Đề thi tuyển sinh vào lớp 10 môn toán Đề thi của các trường chuyên, chọn trên toàn quốc

HA
NGHIA ANH
- NGUYEN THUY
Mtl -
TRAN
KY
TRANH
(Tuyen
chgn
vd
gidi thi$u)
DE
THI TUYEN SINH VAO LdP
10
MON TOAN
OE THI CUA CAC TRl/CfNG CHUYEN, CHQN TREN TOAN QUOC
(Tdi
ban Idn
thiiC
ttC,
c6
8v£a
chUa
bo
sung)
Tm
Vi£N
Tff,'H
EINH THUAN
OWL
f/iiy^e /

A5
NHA
XUAT BAN DAI HQC
QU6'c
GIA
HA
NQI
NH^
XUfiT
Bf^N
DRI
HQC
QUOC
Gifl NQI
16 H^ng Chu6'i - Hai Trang - Hk NQI
Di6n thoai:
Bien
tap-Chg
ban: (04)
39714896:
H^nh
chinh:
(04^ 39714899: Tona bien tip: (04^
39714897
Fax:
(04^39714899
Chiu
trdch
nhifm xuat ban
Gidm doc -

Tong
bi&n
tdp: TS.
PHAM
THI
TRAM
Biin
tdp Idn ddu:
NGUYEN
VAN
TRONG
Biin tdp tdi ban:
NGUYEN THUY
Chi ban: NHA
SACH HONG
AN
Trinh
bay bia:
THAI
VAN
B6i tdc Mn ket
xudt
ban:
NHA SACH HONG
AN
SACH LIEN KET
DE
THI
TUYEN
SINH VAO L6P 10 MON TOAN

Ma s6': 1L-351DH2012
In 1.000
cuon,
klio 16 x 24cm lai
Cong
ty Co phan VSn lioa VSn
Lang.
So
xuat ban: 1557 - 2012/CXB/02 - 253/DHQGHN, ngay 26/12/2012.
Quyet djnh xuat ban so: 355LK-TN/QD - NXBDHQGHN.
In xong va nop lUu
chieu
quy I nam 2013.
LCFl
NOI DAU
Cac em hpc sinh than men!
HQC
la qua trinh ren luyen vat va nhat cua ddi ngudi. Dieu gl
vat va mdi c6 dupe thi do chi'nh la dieu minh quy nhat.
Nhim giup cac em chuan bj thi vao Idp 10, giup cac em c6
them tu lieu va c6 mot cai nhin tong quat ve nhQng van de trong
cac de thi tuyen sinh vao Idp 10 trong nhOng nam gan day,
Chung toi bien sogn cuon:
"DE
THI
TUYEN
SINH VAO
LClP
10 MON TOAN".
Sach gom de thi cua cac trudng chuyen, chpn tren toan quoc

tC/
nam hpc 2000 den nay.
Chung toi hi vpng cuon sach se rat hufu ich cho cac em tren
con dudng hpc tap.
Trong qua trinh bien soan kho tranh khoi nhufng thieu sot, rat
mong nhan dUpc
S[jl
gdp y cua bgn dpc gan xa.
Chuc cac em thanh cong trong ki thi sap tdi.
CAC TAC GlA
DE
1
TRl/dNC
PnH CHUYEN H6NG PHONG -
TRXN
DAI
NGHTA
L6P CHUY^N
NGUYiN
THl/ONG
HIEN
- L^P
CHUYEN
GIA
DjNH
De thi tuyen
sinh
vao I6p 10
PTTH
chuyen tqi

TP.HCIVI
nam hoc
2006
-
2007
Cau
1. Thu gon cdc
bilu
thiJc sau :
A = (2V4 + Ve - 2^)(VlO - V2),
B
=
-y/a
- 1 Va + 1
Va
+ 1 Va - 1
a + 1
vdi
a > 0, a
7^
1.
Cau
2. Vdi gid tri nao ciia m thi dudrng thang (d) : y = — x + 2m cit
2
3
Parabol (P) : y = — tai hai
di§m
phan biet ?
Cau
3.

Giai
cac phiTcfng
trinh
va he phuomg
trinh
:
b) a)
Vs - x^ = X - 1
X y
X y
c) V-x^ + 4x - 2 + V-2x2 + 8x - 5 = V2 + A/3.
Cau
4.
a)
Cho hai so' dirong x, y thoa : x + y =
37xy.
Tinh
—.
y
b)
Tim cac so nguyen dufong x, y thoa : — + i = i.
X y 2
Cau
5, Cho tam gidc ABC c6 ba goc nhon (AB < AC), c6 dudng cao AH. Goi
D,
E Ian
lucrt
1^
trung
diem ciia AB va AC.

a)
Chijfng
minh
DE la tiep tuyen chung ciia hai du^ng
tron
ngoai tiep tam
gidc DBH vk ECH.
)
Goi F la giao diem
thuT
nhi cua hai dudng
tron
ngoai tigp tam
gidc
DBH
va ECH. Chilng
minh
HF di qua
trung
diem cua DE.
'
c)
Chiifng
minh
rkng
diibng
tr6n
ngoai tiep tam gidc ADE di qua diem F.
BAIGIAI
caul.

A =
V16 + 4V(V5 - 1?1 (Vio -
A/2)
=
(VI6T4(X^(A^
-
5
A
=
(V2A/6
+
2V5)(A/10
-
V2)
=
(A^VCVS
+
1)^
(VlO
- N^)
=
A/2(A/5
+ -
V2)
=
(VlO
+
V2)(VrO
-
V2)

=8.
B
=
Va-1 >/a +
l
-=
+ —=
\
(
2 ^
2
1
=
/
I
a + l>
a
-
2Va + 1 + a +
2^3
+ 1
a-1
a-1
a
+
lj
2(a
+1)
a-1
^a-l^'

a
+ lj
2(a
-1)
a
+
1
Cau
2.
Phuang
trinh
ho^h dp
giao
dilm
cua (d) (P)
\k
:
o
3x^
+ 6x + 8m = 0
3
o 3 2
— X
+
2m
= —
X
2
4
(*)

Dilu
ki$n
de (d) c^t (P) tai 2
diem phan bi|t 1^
(*) c6 2
nghiem phan
3
<=>
m < —.
8
bi|t
<=> A' = 9 -
24m
> 0
Cfiu
3.
x^
= X - 1
X
-
1
> 0
5
-
x^
= (x -
if
X>1
X
= -1

hoac
X = 2
X
> 1
2x2
- 2x - 4 = 0
x
= 2.
b)
Dieu ki^n
: x ^ 0, y 0.
Dat
u = i; V = —. H? da cho c6
dang
:
X
y
Vdi
u
= 2, ta c6 : - = 2 => x = i.
X
2
Vdi
V = 1, ta c6 :
—
= 1 => y = 1.
y
So vdi dieu
ki$n
ta

nhan
x = —; y = 1.
2
3u
- 4v = 2
4u
- 5v = 3
<=>
fu
= 2
V
= 1
V$y
h§ da cho c6
nghi$m
(x; y) la
c)
Khi
cdc
dieu
ki$n
xdc
dinh thoa
ta c6 :
V-x^
+ 4x - 2 =
V2
- (x -
2)2 < V2
V-2x2 + 8x - 5 =

V3
-
2(x
-
2)2
< ^3
Do
do : V-x2 + 4x -
2
+ V-2x2 + 8x - 5 <
+
S
Dau "="
xdy ra o x = 2.
V^y phifcfng
trinh
c6
nghiem
duy
nha't
x = 2.
Cfiu
4.
a)
Ta c6 :
Dat
(1)
x
+ y =
at

u = }-, u > 0.
u^
- 3u + 1 = 0
<=>
^ +
l = 3jii
(1)
u
=
3±
VS
x
_ (3 +
V5)2
y
4
b)
Vi vai
tr6
cua x va y la
nhu nhau
n§n ta c6 the gia
suf:
x > y.
x>0
o —<— => y>2.
m
. Ill
Ta
c6 : - + - = -;

X
y 2
x>y>0
=> —<— =>
—
=
—
+
—
<— => y<4.
X
y 2 X y y
Do
do : 2 < y < 4. Ma y la so'
nguyen dUdng
{gik
thiet)
nen y = 3
hoac
y
= 4.
Vdi
y = 3 X = 6. Do
tinh
doi
xuEng
ta
cung
c6 x = 3 va y = 6.
Vdi

y = 4
=>
X = 4.
Cdch
khdc
:
1 1
— <
— =>
y
2
i-i
L<1
2
" X ^
X
y 2
o 2(x + y) = xy o 2x - xy + 2y = 0
o
x(2-y)-2(2-y)
= -4 o (2 - x)(2 - y) = 4
Vi
x, y > 0 nen 2 - x < 2, 2 - y < 2. Do d6 ta c6 cdc
tru6ng
hap sau
•
2-x =
-l
va 2-y = -4 o x = 3 va y = 6
(nhan)

•
2- x = -4va 2-y = -l<» x =
6vay
= 3
(nhan)
•
2-x = -2 va 2-y = -2 o x = 4 va y = 4
(nhan).
Vay
ta c6 ckc cap so
nguyen dUdng
(x; y) la : (3; 6), (6; 3), (4; 4).
Cliu
5,
(Hinh
1)
a)
Taco:
DE//BC
nen
HDE
=
BHD.
Tam
giac
vuong
ABH c6 HD la
trung
tuyen
nen DH = DB, suy ra

BHD
=
DBH.
Do d6
HDE
=
DBH.
Suy
ra DE
tiep xiic diTdng
trbn
ngoai tiep tam
gidc
DBH tai
D.
Tucfng
tir nhu
tren,
ta c6 DE
tiep
xiic
difcfng
tron
ngoai tiep
tam
gidc
ECH
tai
E.
Vay

DE la
tiep tuyen chung
cua
hai
difdng
trbn
ngoai tiep tam
gidc
DBH
va
ECH.
b)
Ta
CO
:
IDF
=
IHD
va
IFD
=
IDH
(=
DBF).
Do
d6 hai
tam
gidc
IDF
va

IHD dong dang.
Suyra:
— = — =>
ID^
=
IF.IH
(1)
_
IH ID
Tuong tu
ta c6 :
lE^
=
IF.IH
(2)
Tif
(1) va (2), suy ra ID = IE hay HF qua
trung
diem
I cua DE.
c) Vi cac
tuf
giac
BDFH
va
CEFH npi tiep
nen :
DFH
+
DBH

=
180°; EFH
+
ECH
= 180°.
Trong tam
giac
ABC
c6 :
BAC
+
DBH
+
ECH
= 180°.
^
~ ^ b) 2x2 ^ 2V3^ -3 = 0 c) 9x^ + 8x2 - 1 = 0.
5x + 3y = -4
Lai
CO
:
DFH
+
EFH
+
DFE
=
360°.
Suy
ra

:
BAC
+
DFE
=
180''.
Vay dirdng
tron
ngoai tiep tam
gidc
ADE
qua F.
DE
2
THI
TUY^N
SINH
VAO
l6P 10
TAI TP. HCM NAM HOC
2006
-
2007
Cau
1.
Giai
cac
phifOng
trinh
va he

phucfng
trinh
sau :
a)
Cau
2.
Thu
gon cac
bilu
thiJc
sau :
^1-2
Va + 2^ r
_
_ . Va -
-V3'
{yla+2
Va-2j
(vdi
a > 0 vk a 7t 4).
Cau
3. Cho
manh
dat
hinh
chO
nhat
c6
dien
tich

SeOml Neu
tang chieu
rpng
2ra va
giam chieu
dai 6m
thi dien
tich
manh
dat
khong doi.
Tinh
chu
vi
manh
dat
Iiic
ban dau.
Cau
4.
a) Viet phuong
trinh
di/&ng
thing
(d)
song song
vdi difcmg
thing
y = 3x + 1
va

cit
true tung tai diem
c6
tung
dp
bkng
4.
b)
Ve do
thi
hkm so y = 3x + 4 v^ y =
tren
cung
mot he
true
tpa dp.
2
Tim
tpa
dp cac
giac
diem
cua
hai
do
thi
ay
bing
phep
tinh.

