI
a+b+d=0
Do
do,
phtfcfng
trinh
ciia
(Q)
c6

dang:
ax +
by
-f (a +
b)z
- (a + b) = 0.
Mat phang
(Q) c6
vectcf phap tuyen
ii
=
(a; b; a
+
b). mat
phang
Oxy
CO
vectcf phap tuyen
k =
(O;
0;
l).
Vi
goc
giufa
(Q) va
Oxy
la a ma
cosa
=

-l=r
nen
yf€'-
cos
(H,k)
Gi6i
thieu
& hu6ng
din
giai
chi tiet de
tni nH khrii A -
1
17
n\K
J
aiA vio
jonb NaiAjiHi
Nil OMOHJ. - VOH HVA 6a
^a^+b^+(a
+
b)-^
^ V ^
<=> a = -2b
hoac
b = -2a.
Vdi
a = -2b,
chon
b = -1,

dtfcfc
mat
phSng
(Qi):
2x-y + z- l = 0.
Vdi
b = -2a,
chon
a = 1,
dtfoc
mat
phSng
(Q2):
x-2y-z
+ l = 0.
Cau
IV.
-f
sin2x
-f
sin2x
1. Ta co: I = J , ^ -z==:rdx = I . rdx.
0
vcos'x
+
4sin'X
nvl
+
3sin"x
Dat

t = 1 +
Ssin^x
=> dt =
3sin2xdx.
Vdfi
X = 0
thi
t = 1,
vdri
x = -
thi
t = 4.
2
,
1 'fdt 2 r
buy
ra:
1
=
—
—p =
—vt
3
y/i 3
2
3'
2. TCr
gia
thiet
suy ra: —+ —= ^ + ^- —.

X
y X" y* xy
Dat
a = —. b = - ta c6: a + b = a^ +
b^
- ab
X
y
A
= a^ +
b^
= (a + bXa" + b' - ab) = (a + b)'.
TCr
(1) suy ra: a + b = (a +
b)^
- 3ab.
^a
+
b^'
(1)
Vi
ab<
V
2 y
nen
a + b > (a + b) (a +
b)'
=^(a
+
b)'

-4(a + b)<0=>0<a + b<4.
Suy
ra: (a +
b)'
<16.
Vdfi
x = y = ^
thi
A = 16.
Vay
gia
tri Idn nhat
cua A la 16.
Cau
V.a
1.
Vl M e
dg
nen
M(2y;
y).
,
2y +
y
+
3
3y +
3
Taco:
d(M,d,)=

^, \
V2
.
d(M,d,)-
2y-y-4
|y-4
d(M.d,)
=
2d(M.d,)«l^
= 2^-
Vo'i
y = -11
di/tfc diem
Mi(-22;
-11).
Vdri
y = 1
dugc diem
M2(2;
1).
y
=
-ll
y:=l
18
GI6i thIeu
&
hubng
din
giai chi

tiet
de thi DH khoi A
2. Tif gia
thie't
suy ra: C°„,, +C;„,, + + C^„,, = 2''
Vi
CL,=Ct:r\Vk,0<k<2n + l nen:
TCf
khai
trien
nhi thufc
Niuttfn
cua (1 + 1)^"^^ suy ra:
TU
(1), (2) va (3) suy ra: 2^" = 2^° hay n = 10.
(1)
(2)
(3)
Ta c6:
J
10
•
+ x
,nk-40
k=0
k-0
He so cua x^^ la
C\Q
vdri
k thoa man: Ilk - 40 = 26 <=> k - 6.

Vay he so cua x^*^ la: Cj; = 210.
Cau
V.b
1. PhiTong
trinh
da cho tifOng di/otng
vdti:
v3y
Dat
t =
+
4 -2 = 0
(1)
(t
> 0), phifcfng
trinh
(1) trd thanh:
3t^
+ 4t^ - t - 2 = 0 » (t + l)"(3t-2) = 0=>t = - (vi t > 0).
2 fix 2
Vdit
= -thi -
l3j
=
—
hay X = 1.
2. Ke dtfofng sinh AA'. Goi D la diem doi xufng vdi A' qua O' va H la
hinh
chieu cua B
tren

difcfng
thing
A'D.
Do BH 1 A'D va BH 1 AA
nen BH 1 (AOO'A').
Suy ra: VOOAB = ^.BH.SAOO
Ta c6: A'B = A/AB" - A'A" = aj3
BD
=
VA'D'-A'B'
=a
=^
ABO'D deu BH=^.
Vi
AOO' la
tarn
giac
vuong can canh ben
bSng
a nen:
SAOQ'
= -^a^
Vay the
tich
khoi til dien OO'AB la: V = = (dvtt).
3 2 2 12
Gidi
thiSu
&
hadng

dSn
gial
chi
tiet
de thi DH
khoi
A
19
3VD OHO Omi
mi
Nil
ONOHJ. - yOH NYA 08
DE THI TUYEN SINH DAI HQC NAM 2007 - 2008
M6n:
TOAN; Khoi A
Thdi gian lam bai: 180 phut
I. PHAN CHUNG CHO TAT CA CAC THl' SINH
Cau
I (2
diem)
X"+2(m
+
l)x
+
m'+4m
Cho
ham so y = ^ (1), m la
tham
so.
x