Cau
5. Cho tam
gidc
ABC
c6 ba g6c
nhpn
va AB < AC.
Dudng
tron
tam O
difdng
kinh
BC
cit cdc
canh AB,
AC
theo
thil
tu tai
E va D.
VIH-A/12
__J_.
V5-2
2-V3'
8
a)
Chu'ng
minli
AD.AC
=

AE.AB.
b)
Goi H la
giao
diem ciia BD
va
CE,
goi K la
giac
diem
cua
AH
va B(
ChuTng minh
AH
vuong
goc vdi BC.
c)
TLT
A ke cac
tiep tuyen
AM,
AN
den
dudng
tron
(O) vdi
M,
N la ca
tiep

diem. ChiJng minh
ANM
=
AKN.
d) Chu'ng minh
ba
diem
M,
H,
N
thSng hang.
BAI
GIAI
Cau
1.
a)
= ^ c H-6y = -3 |x =
-ll
[5x
+ 3y = -4
[l0x
+ 6y = -8 [y = 17
Vay
he c6
nghiem
(x; y) la :
(-11;
17).
b)
2x-+

2V3x-3
= 0 A'= 9 ^ >/A^ = 3.
Phuong
trinh
c6 hai
nghiem phan biet
: x, = ^ ^ •
x.>
= " ^
2
' - 2
c)
9x' + 8x- - 1 =»0.
Dat
t =
X-,
(t > 0).
Phuong
trinh
da cho c6
dang
:
91"
+ 8t - 1 = 0
Vi
a-b + c= l
ti
= -1; tv =
9
So vdi

dieu kien
t > 0 ta
nhaii
t = -
9
Vdi
t=i,tac6x-=-
=> x = ±-
993
1
1
3' "3-
V15-V12
1
V3(V5~2)
1.(2+
V3-)
Vay phuong
trinh
da cho c6
nghiem
la :
Cau
2. A =
V5-2 2-V3 V5-2 (2-V3)(2 +
V3)
= Vs - 2 - x/3 = -2.
B
=
V^~2

V^ + 2^
Va+2
Va-2
-aVa a - 4
_
(V^-2)2
^(4a+2f a -4
4a)
(Va+2)(Va-2)
Va
a-4
•
Cau
3. Goi
chieu rong ciia manh
dat luc dau la : x
(m),
(x > 0).
Chieu
dai
manh
dat luc dau la :
360
(m),
Chilu
rong manh dat sau khi tang 2m la : x + 2 (m).
OCA
Chilu
dai manii dat sau khi giani 6m la : 6 (m).
X

Tir
dieu kien dau bai ta c6 phu'cfng
trinh
: (x + 2)
360
-6
= 360.
Giai phucfng
trinh
nay ta dMc x, = 10 (nhan) va
Xv
= -12 (loai).
Vay : Chieu rong manh dat liic dAu la : 10 (m).
Chi6u dai manh dat luc dau la : = 36 (m)
10
Chu vi manh dat luc dau la : (36 + 10) x 2 = 92 (m).
Cau 4.
a)
Phaong
trinh
(d)
song song
vdi ducyng th^ng y = 3x + 1 nen c6 dang
y
= 3x + b, (dj cat true tung tai diem c6 tung dp bang 4 => b = 4.
Vay (d) : y = 3x + 4.
b)
Bang
gia tri :
X

-4
-2 0
2
4
-8
-2 0
-2
-8
X
0
-2
y
/
y
= 3x + 4
4
-2
4
Phuong
trinh
hoanh do
giao
diem ciia
2
-4 -2 /
0 2 4
(P) : y = - — va (d) : y = 3x + 4 la :
= 3x + 4 c:> x%6x + 8 = 0
<^ xi = -2 va X2 = -4
Vdi

x, = -2 y, = -2
Vdi
Xv = -4 =o y, = -8.
Vay toa do
giao
diem ciia (P) va (d) la : (-2; 2) va (-4; 8).
Cau 5.
(Hinh
2)
a) Ta
CO
: ABD = AEC (cung chdn cung DE).
Suy ra hai tam
giac
vuong ABD va ACE dong dang.
AB AD
Do do
AC
AE
AD.AC
= AE.AB.
LO
b) Ta CO ; BEC = BDC = 90" (goc aoi
tiep
chan
niifa dudng tron).
Suy ra : BD 1 AC va CE 1 AB,
Hay BD va CE la hai dudng cao ciia
tam
giac

ABC.
Ma H la
giao
diem ciia BD va CE nen
H
la trirc tam ciia tam
giac
ABC.
Suy ra AH 1 BC.
1
c) Ta CO : ANM = - MON (goc tao bdi
2
Hinh
2
tia
tiep tuyen va day vdi goc d tam ciing
chan
MN).
Ma OA la tia phan
giac
ciia goc MON (tinh
chat
hai tiep tuyen cat
1
nhau) nen AON = - MON, suy ra : AMN = AON
2
(1)
Ta lai c6 : ANO = AMO = 90°, nen
tiif
giac

ANKO noi tiep (tuf
giac
c6
hai dinh ke cung nhin mot
canh
dudi
hai goc bang nhau), suy ra :
AKN = AON (2)
Tif
(1) va (2), suy ra ; ANM = AKN.
d) Ta
CO
: ANE = ABN (goc tao bdi tia tiep tuyen va day vdi goc npi tiep
ciing
chan
cung NE). Va NAB la goc chung ciia hai tam
giac
ANB va
ANE.
AN
AE
Suy ra AANE ^ AABN (g-g)
Lai
CO : AEH = AKB = 90^
Suy ra AAEH ^r, AAKB (g-g)
Tif(3)va(4) =>
AB AN
AE AH
AK
AB

AN-
=
AH.AK.
AN
AH
AN-
= AE.AB
AE.AB =
AH.AK
3)
(4)
Xet
AANH
va
AANK
c6 : NAK
chung
va ^ ^ .
AK
AN
Suy ra
AANH
or,
AAKN
(c-g-c)
=> ANH = AKN.
Tren ciing mot niia mat phang c6 bd chufa tia
NA,
ta c6 :
ANH

=
ANM-
Suy ra hai tia
NH
va
NM
trung nhau.
Vay 3 diem M, H, N thang hang.
1]
DE
3
TRUdNG
PTTH
CHUYEN
LE
HONG
PHONG,
TPHCM
De thi tuyen sinh vdo I6p 10 chuyen
Todn
nam hoc 2005 - 2006
Cau 1.
a)
Dinh m de hai phifo'ng
trinh
: x' + x + in = 0 va x" + mx +1 = 0
CO it nhat mot nghiem chung.
b) Cho a, b, c la do dai ba canh ciia mot tam
giac.
ChiJng minh r&ng

phirong
trinh
: b'"x" + (b'^ + c" - a")x + c" = 0 v6 nghiem.
Cau
2. Giai phifong
trinh
va he phuong
trinh
:
-y^ =3{x^y) 13x
a)
\) — + = 6.
i
X + y = -1 3x- - 5x + 2 3x^ + x + 2
Cau 3.
a)
Chiing
niinh
2(a' + b"*) > ab' + a"b + 2a'b" vdi mpi a, b.
^ b) ChiJng minh Va^ - b" +
V2ab
- b^ > a, vai a > b > 0.
Cau
4. Tim cac so' nguyen diTOng c6 2 chuf so', biet so' do la boi ciia
tich
2
chCf so' ciia chinh so do.
Cau
5. Cho
hinh

binh hanh
ABCD
c6 goc A nhon, AB < AD. Tia phan
giac
ciia
goc BAD cat BC tai M va cat DC tai N. Goi K la tam ciia dudng
tron
ngoai tiep tam
giac
MCN.
a)
Chufng minh DN = BC va CK 1 MN.
b) ChuTng minh rang
BKCD
la mot tii
giac
noi tiep.
2au
6. Cho tam
giac
ABC c6 A = 2B. Chiifng minh rang :
BC- = AC- +
AB.AC.
BAI GIAI •
:au 1.
a)
Goi Xii la nghiem chung ciia hai phuong
trinh
ta c6 :
xl

+ x,, + m = 0 (*); XQ + mx,, + 1 = 0.
Tirdo (xg+Xo+m)-(x5 + mx„ + 1) = 0 ^ (1 - m)(x„ - 1) = 0.
Voi m = 1, ca hai phUong
trinh
deu c6 dang : x" + x + 1 = 0 (vo nghiem).
Vd-i
x„ = 1 tCr (*) CO m = -2, khi do phuong
trinh
x" + x + m = 0 c6 hai
nghiem la 1 va -2.
PhUong
trinh
x" + mx + 1 = 0 c6 nghiem kep la 1.
Vay m = -2 thi hai phuong
trinh
da cho c6 it nhat mgt nghiem chung.
b) b'-x- + (b- + c'-^ - a^)x + c" = 0
Ja
CO : A = ih- + - a'f - Ah'c' = (b" + c'' - a")" -
(2bc)"
= [(b + c)- - a-ll(b - c)- - a-|
= (b + c + a)(b + c - a)(b - c + a)(b - c - a).
Do a, b, c la do dai ba canh cua tam
giac
nen A < 0.
Vay phaong
trinh
da cho v6 nghiem.
'Cau 2.
x^-y^=3(x-y)

^ |(x - yKx^ + xy + y^ - 3) = 0
x + y = -i .[x + y = -i
x-y = 0
a)
<=>
Giai he (I) ta c6 nghiem (x; y) =
X
+ y = -1
x^ + xy + y^ - 3 = 0
X + y = -1
r
1 _i)
2'"2j-
(I)
(II)
Xet he (II) tir phirang
trinh
cuoi x + y = -1 => y = -1 - x.
Thay
vao
phirong
trinh
dau ta difoc x" + x - 2 = 0. Suy ra he (II) c6 hai nghiem
(x; y) =
(1;-2),
(-2; 1).
( 1 1'
Vay he da cho c6 3 nghiem la - - ; - -
V 2 2,
b) Vi X = 0 khong la nghiem ciia phuong

trinh
nen ta chia ve
trai
ciia
phuang
trinh
cho x ta difcfc :
2 13 2 13
, (1; -2), (-2; 1).
3x^ - 5x + 2 3x^ + X + 2
= 6
2 2
3x + 5 3x + - + l
x X
= 6 (*i
Dat y = 3x + - - 5 (*)
dugc
viet lai :
x
- +
-1^
= 6 o 2y''* + 7y - 4 = 0
y
y + 6
Giai phiicfng
trinh
nay ta diTOc : yi = - ; ya = -4-
Vdi
yi = - c=. 3x + - - 5 = i C5. 6x' - llx + 4 = 0 => x, = ^ , - -
2 X 2 3

Vdi
y2 = -4 3x + - - 5 = -4 c:> 3x- - x + 2 = 0 (v6 nghiem).
X
4 1
Vay phifcfng
trinh
da cho c6 hai nghiem x, = — , Xv = — .
3 ' 2
Cau
3.
a)
Ta CO
<::>
2(a' + > ab-' + a''b +
2a-b
4(a' + b') > 2ab'' + 2a'b + 4a"b'
(b' - 2ab'^ + a'b-) + (a'' - 2a-'b + a-b") + (Sa" + 3b'' -
6a-b^)
> 0
(b- - ab)- + (a^ - ab)^ + 3(a^ - b^)' > 0 (diing)
Vay bat
dang
thufc da cho diing.
b)
Vdi a > b > 0 thi Va" - b^ +
V2ab
- b^
>
a
ta' - b-) + (2ab - b") + 2yl(a^

-b^)(2ab-b^)
>
<^ 2b(a - b) + 2V(a^
-b^)(2ab-b-)
> 0
(dung)
Vay bat dang thiJc da cho dung.
Cau
4. Gpi so can tim la ab (a, b
khac
0). TCf gia thiet c6 ab = m .ab.
Suy ra : 10a + b = mab, hay b = a(mb - 10), suy ra b chia het cho a.
Dat b = na (n < 9).
Tir
na = a(mna - 10) c6 n = mna - 10 o n(ma - 1) = 10 nen chia het
cho n => n e
|1,
2, 5).
- Vdi n = 1 thi ma - 1 = 10 ma = 11, suy ra a = b = 1.
- Vdi n = 2 thi ma - 1 = 5 r=> ma = 6 => a = 1, 2, 3, tUcfng
iifng
c6 b = 2, 4, 6
- Vdi n = 5 thi naa - 1 = 2 <=> ma = 3 => a = 1, tu'Ong
iifng
c6 b = 5.
Thii
lai ta c6 cac so can tim la : 11, 12, 15, 24, 36.
Cau
5. (Hinh 3)
a)