+ 2
1.
Khao
sat
sii bien
thien
va ve do
thi
cua
ham
so (1)
khi
m =
-1.
2. Tim
m de ham so (1) c6
cifc
dai va
cxic
tieu,
dong
thc/i
cac
diem cxic
tri
cua do
thi cung vdi
goc toa dp O tao
thanh
mot

tam
giac
vuong tai
O.
Cfiu II (2
diem)
1.
Giai
phtfOng
trinh:
(1 +
sin^x)cosx
+ (1 +
cos^x)sinx
= 1 +
sin2x.
2. Tim
m de
phiTOng
trinh
sau c6
nghiem
thiic:
sVx^
+
mVx
+ l
=2Vx'-l.
Cau
III (2

diem)
Trong
khong gian
vori
he toa do
Oxyz,
cho
hai
dLforng
thang:
'x
=
-l
+
2t
x
y-1 z+2
d,:
- =
^^
= wk d,:
\
'2-1
1 - ^
y
= l + t
z = 3
1.
Chiing
minh

rkng
di va
d2 cheo
nhau.
2. Viet
phiTcfng
trinh
ducfng
thing
d
vuong
goc
vdi
mat
ph^ng
(P):
7x
+ y - 4z = 0 va c&t
hai
difcfng
thing
di, d2.
Cau
rv (2
diem)
1.
Tinh
dien
tich
hinh

phSng
gidri
han
bdi cac
difdng:
y = (e + l)x,
y
= (1
+.e'')x.
2.
Cho x, y, z la cac so
thifc
diiomg
thay doi
va
thoa man
dieu
kien
xyz
= 1.
Tim gia tri
nho
nhat
cua
bieu
thufc:
x-(y
+ z) ^ y-(2 + x) ^ z-(x + y)
y-7y+2zVz
zVz +

2xVx xy/x+ly^jy
II.
PHAN TL/CHQN
Thf sinh
chi
dU(?c chgn
lam can V.a
hoac
cau V.b
Cau
V.a.
Theo
chifong
trinh
THPT khong phan
ban (2
diem)
1.
Trong
mat
phang
vdi he toa do Oxy, cho tam
giac
ABC c6
A(0;
2), B(-2; -2) va C(4; -2). Goi H la
chan difdng
cao ke
tCr
B; M va

N
Ian
luat
la
trung
diem
cua cac
canh
AB va BC.
Viet phi/cfng
trinh
difdng
tron
di qua cac
diem
H, M, N.
20
Gidi thigu
&
hKcing
dan
giai
chi
tigt <i4
thi DH
khdi
A
2. Ch,'rrg minh
rkng:
+\ci +

+
:^CL"-'
2 4 6 2n 2n + l
(n
la so nguyen di/dng, Cj; la so to
hap
chap
k cua n phan
tijf).
Cau
V.b.
Theo
chifdng
trinh
THPT phan ban thi diem (2 diem)
1.
Giai bat
phLfomg
trinh:
2log,
(4x-3)
+log, (2x + 3)<2.
3
2. Cho
hinh
chop
S.ABCD
c6 day Ik
hinh
vuong canh a, mat ben

SAD la
tarn
giac
deu va nkm trong mSt phang vuong goc vdi day. Goi
M,
N, P Ian
liTcrt
la
trung
diem cua cac canh SB, BC, CD. Chufng minh
AM
vuong goc
vdri
BP va
tinh
the
tich
cua khoi tuf dien CMNP.
DAP
AN
Cau
I.
x + 2
X 3 1
1.
Khim
= -],tac6 y = = x-2 +
—
x + 2
*Tapxacdinh: D =

R\|-2}.
1
* Su bien thien: y' =
1
-
x^ +4x + 3
(x + 2)^ (x + 2)^
; y' = 0«
x = -3
x =
-l
Bang
X
bien thien:
-00 -3 -2 -1 +00
y'
+
0 -
- 0 +
y
-6
-00
''^^^^^^
-00
+00 +00
ycD = y(-3) = -6; ycr = y(-l) = -2.
* Tiem can: Tiem can dufng x = -2,
ti§m
can xien y = x - 2.
* D6 thi:

^
3
ZL
DVD OHD
ONrtt
iNva
Gidi
thi#u &
hi/dng
dSn
giii
chi tigt 66 thi DH khfii A
21
wvM
laiA
vio joal) MaiAjiHi
Nil
DNOHl-VOH MYA09
2.
Ta
c6:
y'=
X" +
4x
+
4-m"
Ham
so (1) c6 cUc dai va
ciic
tieu

g(x) = + 4x + 4 -
m^
c6 2
A'
=
4-4
+
m'
>0
g(-2)
=
4-8
+
4-m-
^0
A(-2
-
m; -2),
B(-2 +
m; 4m
- 2).
nghiem phan biet
x ;^ -2
Goi
A, B la cac
diem cifc tri
<=>
m 0.
Do
OA =

(-m-2;-2)9^0,
OB =
(m-2;4m-2)^0
Nen
ba
diem
O, A, B tao
thanh
tarn
giac
vuong tai
O
Khido:
OA.OB
=
0«-m 8m
+
8
= 0
<=>m
=
-4±2\/6
(thoa man
m ^ 0).
Vay
gia
tri
m
can tim la: m
=

-4
+
Cau
II.
1.
PhLforng
trinh
da
cho ti/ong diiofng vdi:
(sinx
+
cosxXl
+
sinxcosx)
=
(sinx
+
cosx)^
(sinx
+
cosxXl
-
sinxKl
-
cosx)
= 0.
<=>
x = -•^ +
k7i,
X = ^ +

k.27r,
x =
k2jr
(k e Z).
2.
Dieu kien:
x > 1.
PhiXomg
trinh
da
cho tirang di/orng vol:
-3
J
+
24/
=m
x
+
1
x
+
1
x-1
Dat
t = 4 ,
khi
do (1)
tro" thanh -3t'
+ 2t = m
x

+
1
(1)
(2)
va
X > 1
nen
0 < t < 1.
Vx+1
V x+1
Ham
so
fi:t)
=
-3t^
+
2t,
0 < t < 1 c6
bang bien
thien:
m
0
3
0
1
-1
22
PhUcfng
trinh
da cho c6

nghiem
<=> (2) c6
nghiem
t e [0; 1)
<=>-l<m<
3
Gidi
thieu & hu6ng dan glal chi tiet de thi DH khoi A
CauIIL
1. di qua M(0; 1; -2), c6
vecto' chi phLTo'ng
u, = (2;
-1;
1),
d-i
qua N(-l; 1; 3), c6
vector chi phiio'ng
u, = (2; 1; 0).
U|.U,
= (-1; 2;
4) va MN =
(-1; 0;
5).
Vi
U|.U,
.MN
=
21 yt 0 nen
di
va

do
chec
nhau.
2.
Gia
siif
d cht
di
va
d2
Ian
liiot
tai
A,
B.
Vi A e
di,
B e
d2
nen:
A(2s; 1
- s; -2 +
s), B(-l
+
2t;
1 +
t;
3).
=> AB
=

(2t
-
2s
-
1;
t + s; -s + 5).
(P)
CO
vecto" phap tuyen
li =
(7; 1; -4).
AB
_L (?)
nen
AB
cCing phiTong vori
ri
s
= l
lt
= -2
2t-2s-l
t
+
s -s
+
5
5t + 9s + l=0
<=> = = —• <=>
7

1 -4
[4t + 3s +
5
=
0
A(2; 0; -1), B(-5;
-1;
3).
T^i
. , •. , ,.
X" 2
y
z +1
Phtforng
trinh
cua
d
la:
= - = .
7
1-4
Cau
IV.
1.
PhUong
trinh
hoanh
do
giao
diem ciia hai ducfng da cho la:

(e
+ l)x =
(1
+ e'')x
<=>
(e" - e)x = 0 X = 0
ho.^c
x = 1.
Dien
ticli
ciia
hinh
phSng c^n tim la:
S
=
|xe'
-
ex
I
dx
=
e
xdx
0
xe'dx.
Ta c6:
e
Jxdx =
ex"
e

2'
xeMx
-
xe''
Vay
S = Y - 1
(dvdt).
eMx
=
e-e''
= 1.
2.
Ta
c6: x"
(y + z) >
2xVx.
Ti/ofngtiT, y-(x
+
z)>2y7y;
z" (x + y)
>2zVz.
2y^/y
2zv^
Suyra:
P>—f——p+
r-
—7=
+
—r=
7=.