Ta
CO
BAN = DNA, ma BAN = DAN (gia thiet),
suy ra DAN = DNA hay ADNA can tai D
DN
= AD = BC.
Nhan tha'y ACMN can tai C nen
CM = CN, ket hop vdi KM = KN
==> CK 1 MN.
b) Xet hai tam
giac
KBC va KDN c6 :
Hirili
3
BC = DN, KC = KN, KCB = KND (= KMC)
AKBC = AKDN, suy ra KBC = KDC hay tu^ac BKCD la tiJ
giac
noi tiep
(dpcm).
14
C&u
6. (Hinh 4)
Tren tia do'i ciia tia AC
chon
Bi sao cho ABi = AB
luc do BKC =
2B'B7C,
ma BAC =
2ABC
B,

TCf do :
BBjC = ABC, suy ra ABB,C oo
AABC.
BjC BC
BC AC
Hay BC- = AC(AB + AC).
DE
4
Hinh
4
TRl/CfNG
PTTH
CHUYEN
LE HONG PHONG,
TPHCM
De thi tuyen
sinh
vdo I6p 10 (Ban A, B) nam hoc
2005
-
2006
Cau
1. Cho phirong
trinh
(c6 an so la x) : 4x- + 2(3 - 2m)x + m" - 3m +2 = 0.
a)
ChuTng to r&ng phirong
trinh
tren luon c6 nghiem vdi moi gia tri ciia
tham

so m.
b) Tim m de c6 tich ciia hai nghiem dat gia tri nho nhat.
Cau
2. Giai cac phuong
trinh
va he phuong
trinh
:
a)
x^ + y^ = 2(xy + 2)
X
+ y = 6
b) x' +
25x^
(X
+ bf
=
11.
Cau
3.
a)
Cho a > c, b > c, c > 0. Chiirng minh : Vc(a - c) + Vc(b - c) < 4ah.
b) Cho a > 0, b > 0.
Chiirng
minh :
2Vab
<
VVab.
Va + Vb
Cau

4. Tim so chinh phuong c6 4 chC so biet rang khi tang them m6i chijf
so mot don vi thi so mdi
dugc
tao thanh cung la mot so chinh phuong.
Cau
5. Cho tam
giac
ABC c6 ba goc nhon noi tiep trong dudng tron (O; R),
so do goc C b^ng 45". Dudng tron dudng
kinh
AB c^t cac canh AC va
BC Ian \mt tai M va N.
AB
a)
Chiifng minh MN vuong goc vdi OC. b) ChiJng minh MN = -j=r.
Cau
6. Cho tam
giac
ABC cd ba goc nhon noi tiep dudng tron (0; R). Diem
M
di
dong
tren
cung'nho
BC. Tii M ke cac difdng th^ng MH, MK Ian
lu'ot
vuong goc vdi AB, AC (H
thuoc
dudng thang AB, K
thuoc

dirdng
thang AC).
a)
Chirng minh hai tam
giac
MBC va MHK
dong
dang vdi nhau.
b) Tim vi tri ciia M de do dai
doan
HK Idn nhat.
15
BAI
GIAI
Cau
1.
a) 4x- + 2(3 - 2m)x + m'' - 3in + 2 = 0
Ta CO : A' = (3 - 2m)'' - 4(m- - 3m + 2) = 1 > 0.
Suy ra phuong
trinh
da cho c6 hai nghiem phan biet Xj, X2 voi moi gia
tri
ciia tham so m.
b)
Theo
dinh
li Viet : x,Xv =
m^
- 3m + 2 1
3^

m
—
2
2
— > - —
16 " ~16
Dau "=" xay ra <:5> m =
2
Vay gia tri nho nhat ciia X|.xv la
Cau
2.
a)
. 16
dat duoc khi m
2
x^ + y'^ = 2(xy + 2)
x^ + y^ = 4xy + 4
X + y = 6
<=>
X + y = 6
Giai
ra ta dUdc he c6 hai nghiem (x; y) la : (4; 2), (2; 4).
b) Dieu kien x ^ -5. Phirong
trinh
da cho tuong dirong vdi :
X + y = 6
x.y = 8
X + = 11 <=>
(x + 5f
X -

5x
X + 5
\
10
X + 5
= 11
X + 5 ,
10
X + 5
= 11.
Dat y =
(1). Phuong
trinh
tren trcf thanh : y" + lOy - 11 = 0
X + 5
Giai
ra ta dugc y, = 1, y;, = -11.
- Vdi yv = -11, thay vao (1) thay phiTcfng
trinh
an x v6 nghiem.
- Vdi yi = 1,
tijf
(1) ta c6 phi/dng
trinh
- x + 5 = 0. Giai ra ta duoc :
X, =
Thiif
lai ta thay x, =
1
+ V2I

2
1
+ V2I
X^; =
1
- V2T
2
1-V2I
la hai nghiem ciia phiTcfng
triiih
da cho.
Cau
3.
a)
Ap dung bat d^ng thiifc
Cosi
cho hai so duang, ta c6 :
16
c(a - c)
ab
fc(b - c) _ 17
\b '
V
b
c
''b-c^
[ a J
a
V
b J

1
^c a-c^
1
(c b-c]
< — — + + —
—
+
2 Kb a ; 2
la b ;
= 1 => dpcm.
Dau "=" xay ra co
be
b-c
= a.
b) Ap dung bat dSng thiire Cosi cho hai so duong, ta c6 :
2vVa-Vb
< Va + Vb =>
Suy ra ^Vab ^
va
+ Vb
<
(Va +
^Ib}.
VV^.Vb
Va
+ Vb
Cau
4. Goi so chinh phuong c6 bon chuT so can tim 1^
abed
(a ^ 0).

Dat n- =
abed
(n E N'). Ta eo n" <
10000
nen n < 100.
Khi
tang m6i
chiJ
so ciia so'
abed
len mot don vi thi dirac so :
(a + l)(b + l)(c + l)(d + l).
Theo
gia thiet (a + l)(b + IKc + l)(d + 1) = m" (m E
N").
Tirong tir nhtf
tren
c6 m < 100.
Suy ra 2 < m + n < 200.
Xet : m- - ir = (a + l)(b + l)(c + l)(d + 1) -
abed
= 1111.
Hay (m - n)(m + n) = 1111.
VI
nil =
1.1111
=
11.101,
nen chi xay ra m - n = 11 va m + n = 101.
suy ra m = 56, n = 45.

Vay
abed
= 45- =
2025.
ThiJr
lai ta thay so can tim la
2025.
Cau
5. (Hiiih 5)
a) Ke tiep tuyen Cx voi dudng
tron
(O), khi do BAG = BCx (cung
ehan
cung BmC);
Mat
khac
BAG = MNC (cung bij
vdi
MNB)
Suy ra MNC = BCx, tiT do MN // Cx
Mat
khac
Cx 1 OC nen MN 1 OC.
b) Ta thay
ACMN
c/^
ACBA.
MN
CN 1
AB AC V2

(vi
AACN
can tai N)
MN
=
AB
V2'
THLH/IENTI.\!HBINHTHUAN
17
Cau 6.
(Hlnh
6)
a) Bon
diem
A, H, M, K
cung
nkm tren
mot
ducfng
tron
dudng
kinh AM.
Ta CO : MBC = MAC = MHK,
MCB = MAB = MKH.
Suy ra tam
giac
MBC
dong
dang
vdi

tam
giac
MHK.
b)
Theo
tren AMBC oo AMHK, suy ra
BC
MB
HK
MH
ma MB > MH
BC
HK
> 1
HK
< BC.
Dang
thilc xay ra <=> H = B, luc do ABM = 90'' <=> AM la
dudng
kinh
ciia dirdng tron (O). Do do khi M la
diem
dol xufng ciia A qua O thi do
dai HK Idn nhat.
DE
5
TRI/ONG
PTTH
CHIYEN
LE

HONG
PHONG,
TPHCM
De thi
tuyen
sinh
vao I6p 10 chuyen
Toan
nam hoc
2004
-
2005
Cau 1. Giai he :
2x - y X + y
1
1
2x - y X + y
= -1
= 0
Cau 2. Cho X > 0
thoa
: x" + A- = 7. Tinh x"' + —
Cau 3, Giai phiforng
trinh
:
3x
=
A/3X + 1 - 1.
V3x + 10
Cau 4.

a) Tim gia tri nho nhat ciia P = Sx" + 9y- - 12xy + 24x - 48y + 82.
X
+ y + z = 3
b) Tim cac so
nguyen
x, y, z
thoa
he :
x^ + y^ + z^ = 3
Cau 5. Cho tam
giac
ABC c6 3 goc nhon npi tiep trong
dudng
tron tam 0
(AB < AC). Ve dirdng tron tam I qua 2
diem
A, C cat
doan
AB, BC Ian
lirot
tai M, N. Ve dirdng tron tam J qua 3
diem
B, M, N cit dirdng tron
(O) tai dii'm H
(khac
B).
a)
Chdng
minh : OB
vuong

goc vdi MN.
18
b) ChiJng minh : tuT
giac
lOBJ la hhih binh hanh.
c) Chijrng minh : BH
vuong
goc vdi IH.
Cau 6. Cho hinh binh hanh ABCD. Qua mot dii'm S trong hinh binh hanh
ABCD
ke
dudng
thing
song song
vdi AB Ian
lirpt
cat AD, BC tai M, P
va
cung
qua S ke dirdng thang
song song
vdi AD Mn
lirpt
cat AB, CD
tai
N, Q. ChiJng minh 3 dirdng thing AS, BQ, DP
ddng
quy. -
BAI
GIAI

3 6
Cau 1.
2x - y X + y
1
1
Dat a =
2x - y X + y
1
2x - y
= -1
= 0
b =
Dieu kien : 2x - y ^ 0; x + y ^0
1
X
+ y
, , .
f3a-6b
= -l
H? da cho
dirac
viet lai : <^ , „ o
^ • • a-b = 0
1
a = -
3 C5
a = b
a = b =
Giai he
Cau

2.
Ta c6 :
2x - y
1
ta
dugc
nghiem
(ji; y) = (2; 1).
Suy ra :
X
+ y 3
X
+ —
X
1
^
X
+ — + — + X
X
1^
X
+ —
X
4 1
X
+ —
X
( 1
+
X

De
dang
tlnh
dirorc cac tong sau :
1
X
+ —
X
1
X
+ —
X
=
X- + 2 +
= x^ +
1
= 9
Vay :
x^+.
1
( 1
+ 3
X
+ -
x^
/
1
^
V
/

X
+ -
x^^
V
v
X
+ - = 3 (vi X > 0)
X
3 1
( 1^
3
' 1^
X
+ —
- 3
X
+ —
I
x; ^ x;
= 18
4 1 ^
X
+ —
( 1 3
V
X
=
3.47-18
= 123.
Cau 3. Dieu kien : x > -

3
i
Phuong
trinh
da cho
diroc
viet dudi
dang
:
19
3x
V3x
+ 10
(V3x
+
1
+ l) = (V3x + l) -1^
X = 0
VSx
+
1
+ 1-
=
1 (*)
VsxTio
(*) o
V3x
+
1
+

1
=
V3x
+ 10 ci>
VSx
+ 1=4 » X = 5.
Vay phifoing
trinh
c6 2
nghiem
la : x = 0; x = 5.
Cau
4.
a)
Ta CO : P = Sx" +
9y-
-
12xy
+ 24x - 48y + 82
= (X- - 8x + 16) +
(4x-
+
9y-
-
12xy
+ 32x - 48y + 64) + 2
= (X - 4}- + (2x - 3y + 8)' + 2 > 2.
fx
= 4
Dau "="

xay ra <=>
X = 4
2x
- 3y
+
8 = 0
16
b)
16
Vay
gia
tri
nho
nhat ciia
P la 2,
dat d\Xac khi
x = 4 va y = — .
3
X + y + z = 3 (1)
x^
+
y^
+ = 3 (2)
(1)
o (x + yf = {3-zf
<=>
x" + y* +
3xy(x
+ y) = 27 - 27z + 9z- - z'*
Tir

(2) => 3 - z'' +
3xy(3
- z) = 27 - 27z + 9z' - z'*
o 1 +
xy(3
- z) = 9 - 9z + 3z- o (3 -
z)(xy
+ 3z) = 8
Vay
cac
nghiem nguyen ciia phufang
trinh
la :
(1;
1; 1), (4; 4;
-5),
(4; -5; 4), (-5; 4; 4).
Cau
5.
(Hiiili
7)
a)
Ve
tiep tuyen
(d)
tai
B
ciia diidng tron
(O).
Ta CO

: Bi = Ci (goc tao
bcfi
mot
day va
tiep tuyen ciJng
chan
AB
ciia
(O)).
Ma
Ml = Ci ^ Bi =
Ml
=> MN
// (d)
Ma
OB 1 (d) tai B (d la
tiep
tuyen tai
B
ciia dudng tron
(0)).
Suy
ra OB 1
MN.
b)
Ve
tiep tuyen
(d')
tai
B

ciia dudng tron
(J).
Ta
CO B2 =
Ml
. Ma Ci =
Mi.
Suy
ra B2 =
Ci
=>
(d")
// AC
20
Ma
(d) 1
JB tai
B. Suy ra AC ±
JB.
Ma
AC 1 01 (01 la
ducrng thSng
noi
tam)
=> JB // OI
Mat
khac
OB 1
MN (chufng minh
cau a)).

Ma IJ
1
MN
=>
OB
//
IJ
Tif
(1) va (2) suy ra
OIBJ
la
hinh binh hanh.
c)
Ta
CO
OJ 1 BH
tai trung dig'm
K
ciia
doan
HB.
Goi
P la
trung diem ciia
OJ
thi
P
cung
la
trung diem ciia

doan
BI.
Vay
PK la
dudng trung binh ciia
tain
giac
BHI.
Suy
ra PK //
IH
=>
IH
1 HB
(dpcm).
Cau
6.
(Hinh
8)
Goi
I la
giao
diem ciia
DP
vdi
NQ.
Goi
K la
giao
diem ciia

AS
vdi
BC.
KP
KP SP SP IS
Ta
CO
Vay
;
BP
KP
AM
IS
SM
DQ IQ
(1)
(2)
BP
IQ
Suy
ra
SK, IP,
BQ
dong
quy.
Hinh
8
DE
6
TRI/ONG

PTTH CHUYEN LE HONG PHONG, TPHCM
De thi tuyen
sinh
vaa idp
10
(Ban
A,
B) nam hoc
2003
-
2004
Cau
1. Cho
phucfng
trinh
x^
-
2mx
- 6m - 9 = 0 (c6 an so la x).
a)
Tim
m de
phuong
trinh
c6
hai nghiem phan biet
deu am.
b)
Goi Xi, X2 la
hai nghiem ciia phaang

trinh.
Tim
m de c6 Xj + =13.
Cau
2.
a)
Cho x>0,
y>0vax
+
y<l. ChiiTng minh
:
b) Tim
gia
tri
nho
nhat ciia
bi§u
thufc A' =
x^
+ xy y^ + xy
>
4.
Cau
3.
Giai
he
phirong
trinh
:
X + y + xy = 11

a)
x^y
+
xy^
= 30
b)
2
+
V2x
-
x^
+ 7
xy
= -64
i_l
- i
X y 4
Cau 4. ChiJng minh
rhng
neu a + b > 2 thi it nhat mot
trong
hai
phuang
trinh sau c6
nghiem
: + 2ax + b = 0; + 2bx + a = 0.
Cau 5. Cho diTcrng tron tarn O
du&ng
kinh AB. Goi K la trung
di§m

ciia
cung AB , M la di6m di
dgng
tren cung nho AK (M khac
diem
A va
K). Lay
diem
N tren
doan
BM sao cho BN = AM.
a) Chufng minh AMK = BNK .
b) ChiJng minh tam
giac
MNK Ik
tarn
gidc vuong can.
c) Hai dudng
thSng
AM va OK c^t nhau tai D. Chufng minh MK la dudng
phan
giac
ciia
goc DMN.
d) Chufng minh
rang
diTdng
thang
vuong goc vdi BM tai N luon luon di
qua mot diem co'

dinh.
Cau 6. Cho tam
giac
ABC c6 BC = a, CA = b, AB = c va c6 R la ban
kinh
dudng
tron
ngoai tiep thoa man h$ thiJc : R(b + c) =
aVbc
. Hay
dinh
dang tam
giac
ABC.
BAIGIAI
Cau 1.
a) Phircfng
trinh
c6 hai nghiem phan
biet
deu Sm khi va chi khi :
A' > 0
S <0
P > 0
m^
+ 6m + 9 > 0
2m
< 0 o
-6m
- 9 > 0

m
< — va m 9i -3.
2
b)
Theo
dinh
Viet ta c6 : Xi +
X;,
= 2m; XiX^ = -6m - 9.
Do do : X? + = 13 o (x, + x^)' - 2x,X2 = 13
<=> 4m- + 12m + 5 = 0 =>
1
m
= ;
5
m
= —.
2
Cau 2.
a) Nhan xet rang v6i a, b la cac so ducfng thi i + - > .
a
h a + b
TCr do ta CO :
1
x^ + xy y^ + xy (x^ + xy) + (y^ + xy) (x + y)^
Vi
x, y > 0 va x + y < 1 nen
(x + y/
> 4.
TCr (*) suy ra :

1
x^ + xy y^ + xy
> 4.
Dieu
ki$n
: -x" + 2x + 7 > 0 c=. 1 - 2V2 < x < 1 + 2^2
Taco:
-x'+ 2x + 7 =-(x - 1)'+ 8 < 8.
3
3(V2-1)
Do do : A =
2 + V2x - x^ + 7
3(V^-l)
V$y gia tri nho nhat cua bieu thufc A la , dat duofc khi x = 1.
Cau
a)
3.
Dat u = X + y, V = x.y. H| da cho c6 dang
Giai
h$ nay tim
diTOc
(u; v) = (6; 5), (5; 6).
x
+ y = 6 _^ |x + y = 5
u
+ V = 11
u.v = 30
Giai
he : va
xy =6

b)
xy = 5
ta
diroc bon nghifm (x; y) sau : (1; 5), (5; 1), (2; 3), (3; 2).
Di§u
kif
n
: x 0 va y * 0.
xy = -64
x
y 4
f(y + 16)y = -64
<—> J <=>
lx-y
= 16
xy = -64
y
- X _ 1 <=>
xy 4
y^ + 16y + 64 = 0
xy = -64
y - X _ 1
-64 ~ 4
<=>
xy = -64
X
- y = 16
x
= 8
y

= -8
Cau
Cau
a)
b)
X
- y = 16
Vay nghiem (x; y) ciia he phirang
trinh
la (8; -8).
4. PhiTdng
trinh
x'^ + 2ax + b = 0 c6 biet so Aj = a^ - b .
Phirang
trinh
x^ + 2bx + a = 0 c6 biet so Ag = b^ - a .
Ta CO : Al + A2= (a^ - b) + (b'^ - a) = (a - 1)' + (b - if + (a + b - 2) > 0.
Do do trong hai so A;, A^ C6 it nhat mot so khong am nghia la c6 it
nha't
mot trong hai phirgmg
trinh
da cho c6 nghiem.
5. (Hinh 9)
Ta CO : AAKM =
ABNK
(c.g.c).
Suy ra AMK = BNK .
Tir
ket qua cau a)
AAMK

=
ABNK.
Ta CO : KM = KN va AKM = BKN.
Ma BKN + AKN = 90'^ ^ AKM + AKN = 90°.
Do do tam
giac
KMN vuong can tai K.
22
23
c)
TCr
ket qua cau b)
tam
giac
MKN vuong
can tai
K
nen
NMK
=
45°
,
D
Do DMN
=
90"
DMK
=
45".
Suy

ra
MK
la
ducfng phan
giac
ciia DMN.
d)
Gia
siif
du'cfng thSng vuong
goc
vd'i
BM
tai
N cat
ducfng
thflng
AK
tai
E.
Nhan
thay
rang til
giac
BEKN
npi
tiep,
suy ra
AEB
=

MNk
=
45°.
A
\ /
Hinh
9
Mat
khac BAE
=
45°
nen
tam
giac
ABE
vuong
can
tai
B,
dfin
tdi
E co
dinh
(dpcm).
Cau
6.
Tij'
moi quan
he
giuTa

do
dai
dudng
kinh
va
day
cung
ta
c6 :
2R
>
BC
= a va b +
c>2Vbc,
suy ra
R(b + c) >
aVbc
.
Dang thiJc
xay
ra
khi
va
chi
khi tam
giac
ABC
vuong
can
tai

A.
TCr
do
R(b
+ c) =
aVbc
thi tam
giac
ABC vuong
can
tai
A.
DE
7
TRUONG PTTH CHUYEN
LE
HONG
PHONG,
TPHCM
De
thi
tuyen
sinh
vdo
I6p
10
chuyen
Todn
nam hoc
2003

-
2004
Cau
1.
a)
Thu
gon
bieu thiJc
A =
nghiem thi phuong
trinh
sau
luon
c6
nghiem
;
^yttk
(an
~
mb)x-
+
2(ap
-
mc)x
+
bp
-
nc =
0.
^^^u

5.
Cho
tam
giac
ABC
vuong tai
A
(AB
<
AC)
c6
dUcfng
cao
AH
va
trung
tuyen
AM.
Ve
ducfng
tron
tam
H
ban
kinh
AH,
cat AB d
diem
D, cat
AC

d
diem
E (D va E
khac diem
A).
a)
ChiJng
minh
; D,
H,
E
thang hang.
b) ChiJng
minh
:
MAE
=
ADE
va
MA
vuong
goc
vdi
DE.
c) ChiJng
minh
bon
diem
B, C, D, E
cCing thuoc

mot
dudng
tron
tam
la O.
Tuf
giac
AMOK
la
hinh
gi ?
d)
•
Cho
goc
ACB
= 30° va
AH
- a.
Tinh
dien
tich
tam
giac
HEC.
Cau
6.
Cho
hinh
thang ABCD

c6
hai
dudng
cheo
AC
va
BD
ciing bang canh
day Idn AB.
Goi
M la
trung
diem ciia
CD. Cho
biet MBC
=
CAB
.
Tinh
cac
goc
ciia
hinh
thang ABCD.
3V2-2V3
V2-V3
\3V2
+
2V3
b) Tim

gia
tri nhd nhat ciia
: y =
Vx^^l~-"2Vx^
+ Vx + 7 -
6Vx
- 2
Cau
2.
Giai
cac
phucfng
trinh
va he
phuforng
trinh
:
x
+
y +
xy
=
2 + 3-72
y^
+ = 6
a)
b)
•
+
x^

- 4 = 0.
Cau
3.
Phan
tich
thanh
nhan tuT
: A =
x"*
-
5x-'
+
lOx
+ 4.
4
.
Ap dung
:
Giai
phiTcfng
trinh
:
x2-2
=
5x.
Cau
4.
Cho
hai phuang
trinh

:
ax^
+ bx + c = 0, a ^ 0
mx^
+
mx
+ p = 0, m ^ 0
(1)
(2)
ChiJng
minh
r^ng
neu It
nhat
mot
trong
hai
phuomg
trinh
tren
v6
BAI
GIAI
Cau
1.
a)
A =
3V2 - 2V3 1
V2-V3
V3V2

+ 2V3
V2-V3
^
76(73
+ 72
^
i75-75U
73 + 72 72 -73 \
1
p^7i)(73->^)
72-73
-(73-72
=
-1 (vi 73 - 72 > 0 )
Vay
A =
-1.
b) Dieu
kien
: x > 2.
3-2
73
=
7x
- 1 - 27x'^ +
7x + 7
- 67x - 2
=
7x
- 2 - 27x - 2