y^/y
+
2zVz
zvz +
2x\'x
xvx+2y^.y
Dat
a =
xVx+2y,/y,b=
yiy'y+2zVz,
c = zVz +
2xVx".
Khi do:
Xyfx
=
4c + a-2b
r
4a + h-2c
r
4b + c-2a
——.
yVy= : -•
zVz=-
Gii5i
thifeu
&
hudng
din
glai
chi

tiet
de thi DH
khoi
A
ONG TIN
VIET NAM
NLOAI
HOA
HOC TON(
:HO VA
JSCfDUNG
HA NOI
-
2002
Do do P > -
9
_2
~9
c a b
(Do - + - + - =
b
c a
4c + a-2b 4a + b-2c 4b + c-2a
•
+ -
cab
—+
- + -
b
c a

+
a
b c
—+ —+ -
b
c a
-6
>-(4.3 + 3-6) = 2.
fc
a^
—+ -
-I-
-
+
1
—
Vb
cj
U
J
-1>2
5
+
2,|5-l>4-l = 3,
b
V
a
hoac
^ + ^ +
^>33iA^

= 3.
Ttrc^ngta,
^
+
^
+
i>
3).
bcaVbca bca
Dau
"=" xay ra
khi
va chi khi x = y = z = 1.
Vay
gia
tri
nho
nhat
cua P la 2.
CauV.a
1. Ta CO
M(-l;
0),
N(l;
-2), AC = (4; -4). Gia ijf H(x, y). Ta c6:
BHIAC
f4(x + 2)-4(y + 2) = 0
fx=l
;
\.

HGAC [4x + 4(y-2)=^0 [y^l ^ ^
Gia sijf phi/orng
trinh
dtfdng tron
can tim la:
x^
+ y^ + 2ax + 2by + c = 0 (1)
Thay
toa do
ciia
M, N, H vao (1) ta c6 he
dieu kien:
2a-c = l
2a-4b + c = -5 <=>
2a
+ 2b + c = -2
1
a
= —
2
b^i
2
c = -2
Vay phtfcfng
trinh
diTofng
tr6n
can tim la: x^ + y^ - x + y - 2 = 0.
2. Ta c6: (l +
x)'"

= + C^„x
+
+ C^x'"
(1-xf
=c»„-cLx
+

+
c^-
=> (1
+ xf -
(1
- xf =
2(CLX
4-
+
CLx^ +
+ Cfx- ).
^
ui^xf-o-xf^^^
(uxf%(i-xr
J
7
?^7TI4.1^
2-"-l
2n
+
l
(1)
24

2 2(2n + l)
'(cLx
+ C^,x^+C^„x'+ +
crx^"-')dx
GWi
thi(|u
& hadng din giai
chi
\i6t
66
thi
DH
khfil
A
V*
V^"
^
c;^.iL+ct—+ct—+ +ct'.—
2n
2 2n 4 2n ^ 2n
TO
(1)
va
(2)
ta
c6:
ic^,
+^0^
+icL
+ + :^er'

2
4 6 2n
2n
+
l
(2)
Cau
V.b
1.
Dieu kien:
^ > ^- Bat
phifcfng
trinh
da
cho
tUcfng dUofng vdi:
log3
(4x-3)^
2x
+
3
<2»(4x-3)'
<9(2x
+ 3)
«16x'-42x-18<0» <x<3.
8
Ket hcfp dieu kien
ta
di/cfc nghiem
cua bat

phi/orng
trinh
la:
—
<x<3.
4
2.
Goi
H la
trung diem ciia AD.
Do
ASAD
deu nen
SH X
AD.
Do (SAD)
±
(ABCD)
nen
SH 1
(ABCD)
^ SH 1 BP (1)
Xet hinh vuong ABCD
ta c6:
ACDH
=
ABCP
=:> CH 1 BP (2)
TCr
(1)

va
(2)
suy
ra BP 1
(SHC).
Vi
MN
// SC va
AN
//
CH
nen
(AMN)
//
(SHC).
Suy
ra BP ±
(AMN)
BP 1
AlVI,
S
Ke
MK
1
(ABCD),
K €
(ABCD).
Taco:
VcMNP=
^MK.SCNP.

Vi
MK=lsH
=
-^,S,.^p=icN.CP
= ^
nenVcMNP
=
Z 4 Z o
Gidi
thigu
&
hi/dng
din
giai chi
tiet
de
thi
DH khfii
A
[NG
TIN
VIET
NAM
LOAI
HOA
HOC TONC
:HO VA
fsCfDUNG
HA
Ndl

-
2002