+
1
+
7x
- 2 -
67^r^ +
9
(71^-if
+ ^(3 - 7^) =
l7^-il
+
l3-7^
Ap dung
:
iA|>A.
Dau "=" xay ra
<=> A > 0.
Ta
CO : y > 7x - 2 -
1
+ 3
- 7x - 2 = 2
24
25
Dau "=" xay ra <=>
Vx - 2 - 1 > 0
3 - Vx -2 > 0
1
< Vx - 2 < 3
c:>

l<x-2<9
<=>
3<x<ll
Vay gia tri nho nhat ciia yla2 c? 3<x<ll.
Cau2.
[x
+ y + xy = 2 + 3V2 (1)
(2)
a)
y^
+ = 6
Ta CO
<=>
X + y = 2 + V2 (3)
X + y = -4 - V2 (4)
y^
+ x^ + 2(x + y + xy) = 6 + 2(2 + 3A/2)
(X- + 2xy + y^) + 2(x + y) = 6 + 4 +
(X
+ y)^ + 2{x + y) + 1 = 3- +
2.3V2
+ (72)^
(x + y + = (3 + V2)^
^x + y + l = 3 + >^
x + y + l =
-3->/2
Tir(l)
va (3)
CO
xy = 2V2

xy
= 2V2
X + y = 2 + A/2
Nen
X, y la nghiem cua phuong
trinh
: - (2 + A^|X + 2V2 = 0
Nghiem
cua (*) la : X, = 2; = V2 .
X = 2 fv - ./o
Taco:
^; ^ -
y
= V2 [y = 2
Tii (1) va (4) ta c6 : xy = 6 + 4V2
(X
- y)^ = y- + X- - 2xy = 6 - 12 - 8 V2 = -6 - 8A/2 < 0 (loai).
Vay nghifm ciia he phuang
trinh
da cho la :
X = 2
y
= >/2'
X = V2
y
= 2
b)
+ x^ - 4 = 0 (Dieu kien : -2 < x < 2)
0
X > 0

x^ = 4 - x'^
X > 0
2x^ = 4
X
= V2 (thoa dk -2 < x < 2)
Vay nghiem cua phuomg
trinh
da cho la x = -\/2 .
Cau 3. A = x' - 5x' + lOx + 4
= x"* - X'* - 2x^ - 4x-' + 4x' + 8x - 2x' + 2x + 4
= X- (x^ - X - 2) - 4x(x^ - X - 2) - 2(x' - x - 2)
= (x- - X - 2)(x^ - 4x - 2) = (x' - 2x + x - 2)(x- - 4x - 2)
= [x(x - 2) + (x -
2)|(x^
- 4x - 2)
Vay A = (X - 2)(x + IKx" - 4x - 2).
Ap
dung : Dieu ki#n - 2 ^ 0 c=>
x^±^/2
+4
x^ -2
= 5x
x" + 4 = 5x(x^ - 2)
x" - 5x' + lOx + 4 = 0 (X - 2)(x + IXx^ - 4x - 2) = 0
x-2 = 0
X + 1 = 0
x^ -4x-2 = 0
X = 2
X = -1 <=>
x^ - 4x + 4 = 6

x = 2
X = -1
(X
- 2f =
(yfef
<=>
(thoa man dieu kien x # ± V2
"x = 2 X = 2
X = -1 X = -1
x-2 = Ve ^ x = Ve+ 2
x-2 = ->/6 [x = -V6 + 2
Vay nghiem ciia phucfng
trinh
da cho la :
X, = 2; X, = -1; x,-, = V6 + 2; x, = -V6 + 2.
Cau 4. Xet cac phuofng
trinh
: ax" + bx + c = 0 (1)
mx' + nx + p = 0 (2) (a, m ^ 0)
va (an - bm)x- + 2(ap - mc)x + bp - nc = 0 (3)
Do gia thiet bai toan it nhat mot trong hai phuang
trinh
(1)
hoac
(2)
v6 nghiem, gia sijf (1) v6 nghiem nhu vay b^ - 4ac < 0.
Dat A = an - bm, B = ap - mc, C = bp - cn, ta dUcfc :
cA - bB + aC = acn - bmc - abp + bmc + abp - anc = 0 (*)
(3) trd thanh : Ax' + 2Bx + C = 0 (4)
1. Neu A = 0 tCr (*) ta c6

aC + bB = 0 <^
a
27
(4)
tro
thanh
: 2Bx + C = 0 ci>
2Bx
+ -B = 0
a
-
B = 0 ta
CO
X
ttiy
y
-
B 7^ 0 ta
CO
X =
2a
(3)
CO
nghiem.
(3)
CO
nghiem.
2. Neu
A ^ 0. Ta
CO

:
A'
= B" -
AC
-AC<0
^ A'>0
-
AC
> 0
do b-
- 4ac < 0 ^
Tu-
(*)
CO
bB =
cA
+ aC ==>
Suy
ra
b-(B'
-
AC)
> 0
(3)
CO
nghiem.
b"
< 4ac =>
b"AC
<

4acAC
b"B"
=
(cA
+ aO' >
4acAC
>
b'AC
b^
> 0
B^
-
AC
> 0
Vay
A'
> 0 => (3) c6
nghiem.
Trircfng
hop
phirong
trinh
(2) v6
nghiem, chiing minh tirong tiT nhiT tren
duoc
(3)
CO
nghiem.
Tom
lai

vdi
gia
thiet
bai
toan thi phirang
trinh
(3)
luon luon
c6
nghiem.
Cau
5.
(Hinh
10)
a)
Ta
CO
:
DAE
=
90°
z=>
DE la
dirdng
kinh
ciia dudng tron (H; AH)
=> D,
E, H
thang hang.
b) AABC vuong

tai
A c6
AM
la
trung tuyen
(gia
thiet)
AM
=
MB
=
MC
= — .
AMAC
can tai
M ^
MAE
=
MCA
(1)
HA
=
HD (vi
A,
D e
(H; HA))
AHAD
can tai
H =>
ADE

= BAH (2)
Ma MCA
=
BAH
(3)
(hai
goc
nhon
c6
canh
tuong ufng vuong
goc)
Tir
(1),
(2) va (3)
ta c6 :
MAE
=
ADE
Goi
I la
giao
diem ciia AM
va
DE.
Ta CO
:
MAE
+
DAI

=
DAE
.
Do
do
ADE
+
DAI
=
90°
^
Hinh
10
AID
=
90°
MA
1
DE.
c)
Tir
(2) va (3)
ta c6 :
ADE
=
MCA
=>
B,
C,
D,

E
cung thuoc mpt dudng tron
tam O.
I
28
M
la
trung diem
day
BC
nen
OM
±
BC.
H
la
trung diem
day
DE
nen
OH
1
DE.
Ta
CO
:
AH
1
BC
(gia

thiet), OM
1
BC
=>
AH
//
OM.
va MA
1 DE
(cau
b),
OH
1 DE
MA
//
OH.
TiJ giac
AMOH
c6
AH
//
OM
va
MA
//
OH
nen tuf giac
AMOH
la
hinh

binh
hanh.
d)
Ve
HK vuong
goc
vdi AC.
AHAC
vuong
tai H c6
ACH
= 30° (gia
thiet)
=>
AHAC
la
nufa
tam giac deu =>
HAE
= 60°
AC
=
2AH
= 2a.
AHAE
can tai H
(HA
=
HE)
c6

HAE
= 60°
AHAE
deu
AE
=
HE
= a,
HK
=
AHVS
aV3
Do
do
Vay:
EC
=
AC
- AE = 2a - a = a.
1
aVs
V3 .
S,i, c
= -
EC.HK
= -
•
a
•
= —

a^
(dvdt).
2
2 2 4
Cau
6.
(Hinh
11)
Hinh
thang ABCD
(AB
//
CD)
c6 : AC = BD (gia
thiet)
nen la
hinh
thang
can
^
BAD
=
ABC
Ma
Do
do
BXD
-
CAB
=

ABC
-
MBC
Goi
N la
trung diem
canh
AD
=D
MN
la
dudng trung binh AADC
=>
MN
//
CA
=> DXC
=
DNM
(2)
CAB
=
MBC (gia
thiet)
Hinh
11
Tir
(1) va (2)
c6
ABM

=
DNM
=>
Tuf
giac
ANMB
noi
tiep duoc.
Mat
khac
ABAD
can tai
B
(AB
=
BD)
c6
BN
la
trung tuyen.
=> BN
la
duo'ng
cao
ciia
tam
giac
BAD
=>
ANB

=
90°.
=> AB
la
ducyng
kinh
ciia ducmg tron ngoai tiep AMAB
=> AMB
=
90°
. Ma
MA
=
MB
(tinh
chat
doi xufng true).
Do
do
AAMB
vuong
can tai M s>
Ta
CO
:
AABC
can tai A
(AB
=
AC).

MAB
=
MBA
=
45°
.
29
Ta CO
ACB
= ABC = MBA + MBC = 45° + MBC
BXC + ACB + ABC = 180°
MBC
+ 45° + MBC + 45° + M'BC = 180°
3MBC = 180° - 90° hay MBC = 30°
Do do
Vay
BAD = ABC = MAB + MBC = 75° .
ADC = BCD = 180° - ABC = 180° - 75° = 105°.
DE
8
TRl/CfNG
PTTH CHLYEN
LE HONG
PHONG,
TPHCM
Mm
De thi
tuyen sinh
vao Idp 10
chuyen Todn

nam hoc
2002
-
2003
Cau
1, Tim gia tri cua m de phifang
trinh
sau c6 nghi^m va
tinh
cac
nghiem ay theo m :
X + I - 2x + m
I
= 0.
Cau
2. Phan
tich
thanh nhan
tilf
: A = x'" + x'' + 1.
Cau
3. Giai cac phucfng
trinii
va he phirong
trinh
:
b)
Vx + 2 +
3V2x
- 5 + = 2A/2

1
. x^ 48 ^Jx A\
a)
— + — = 10
3 x;
c)
xy- - 2y + 3x^ = 0
y^ + x'^y + 2x = 0
Cau 4. Tim gia tri
Idtn
nhat va gia tri nho nhat ciia : y =
x^ - 5x + 7
Cau 5. Cho tam
giac
ABC c6 3 goc nhpn noi tiep trong dudng
tron
(0) va
CO
AB < AC. Lay diem M thuoc cung BC khong chiifa diem A ciia dudng
tron
(O). Ve MH vuong goc BC, MK vuong goc CA, MI vuong goc AB
(H
thuoc BC, K thuoc AC, I thuoc AB).
Chufng
minh
:
BC
AC AB
+
MH

MK MI
Cau
6. Cho tam
giac
ABC. Gia
sijf
cac
dacrng
phan
giac
trong
va phan gi^c
ngo^i cua goc A cua tam
giac
ABC Ian
liTot
dt
duemg
th^ng BC tai D,
E
va CO AD = AE.
ChiJng
minh
: AB" + AC^ = 4R- vdi R la ban
kinh
diTcJng
tron
ngoai
tiep
tam

giac
ABC.
30
Cau
1.
BAI
GIAI
X +
I
x^ - 2x + m
I
= 0 «
i
x- - 2x + m
I
= -x
-x > 0 X < 0
x^ - 2x + m = -X
x^ - 2x + m = X
x'-^ - X + m = 0 (1)
x^ - 3x + m = 0 (2)
+
Xet m > 0 : Neu (1), (2) c6 nghiem thi cac nghiem deu duong (S > 0,
P
> 0) khong thoa x < 0.
+
Xet m = 0
•
(1) CO
ngliiem 0 va 1; (2) c6 nghiem 0 va 3. Nhan x = 0.

+
Xet m < 0 : phUdng
trinh
da cho c6 bon nghiem, hai nghiem duang. hai
nghiem am (do P < 0) thoa dieu kien x < 0.
1
- Vl - 4m 3- V9-4m
Hai
nghiem am la : x = ^ ; x = ^ .
Vay m < 0 thi phifong
trinh
c6 nghiem :
X = 0; X =
1
- Vl - 4m
X =
3 - V9 - 4m
2 • 2
Cau
2. A x'" + x' + 1
=
x'° + x% x« - x" - x« - x^ + x^ + x« + X'' - x« - X'' - x^ + x^' +
+
x'* + x'* - X-' - x'-^ - X + X^ + X + 1
=
x'\x- + X + 1) - x^(x' + X + 1) + x'^(x' + X + 1) - x^x- + X + 1) +
+
x''(x- + X + 1) - x(x- + X + 1) + l(x- + X + 1)
Vay A =
(X-+

X +
IXx**
- x" + x" - x" + x'- X + 1).
Cau
3.
a)
Di^u ki|n x 0.
X 4
Dat
y = .
3 X
Ta CO
16
8
x^ 48
Do do 3y- + 8 = — + —
3 X"
Ta
CO
phuang
trinh
: 3y' + 8 = lOy <=> 3y' - lOy + 8 = 0
A'
= 25 - 24 = 1
5+1
^5-14
* Vdi y = 2, ta CO : - - - = 2 x'^ - 12 = 6x x^ - 6x - 12 = 0
3 X
31
A'

= 9 + 12 = 21 => VA^ = V2T.
X|
= 3 + V2I ; xv = 3 - V2T .
4 X 4 4 •>
* Vdi V = - , ta CO : = - <=> x" - 12 = 4x <=> x" - 4x - 12 = 0
3 3x3
A'
= 4 + 12 = 16
VA^ = 4 .
x,( = 2 + 4 = 6; X4 = 2 - 4 = -2.
PhUong
trinh
c6 bon nghiem la :
X, = 3 + A/2T ; Xv = 3 - V2T ;
x.) = 6;
X4
= -2.
b) Dieu
kien
x > - . Ta c6 :
Vx
+ 2 + 3/2x - 5 +
-y/x
- 2 - V2x - 5 = 2V2
2
o
o
o
o
<=>

<=>
<=>
V2 (Vx + 2 +
3V2X
- 5 +
Vx-2-A/^^^^
= 2V2 V2
V2x + 4 + 6V2x - 5 +
V2X-4-2V2X-5
= 2.2
V2x - 5 + 6V2x -5+9 + V2x - 5 -
2V2x
-5+1 = 4
^(V2x - 5
+
sf
+ ^(
V2x - 5 ^ = 4
V2X-5+3 + lV2x-5-l| =4 c=. |V2x-5-l| = 1 - V2x - 5
V2x - 5 - 1 < 0 (vi
I
A| =-A c=. A < 0)
V2x - 5 < 1 o 0<2x-5<l-
5<2x<6 <=>
-<x<3
2
Vay nghiem ciia phuong
trinh
la : - < x < 3 .
2

c)
xy^
- 2y + Sx^ =0 (1)
y-
+ x-y + 2x = 0 (2)
* Xet X = 0 thi y = 0, he c6 mot nghiem
X = 0
[y = 0
* Xet X
7i
0,
nhan
hai ve ciia (2) vdi x roi
trCr
cho (1) ve theo ve ta c6 :
x'V +
2y - X" = 0 y(x'* + 2) = x'
y
=
(vi
x'* + 2*0).
Thay vao (1) ta c6 : x
(
2 N
U
+2
-2
^ x^
x^
+2j

x^
+2
+ 3x^ = 0
32
+ 3 = 0 CO X-' - 2(x'' + 2) + 3(x-' + 2)' = 0
(x=^ + 2f x^ + 2
CO X'' - 2x'' - 4 + 3x" + 12x'' + 12 = 0
3x"
+ llx'' + 8 = 0 CO (x'' + l)(3x''' + 8) = 0
x'^
+ 1 = 0
3x^
+ 8 = 0
x^
=-1
x3=-«
^
X = -1
* X =
-1,
ta CO : y =
(-ir
+ 2
= 1.
f-2^
* X = 3/— = — , ta CO : y =
-2
+ 2
Nghiem ciia phuong
trinh

la :
X = 0
y
= 0
X = -1
y
= 1
X =
-2
Vs
-6
Cau 4. Ta CO : x^ - 5x + 7 =
5
X —
2
+ — > 0. Do do y xac
dinh
vdi moi x.
4
y
= ^
x^
- 5x + 7
c:> yx" - 5xy + 7y = x' c:> (y - l)x' - 5xy + 7y = 0
* Xet y = 1, ta CO : -5x + 7 = 0 o x = -
5
* Xet y ^ 1, ta c6 : A = 25y' - 28y(y - 1) = 25y' - 28y- + 28y
= -3y- + 28y = y(-3y + 28)
De CO x thi A > 0 co
y

> 0; - 3y + 28 > 0
y
< 0; - 3y + 28 < 0
28
CO
28
y^O; y< —
o
y
^ 0; y >
28
CO 0 < y <
Vay : Gia tri nho nhat ciia y la 0 v6i x = 0.
Gia tri Ion nhat ciia y la — vdi x = — .
3 5
Cau 5. (Hinh 12)
Goi D la diem tren canh BC sao cho DMC = BMA
Xet
AMBA
va
z\MDC
c6 :
BMA = DMC ; BAM = DCM
(hai goo noi tiep cijng chan cung BM).
Do do
AMBA
^
AMDC.
MI,
MH Ian luot la diro'ng cao dng vdi

canh AB, DC cua tam
giac
MBA, MDC.
AB DC
Ta CO :
MI
MH
(1)
Mat khac : BMD = BMA + AMD ,
AMC = DMC + AMD va BMA = DMC ^ BMD = AI
Xet AMBD va
AMAC
c6 : BMD = AMC ; MBD = MAC (hai goc noi
tiep Cling chan cung MC).
Do do : AMBD co
AMAC.
MH,
MK Ian luot la dudng cao ufng vdi canh BD, AC ciia tam
giac
RD AC
MBD, MAC. Ta CO : — = (2)
TO (1) va (2) ta c6 :
Vay :
BC AC
MH
AC
MK
AB
MK
AB BD + DC BC

MI
MH
MH
MI
MH
MK
Cau 6. (Hinh 13)
AD cat cung BC tai F.
Ve du'dng
kinh
AG ciia dudng
tron
(ABC).
Ta CO : ABG = 90° (gdc noi
tiep chdn nua du'dng tron)
sdGB
+
sdFC
+
sdAC
+ sdBF =
sdACG
= 180"
BAF = FAC (AD la phan
giac)
=^
sdBF=sdFC
(1)
(2)
34

AD, AE la hai tia phan
giac
ciia hai gdc ke bii BAG va CAx nen
DAE = 90° .
ADAE
vuong
CO AD = AE (gia thiet) nen la tam
giac
vuong
can
=^ ADE = 45°.
sdADE =
'^^^^'^^^
^
SdAC
+
SdBF'=
90" (3)
2
Tii
(1), (2) va (3), ta CO : GB = AC GB = AC.
ABAG
vuong
tai B nen : AB' + BG' = AG' => AB" + AC' =
(2R)'"
=> AB' + AC' = 4R' (dpcm).
DE
9
TRUCfNG
PTTH

CHUYEN
LE
HONG
PHONG,
TPHCM
De thi
tuyen
sinh
Idp 10 (Ban A, B) nam hoc 2001 -
2002
Cau 1. Cho
phUcfng
trinh
: (m + l)x' - 2(m + 2)x + m - 3 = 0 (c6 x la an so).
a) Dinh m de phucng
trinh
c6 nghiem.
b) Dinh m de
phUOng
trinh
c6 hai nghiem
Xj,
x^. thda : (4xi +
l)(4xv
+ 1^ = 18.
Cau 2. Chufng minh cac bat d^ng thufc sau :
a) a' + b' + c' > ab + be + ca vdi moi a, b, c.
, , a^ + b*^ + 111, nun
b)
—>-

+ - + - (a > 0, b > 0, c > 0)
a^b^c a b c
c) a' + b' + c' + d' + e' > a(b + c +. d + e) vdi moi a, b, c, d, e.
Cau 3. Giai cac phuong
trinh
:
«\ 2x „ , ^ 4x 5x
a) x'+ = 8 b) — + — = -1
X - 1 x^ - 8x + 7 x^ - lOx + 7
Cau 4. Cho tam
giac
ABC cd ba gdc nhon noi tiep trong difdng tron tam 0
va true tam la H. Lay diem M thuoc eung nho BC.
a) Xae dinh vi tri ciia diem M sao cho tuf
giac
BHCM
la mot
hinh
binh
hanh.
b) Vdi M lay bat ki thuoc cung nho BC, goi N, E Ian lugt la cac diem doi
xiing
cua M qua AB va AC. Chilng minh ba diem N, H, E thSng hang.
Xac dinh vi tri cua M thuoc cung nho BC de cho NE cd do dai Idn
nhat.
35
Cau
5. Cho
dLfdng tron
co

dinh
tam O, ban
kinh
bang
1. Tam
giac
ABC'
thay
doi va
luon ngoai tiep dudng tron
(O). Mot
dirdng thang
di qua
tam
O cat cac
doan
AB, AC Ian
lugt
tai M, N. Xac
dinh
gia tri nho
nhat ciia dien tich tam
giac
AMN.
BAI
GIAI
Cau
1.
a)
(m +

l)x^
-
2(m
+ 2)x + m - 3 = 0
* Vdi
m+l =
Oc:>m
=
-l
PhUcfng
trinh
trd thanh
: -2x - 4 = 0 <=> x = -2
Vay khi
m =
-1, phucfng
trinh
c6
nghiem
x = -2.
* V6i
m + 1
7^
0 ci. m
;t
-1
A'
=
(m
+ 2f -

(m
+
l){m
- 3) =
m"
+
4m
+ 4 - +
3m
- m + 3 =
6m
+ 7
PhiiOng
trinh
c6
nghiem
<=> A' > 0.
a^
+ + c^ > ab + be + ca <=> a" + b" + c" - ab - be - ca > 0
c:> 6m + 7 > 0 o 6m > -7
7
m
>
6
Phirong
trinh
c6
nghiem
c:>
m?i-l;m>

— •
6
Vay phUdng
trinh
c6
nghiem
<=> m > —.
6
CaU
2.
a)
^
2a- + 2b' + 2c- - 2ab - 2bc - 2ca > 0
^
(a- + b- - 2ab) + (b" + c" - 2bc) + c" + a" - 2ca) > 0
^
(a - b)- + (b - c)- + (c - a)- > 0
(dung)
Do
do a" + b" + c' > ab + be + ca la bat
ding thufc dung,
b) Ap dung
cau a) ta c6 :
a»
+ b« + c« > a^b^ + b^c^ + c^a^ =
(aV)"'
+
(bV)"
+
(cV)'''

>
>
(a-b-)(b-c-)
+
(b-c')(c-a-)
+
(a-b-Kc'^a")
=
a-b-c-(a"
+ b" + c")
>
a"b"c'
(ab + be + ca)
0
do
+
b**
+
c*"
^
a2b'c''(ab
+
be
+
ca)
a^b^c^
"
a^b^c^
+
b' + c^ +

d"
+ e' > a(b + c + d + e)
a"
+ b" + c~ + d' + e' - a(b + c + d + e)>0
a"
+ b" + c' + d' + e" - ab - ac ad - ae) > 0
a"
+ b" + c" 111
"
a ^ b ^ c
•
+ b^
- ab
+
c" - ac
—
+
d^
- ad
4
a
2
—
+ 8 - ae
4
>
0
b) Dieu kien
: m > — ; m
5^ -1.

6
36
Theo
dinh
li
Viet
ta c6 :
S = + X2 =
P
=
XjXa
=
Do
do : (4x, +
l)(4xv
+ 1) = 18
2(m
+ 2)
m
+ 1
m
- 3
m
+ 1
o
16x1X2
+ 4xi + 4x2 + 1 = 18
<=>
16x,X2
. 4(x, .

X2)
- 17 = 0 «
16(m-3)^8(m.2)_^^^^
m
+ 1
m
+ 1
16(m
- 3) +
8(m
+ 2) -
17(m
+ 1) = 0
16m
- 48 + 8m + 16 -
17m
- 17 = 0
7
7m
-49 = 0 o m = 7
(thoa dieu kien
m
;t
-l;
m > — )
6
Vay
m = 7
thi phucfng
trinh

c6 hai
nghiem Xi; X2 thoa man
:
(4xi
+
l)(4x2
+ 1) = 18.
^-b
2
2
' a
2
'a
2
fa
^
+
.2
-
c
+
v2
-
d
+ e
I2 j
>
0
(bat
dang

thuTc
dung)
Do
do
a" +
b"
+ c" +
d-
+ e'- >
a(b
+ e + d + e) la
bat
dang
thufc
dung.
Cau
3.
a) Dieu kien x-\*Qoxi^\.
2x
Ta
CO
:
X'
+
o
X' - X- + 2x = 8x - 8
X
- 1
X''
- X- - 6x + 8 = 0

»
X'' -
2x''^
+ X- - 2x - 4x + 8 = 0
» x^(x
- 2) +
x(x
- 2) - 4(x - 2) = 0
o
(x - 2)(x- + X - 4) = 0 <=> (x - 2)
2 1
X
+ X + —
4
17
4
=
0
(X-
2)
X
+
—
2
2
17
4
=
0
x-2

= 0
if
X
+
—
2)
-11 = 0
4
37
X
= 2
1
Vl7
X
+ - = ±-—
2
2
<=>
X
= 2
±Vl7
- 1
(thoa dieu kien
x * 1)
X
=
Vay
phucfng
trinh
c6 ba

nghiem
la : 2;
Vl7
-
1 -Vl7
- 1
2 ' 2
b)*
Xet X = 0.
X
= 0
khong
la
nghiem ciia phifang
trinh
vi
0 -1.
*
Xet x*<d.
Chia
ve
trai ciia phuo'ng
trinh
cho x,
phucfng
trinh
trot thanh
4
5
x

-
•I
x-l,.l
X
X
=
-1
Dat
t =
X
- 10 + - .
Phircfng
trinh
trd thanh
:
X
c:>
4t + 5t + 10 =
-t'-
- 2t
t
+ 2 t
o
t-+
lit+10
= 0 c:> t =-1
hoac
t =-10
(vi
a - b + c = 0)

Vdi
t
=-1.
Ta
CO
: x-10 + - =
-l
o
X
o
Xi
=
7
X
9
+V53
X
- 9 + - = 0
X
Xv
=
9- V53
Vdi't
=-10. Ta
CO
: x - 10 + =-10 »
X
7
„
X

+ - = 0
cx>
X- + 7 = 0
Vay phirang
trinh
c6 hai
nghiem
la
Cau
4.
(Hinh
14)
a)
Ta
CO
:
AH
1
BC, CH
1
AB.
Tur
giac
BHCM
la
hinh binh hanh
c:.
BH
//
MC

va
CH
//
MB
CO
AC
1
MC
va AB 1 MB
« AM
la
difcfng
kinh
cua
(O)
^
<=>
M la
diem doi xufng ciia
A
qua
O.
b) AMB
=
ANB (tinh
chat
doi
xiJng
true)
<=>

X e 0
9
+
V53
9 - >/53
Hin/i
14
AMB
=
ACB (hai
goc
noi tiep ciing
chan
mot cung AB)
38
Cau
Suy
ra
ANE
= ACB
.
Ma
AHB
+
A'NB
=
180°
==. tir
giac
NAHB

noi tiep NHB
=
NAB
.
Ma NAB
=
BAM
(tinh
chat
doi xiirng true).
Suy ra
NHB
=
BAM
ChiJng
minh tuong
tu ta c6 :
CHE
=
MAC
B'AC
+ BHC = 180''
Do do
NTTE
= NHB + CHE + BHC = BAM + MAC + BHC
= B'AC
+
BHC
=
180°

=>
N, H,
E
thSng
hang,
NAE
= 2BM: ,
Ve
AK 1 NE, K e NE.
Ta
CO
:
AM
=
AN,
AM
=
AE (tinh
chat
doi xufng true).
=> AN
= AE
AANE
can tai A, ma
AK
la
duo'ng
cao
iz>
AK vCra

la
phan
giac,
vi^a
la
trung tuyen
==> ANE
=
2NAk
,
NE
=
2NK
do do
B^AC
=
N'AK
AKAN
CO
K = 90" nen NK =
AN.sinNAK
Do
do
NE
=
2AM-sin B'AC
Vi
AM
< 2R;
sin

BAC
khong doi.
NE
16-n nhat
CD
AM idn nhat
co
AM
la
dudng
kinh
cua (O) <=> M la
diem doi xiing ciia
A
qua
O.
Vay khi
M la
diem doi xufng ciia
A
qua
O
thi
NE
16'n nhat.
5.
(Hinh
15)
Ta
CO

(0)
tiep xiic AB, AC \An luqt
tai
H,
K.
S
'AMN
=
S
'0.'\
S0 =i0H.AM40K.AN
=
^^^^^^^
2
2
Ve MI
i
AC,
I 6
AC.
Ta c6 :
AM
>
MI
Ap
dung
bat
dang thUc Cosi
cho hai s6
khong am,

ta c6 :
AM
+AN
> VAM.AN
Do do
s,vM^:
>
VAM.AN
>
VMI.AN
Hinh
15
SAMN
= -
MI.AN
Ml. AN
= 2SM
AiMN
39
Ta CO
: S
AMN
AMN
SAMN
^ 2S
AMN
Dau "=" xay
ra o I = A <:=.
MN
1

OA, BAG
=
90°
.
Vay
gia
tri nho nhat ciia dien tich tani
giac
AMN
la 2.
DE
10
TRl/dNG
PTTH
CHUYEN
LE
HONG
PHONG,
TPHCM
De thi
tuyen
sinh vdo
I6p 10
chuyen Todn nam hoc 2001
-
2002
Cau
1.
Tim
tat ca cac

so' nguyen
x
thoa
: x"*
+
8 =
VVSx
+ 1 .
Cau
2.
Cho
n la so
nguyen duong. Chufng minh
ta
luon
c6 bat
dang thufc
12
3 n 3
-
+
—
+
—
+ +
— <—.
3"
4
3
3^

•
3^
Cau
3. Tim
tat ca cac
so' thufc
x
thoa
:
t/(x
-
2)(4
-
x)
+
Vx
- 2 + -^4 -
X
+ 6x>/3x < x^ + 30.
Cau
4. Cho
cac
so' thufc duong
a, b, c
thoa
a" + b" + c" = 27.
Tim
gia
tri nho nhat ciia
S = a' + b' + c'.

Cau
5. Cho tam
giac
ABC
c6 ba goc
nhon.
D la
mot diem tren doan
BC.
Dat
BC = a,
CA
= b,
AB
c,
AD
=
d,
DB =
m, DC
= n.
1.
ChiJng
niinh
: d"a =
b"m
+ c'n -
amn.
2. Cho biet AD
la

phan
giac
trong
goc
A. Chufng minh
:
AD"
<
AB.AC.
BAI GIAI
Caul.
Di§u kien
: 8x + 1 > 0 c=. 8x >-1 <=> x>
8
*
Cdcli
7 ;
Do
X £ Z
nen
x >
0.
Ta c6
:
x* + 8 = 7V8x + 1
c:. x''
+
16x'
+ 64 =
49(8x

+ 1)
(x"
+
8)'
= 7V8x + 1
» x'
+
16x'
- 392x + 15 = 0
c=.
x" -
3x'
+
Sx'
- gx" +
gx'
-
27x'
+
43x'
-
129x'
+
129x'
- 387x
-
5x + 15
«
' x'Hx -
3)

+
3x''(x
- 3) +
9x'(x
- 3) +
43x'(x
- 3) +
129x(x
-
3),-
- 5(x
-
3)
;
<=>
(X
-
3)(x'"
+
3x'
+
9x'
+ 43x- + 129x - 5) = 0
40
X
~ 3 = 0
x^
+
3x*
+

Qx^
+
43x2
^
i29x
-5 = 0
X
= 3
x^ + 3x'* + 9x3 +
43x2
+ 129x - 5 = 0
Ta thay
x = 0
khong
la
nghiem ciia
(*)
vi
: -5 0.
x?^0tac6x>0vax€
Z nen x > 1.
Do do
:
x"'
+ 3x'' + 9x' +
43x^
+ 129x > 5
=>
(*) v6
nghiem nguyen

khac
0.
Vay phifong
trinh
chi c6
mot nghiem nguyen
la x = 3.
Cdch
2 ; x > 0; X
G
Z.
Vdi
X = 0; 1; 2; 4
dang thufc khong thoa.
V<Ji
X = 3
thi dang
thiJc
thoa.
V6i
X > 5. Ta
CO
: 7V8x + 1 < 7V8x + x =
2lVx
< 21x <
x^.x
= x^ < x^ + i
Vay phuong
trinh
da cho

chi
c6
mot nghiem nguyen
la x = 3.
m
+
0,5 m
+
1,5
Cau
2.
V6i
m
nguyen duong,
ta c6 :
m
2.3
m-l
2.3"
Thay
m
Ian lUOt bdi
1; 2; m. Ta c6
1 1,5 2,5
3^2 2.3
2
2,5 3,5
32
2.3
2.32

n
+
0,5 n
+
1,5
3"
2.3
2.3"
1
2 3 n 1.5 n
+1,5
1,5 _ 3
Do do
:
—
+
^
+ —r +
•••
+
— = < ~
3
32
3"
2.3"
(X
-
2X4
- X) > 0
X

- 2 > 0
4
- X > 0
X
> 0
2
< X < 4.
Cau
3. Dieu kien
Ap
dung
bat
dang thufc Cosi
cho
hai
so
khong am,
ta c6 :
1
• ; ,
/x
- 2
+
4 - X
V(x-2)(4-x)
=
VV(x
- 2){4 -
X)
< J ^ = 1

X-2 +
1
—
+1 ,
2
_ X +1
41
4
- X + 1
2
2 4
QxSi
=
2Vx'l27
<
x''^
+ 27
Do
do
:
V(x
-
2)(4
-
x)
+
^(x
-
2)
+

^4
^
x
+
GxVs^
<
x^
+
30
Vay
2 < X < 4 la gia
tri
cin
tim.
Cau
4.
* CdcVi
7 ;
Taco:
(2a + 3)(a - 3)" > 0 <=• (2a +
3)(a^
- 6a + 9) > 0
<=:>
2a' - 12a- + 18a + 3a- - 18a + 27 > 0
CD
2a'' - 9a- + 27 > 0
Dau
"=" xay ra c=> a = 3
TLfOiig
tu, ta CO : 2b'' - 9b' + 27 > 0, 2c' - 9c' + 27 > 0.

Ma
a' + b' + c' = 27 (gia
thiet)
Do
do 2(a'' + b'' + c'') - 9(a- + b' + c') + 27.3 > 0
c=>
2(a''+
b'+ c'') > 6.27 » a''+ b''+ c''> 81
Dau
"=" xay ra o a = b = c = 3.
Vay
gia
tri alio nhat ciia
S = a' + b' + c' la 81.
* Cdch
2 : Ap
dung
bat
dang thiJc
Cosi
cho hai so
khong
am, ta c6 :
a'
+ 9a >
2Va''.9a
= 6a^
a-
+ 9>2V^ = 6a CD -(a'- +
9)>9a

2
3
Do
do : a' + -
(a''
+ 9) > 6a' c=> 2a'' - 9a' + 27 > 0
Tuong
tu
nhu'
each
1,
giai tiep.
* Cdc/i
3 : Bai
toan
phu :
ChiJng minh rang
i
a,b, + avb, + a,b.,
|
< ^(af + a|
+a5)(bf
+ b^ + hj)
(Bat dang thiJc
B.C.S)
Dau,"="xayra
« ^ ^ ^1 ^ ^
bi
b, bg
Dodo:

la.l
+ b.l +
c.l|
< V(a^ + b^
+c-)(l^
+1" + 1^)
=>
a + b + c <9
42
va
i'
+ b' +
c'
=
I
^fa'^
+
Vb=^Vb
+
V^V^
I <
<
V(a^
+ b^ + c^)(a + b + c) <
V9(a^
+
b^
+ c^)
Dodo
a''.+b'+

c''> 81
Dau
"=" xay ra <=> a = b = c = 3.
Vay
gia
tri
nho
nha't ciia
a' + b' + c' la 81.
Cau
5.
(Hinh
16, 17)
1.
Ve AH 1 BC, H e BC
*
Xet D n^m
tren duo-ng thang
HC.
AHAB
CO H = 90° ,
theo
dinh
li
Pitago
ta c6 :
AH'
+
BH'
=

AB'
=5
AH'
= c' -
BH'
AHAD
CO H = 90° ,
theo
dinh
li
Pitago
ta c6 :
AH'
+
DH'
= AD'
AH'
= d' -
DH'
Do
do d' -
DH'
= C' -
BH'
^
d' = c' +
DH'
-
BH'
^

d' = c' +
BD(DH
- BH)
=> d'n = c'n +
mn(DH
- BH)
ChiJng minh tuong
tu ta c6 :
d'm
= b'm +
mn(-DH
- CH)
Ta
CO : d'm + d'n = b'm + c'n +
mn(-DH
- CH + DH - BH)
d'(m
+ n) = b'm + c'n +
nm(-CH
- BH)
d'a
= b'm + c'n - amn
*
Xet D nam
tren duo'ng thang
HB.
Chufng minh tUo'ng tif tren
ta
cung
c6 : d'a = b'm + c'n - amn.

2. A'DC > ABC ( ADC la goc ngoai cua
AABD)
Do
do ve E
tren canh AC
sao cho
ADE
=
ABC
Ta
CO : AE < AC
Xet
AABD
va
AADE
C6 :
BAD
= DAE (AD la
phan giac),
A'BD
=- ADE
AD
AB
Do
do
AABD
o)
AADE
Do
do AD' <

AB.AC.
AE
AD
AD'
=
AB.AE
43
DE
11
TRl/dNG PTTH CHUYEN LE
H6NG
PHONG,
TPHCM
De thi tuyen sinh vdo Idp 10 (Ban A, B) nam hoc 2000 - 2001
Cau
1. Cho phuong
trinh
: x' - 2(m + l)x + m" - 4m + 5 = 0 (c6 an so la x).
a) Dinh m de phiidng
trinh
c6 nghiem.
*
b) Dinh m de phuong
trinh
c6 hai nghiem phan biet deu difOng.
Cau
2. Giai cac phuong
trinh
va he phiTcfng
trinh

:
X
-y = 3
a) x'' + 3x' + 3x + 1 = 0 b)
- y^ = 9
Cau
3. Vdfi a > 0, b > 0, c > 0, hay chiJng minh cac bat dSng thiJc sau :
, ab be " , , ab be ca
a) — + —>2b b) — + — + —>a + b + c
c a cab
+ b^ b^ + c^ c^ + a^
c) + + >a + b + c.
2ab 2bc 2ca
Cau
4. Cho diTong
tron
(O; R) c6 dudng
kinh
AB va mot diem C bat ki
thuoc diJorng
tron
khac A va B. Goi M, N Ian lugt la
trung
diem ciia
cung nho AC va CB.
a) Ke ND vuong goc voi AC (D thupc AC). Chufng minh ND la tiep tuyen
ciia
duofng
tron
(0).

b) Goi E la
trung
dii'm ciia doan BC. Dudng thSng OE cAt diTong
tron
(0)
tai
diem K (khac diem N). Chufng minh
tiif
giac
ADEK
la mot
hinh
binh
hanh.
c) Chufng minh rang khi C di chuyen tren dudng
tron
(O) thi MN luon
luon
tiep xiic voi mot duang
tron
co
dinh.
Cau
5. Cho hai tam
giac
ABC va DEF c6 cac goc deu nhon va c6
ABC = DEF, BXC = EDF , AB = 3DE.
Chufng minh rang ban
kinh
dudng

tron
ngoai tiep tam
giac
ABC bang
ba Ian ban
kinh
dUcJng
tron
ngoai tiep tam
giac
DEF.
BAI GIAI
Cau
1.
a) X" - 2(m + l)x + m' - 4m + 5 = 0
A' = (m + l)''' - (m'- - 4m + 5) = nr + 2m + 1 - m'^ + 4m - 5 = 6m - 4
44
PhifOng
trinh
c6 nghiem «• A' > 0 c?.
<=> 6m > 4 o
6m
- 4 > 0
2
m
> - .
3
b) Phi/ong
trinh
c6 2 nghiem phan biet deu duong

a ^ 0 1*0
j^-
> 0 6m - 4 > 0
S > 0 ^ l2(m + 1) > 0
P > 0 m'^ - 4m + 5 > 0
6m
> 4
c=>
<!
m + 1 > 0
(m-2)^ +1 >0
2
m
> —.
3
Cau
2.
a) x'' + 3x'- + 3x + 1 = 0 (x + l)' = 0
Vay phirong
trinh
c6 nghiem la -1.
x
= -1
b)
x
- y = 3
X
= 3 + y
27
+'27y

+ 9y^ + y^ - =
.2
, o,. o _ n iy = -1
hoac
y = -2
x
= 2; y = -l
x
= l; y = -2
X
= 3 + y
- y^ = 9 ^ [(3 + y)^ - y^ = 9
fx = 3 + y fx = 3 + y
^ ly2 +3y + 2 = 0
« Vay he phuong
trinh
c6 hai nghiem la : (x = 2; y = -1) va (x = 1; y = -2)
Cau
3.
a) Ap dung bat dang thufc Cosi cho hai so duong ta c6 :
ab be
+ — > 2b.
c a
be ca
— + — > 2c.
a b
ab be „ ab be
_ + _ > 2
J
C5

c a \ a
b) ?^ + — >2h
(theo
cau a)
c a
ab ca
Chufng minh tifong tu cau a) ta co : —- + — > /a ;
^ , ab be ab ca be ca , „ „
Do do: — + — + — + — + — + —>2b + 2a + ^c
c a c b a b
ab be ca ^ , „
<;:> — + — + — >a + b + c.
cab
c) Vdri a, b > 0. Ta c6 :
(a + b)(a - b)'> 0 (a +
b)[(a'-
ab + b") - ab| > 0
<=> a'' + b'' - ab(a + b) > 0 c:> a'' + b' > ab(a + b)
*
a + b
a^_+b^
2ab
Tiro-ng tir ta c6 : """" > -^-^
2bc 2
+ ^ c + a
2ca " 2
n
i'
a3+b
Do do : +

2ab 2bc
c+a a+b b+c c+a
> + +
<=>
+h' h' + + a^
2bc
2ca 2 2
> a + b + c .
2ca
2ab
Cau 4. (Hinh 18)
a) ACB = 90° (goc noi tiep chdn nijfa difong tron)
N
la trung diem cung BC =^ ON ± EC.
Ma AC 1 BC.
Suy ra AC // ON.
DN
1 AC (gia thiet)
Do do DN 1 ON.
=> ND la tie'p tuye'n cua dudng tron (0).
b)
N la
trying
diem cung BC,
E la trung diem day BC (gia thiet)
=> O, N, E thang hang.
Hinh
18
Tii
giac

CDNE c6 CBN = DNE = DCE = 90° nen la hinh chuf nhat.
Goi I la
giao
diem cua DE va CN.
Taco:
IE = IN => AlEN can tai I lEN = INE.
Hinh
thang ACNK (AC // ON) noi tiep dudng tron (0) nen la hinh
thang can =- INE = AKN .
Taco:
AKN = lEN (=
INE)
^ AK = DE.
> tiJ
giac
ADEK la hinh binh hanh.
, AK // DE, AC // ON
1
-
1
c) sdMN = sdMC + sdCN = - sdAC + -
sdBC
= 90°
1
2 2 j
MN
la
canh
ciia hinh vuong noi tiep (0; R) '
RV^ ' _ ' _J

OH
=
(H
la hinh chieu ciia O tren MN)
Do do
MN
luon luon tiep xuc voi dudng tron co dinh tam O, ban
kinh
s
RV2 j
46
Cau 5.
(Hinh
19, 20)
Goi (O; R), (I; r) Ian luot cac duong tron ngoai tiep cac tam eia^
ABC, DEF.
AABC ^
.'\DEF
ACB = DFE
ACB,
DFE nhon nen :
ACB = - AOB ,
2
DFE = -
DTE
.
2
Suy ra AOB = DIE
OA = OB (= R)
ID

= IE (= r)
Do do AOAB ^ AIDE
Hinh
20
Hinh
19
AOAB can tai O.
AIDE
can
tai
I.
OA _ AB R ^ 3DE
ID"
" DE ^ r ^ DE
R
= 3r.i
DE
12
TRUONG
PTTH
CHUYEN
LE
HONG
PHONG,
TPHCM
De thi
tuyen
sinh
vao Idp 10 chuyen
Toan

nam hoc
2000
- 2001
Cau 1.
ChiJng
minh :
A
+ a + B + b
B+b+C+c
C + c + A + a
C+c+A+a+b+d
A+a+B+b+c+d
B+b+C+c+a+d
trong
do A, a, B, b, C, c, d la cac so diTcfng.
Cau 2. Cho X,
y>Ovax
+ y<l. Tim gia tri nho nhat ciia bieu
thiJc
:
1
2
A
=
+ 4xy.
x^ + y- xy
Cau 3. Mot so nguyen duong N c6 dung 12 uac so'
(daong)
khac
nhau ke ca

chinh no va 1, nhUng chi c6 3 Ud'c so nguyen to
khac
nhau. Gia
siJf
tong
ciia cac Uo'c so nguyen to' la 20,
tinh
gia tri nho nhat c6 the c6 ciia N.
Cau 4. Tim tat ca cac so' nguyen x, y, z thoa phuong
trinh
:
3x- + 6y' + 2z' + 3y-z- - 18x - 6 = 0.
Cau 5. Dien tich tam
giac
ABC la 1. Goi A,, B,, C, Ian luot la trung diem
BC,
CA, AB. Tren cac doan
AB,,
CA,,
BC, Ian lugt chon cac digm K, L,
M.
Tim gia tri nho nhat ciia dien tich phan chung cua hai
.AKLM
va
AA,B,C,.
47
BAI
GIAI
Cau 1. Bai
toan

phu : Cho 0 < x < y, z > 0.
Chdng
minh
rang
:
Vi
0 < X < y, z > 0 => zy > zx xy + zy > xz + xy
y(x
+ z) > x(y + z) =>
A
+ a + B + b
X
+ z X
>
—.
y
+ z y
X
+ Z X
>
—.
y
+ z y
Ap
dung
bai
toan
phu ta c6 :
Tircfng
tu :

A
+ a
A+a+B+b+c+d A+a+c+d
B+b+C+c
C
+ c
Ma
A
+ a
C
+ c
B+b+C+c+a+d C+c+a+d
A
+ a C + c
A+a+c+d
C+c+a+d C+c+A+a+b+d C+c+A+a+b+d
Do
do :
A+a+B+b ^ B+b+C+c C+c+A+a
A+a+B+b+c+d B+b+C+c+a+d C+c+A+a+b+d'
Cau
2.
a, b > 0
theo
bat
dfing
thiJc
Gosi
cho hai so
du'Ong

ta c6 :
a + b
>
Vab
ab (a + b)^
1
A
=
(1)
(3)
1
x^
+ y^ 2xy
(a +
b)"^
>
ab
i
+
l J-
a b a + b
(2)
Ap
dung
(1), (2), (3) ta c6 :
4xy
+
4xy
^5
4 xy

x^
+ y^ + 2xy
>
4 + 2 + 5=11
-
+ 2
j4xy.
+
2 +
4xy
4 (x + yf (x + yf (x + y>
A
> 11. Dau "=" xay ra x = y = -
2
Vay
A dat gia tri nho
nhat
la 11.
Cau
3.
Goi cac
irdc nguyen
to cua so N la p, q, r va p < q < r.
q
= 5; r = 13
q
= 7; r = 11
=>p = 2; q + r=18=^
Vdi
a, b, c e N va (a + l)(b + l)(c + 1) = 12.

N
= 2^5^13^
N
=
2^7^lr•
48

, 12 = 2.2.3
't)o
do N
CO
the la : 2 5.13; 2.5 13; 2.5.13^; 2".7.11; 2.7 11; 2.7.11'.
N
nho
nhat
nen N =
2 5.13
= 260.
CaU 4- D§
thay
z"
chia
het cho 3 => z'
chia
het cho 9.
•
Xet z- = 0. Ta c6 3x- + 6y- - ISx - 6 = 0
o
3(x - 3)-+ 6y-= 33 (x - 3)" + 2y-= 11
2y-<ll

=> y" < 2- => y"* = 0^ 1'; 2". Ta Ian
lifcJt
xet :
y-
= 0 ==> (x - 3)' = 11. V6 li !
y-
= 1- ^
(X
- 3)- = 3- => X - 3 = ±3 => X 6 hoac x = 0
Co
cac
nghiem
(x = 6; y = 1; z = 0); (x = 6; y =
-1;
z = 0);
(X
= 0; y = 1; z = 0); (x = 0; y =
-1;
z = 0).
y'
= 2' =>
(x-3)'
= 3.
Voli!
'et
z- > 9. Ta c6 3x- + 6y- + 2z- + 3y-z- - ISx - 6 = 0
C5
3(x - 3)- + 6y' + 2z' + 3y^z' = 33
thi
2z^ + 3y-z^ > 2.9 + 3.1.9 > 33

(loai)
thi
3(x - 3)- + 2z' = 33
thi
3(x - 3)- = 15
(loai)
=> z' > 6' = 36 Ta c6 : 3(x - 3)" + 2z- > 33
(loai)
Nghiem
nguyen ciia phucfng
trinh
la : (x = 6; y = 1; z = 0);
(x
= 6; y =-1; z = 0); (x = 0; y = 1; z = 0); (x = 0; y =-1; z = 0).
Cau
5. (Hinh
21)
Goi
S la
dien tich phan chung ciia
hai
tam
giac
KLM
va
AiBjC].
Ta
CO : AjB, // AB
BC,
z^

= 9
z'
> 9
All
BM ^
——
< < = 1
IH
MA CjA
A,l
< IH
•hijfng
minh
tuong
t\i ta cd :
'HB,E
<S
HEG
'
^GC,N
<s,
GNF
•
Do do : SFA,I +
SHB,E
+ SGC,N ^ ^vm + S
HEl
;
+S,
'GNF

<s
SAJBIC, - S < S
ma
'A,B,C,
I
4
49