Bộ đề tự ôn thi violympic môn Toán lớp 8 Tập 1

NGUYEN HA cHAU . NGUYCN DANG cUdNG
ueuy€N NGgc oAM - uE rHdrue runAr

r$ sAcH vIoLYHPIc

TAp m0t

Nsn xuAr anru crAo nuc vlEr NAM


e"-(ffi

Id, Nor oAu
Cdc em hqc sinhyAu qu;t' !

Cu\c thi gidi todn qua Internet (ViOlympic) ld cuQc thi qudc"gia do B0
Gtdo dryc vd Ddo tqo td chttc tir ndm hgc 2008 - 2009. VOi hinh thfic thi
moi lq nAn nhiiu em cdn bonglkhi tham gia cuQc thi, d4c bi|t khi ngdi tru6c
mdn hinh caa mdy tinh. Dd tqo didu hen cho cdc em ldm quen trufu khi rin
luyfut vh thi qua Intentet, Tidu ban nQi dung cfia Ban td chttc cdp Qudc gia
biAru soqn b0 sdch

"Tty luyAn ViOlympic".

Til ndm hpc 2009 - 2010, lu\c thi sC gdm 35 vdng thi, mdi vdng thi di theo
mdi tuin hgc ffAn lop. Cdc ek'cb thd sir dung cuon sach ndy dafui sty hwong
ddn cria thfii cA gido hoqc phtl.huynh dd rin luyAn min Totin. I{hi s* dryng,
cdc em kh|ng nAn nbng v\i md cdn theo fuing tidn dp chuucg trinh furrc hoc
trAn lap.

Dd thudn lqt cho cdt em, vdi mdi lap, sdch dwqc chia ldm hai tdp ftng vhi hai

hpc ki. Cdc em cdn d7c kt phdn Hwang ddn srt dqtng scich dd ldm cdc dqng bdi
thi cho dilngyAu cdu. Cdc em cd thd tO mdu cdc hinh ve theo y minh dd cudn
sdch trArug hdp dnn hon.

IAi

hyAn xong mdi vdng thi trong cudn scich niy, cric em bdt ddu truy cQp
Internet, vdo dla chi : www.violvmpic.vn dd tham *a dry thi. Indy, cdc em cdn dpc ki phdn Trq giilp vd ldm theo hwang dkn dd cd thd thi
tdt. Cdc em chwa ddng ki thinh viAn thi phdi ddng ki thdnh viAn. Cdc em dA
ld thdnh viAn thi cd quyin ddng nhQp dd dry thi.
Cdm an Cong ty Cd phdn Sdch ddn tlc

-

da tqo diiu ki|n dd cudn sdch thi dwgv ctic
Cdc em hay

g*

cdc

y

kidn gbp

y

NhA xudt bdru Gido dttc ViCt Nam
em.

cho cdc tclc gid, qua

dia chi thw dign tt? :

hoqc yt clua Brru di\n vd dia chi : Dt dn
ViOlympic, Visly, FPT -'fdng 12, Tia nhd FPT, Cdu Gidy, Hh NQi.

Chric cdc em hpc mhn Todn nghy ckng say mA han, tidn b0 han vd thdnh dqt

trong hec tQP

!

cAc rAc

cn


I. HUONG DAN CHUNG

Coc em cdn dgc ki y6u cdu c0o tung bdi thi, dgc th6m hu6ng d6n o
muc ll d6i voi mOi s6 hinh th0c thi c6 y6u cdu ph0c top,

D6i voi coc hinh thUc thi dung bong (Tim duong trong m€ cung,
Chgn cap bdng nhou, Sop x6p), cu6n soch s0 drJng quy tdc ddm dd
xoc dinh vi tri 6 trong bong lo t& ttdi sang phdi, l& tr€n xudng dudi
D6i v6i coc hinh th0c thi: Chgn ddp on dlng, Di6n k6t qud voo
ch6 tr6ng, Di6u khidn xe vugt chuong nggivQf, Hoon thiQn phep tinh,
coc em ldm boitruc tiSp trOn d6,

u. HU6NG oAru rr,rQr

s6

Hir.rx TH0c THr

I. Tim dudng lrong mE cung
o Buoc

l:

Tim

duong ditrong m€ cung cho

Th6

Dung b0t v6 mOt duong di quo c6c 6, xudt phot tu vi tri 6 co Tho d5n
vitri 6 chUo c0 cd r6t. ChO'i: KhOng duoc v6 voo 6 c6 tuong ch6n Q

Vidu:

ffi
oJ&
ou@Jf-l{]i,- gol-l

ffi

o-Y&

E [J@.1

DEI-]CACT-jpW O Flf-l
B8][][][]EJf*tQ]r*J E E[]
DLDET][Jf]I'-]I--J [J Dfl

LQI[]qJncoElql
EJ[JC[]C@]r-Jl--lr-l

r*l Eli*l
EJ

EE]

: relrll
Er"-JFTJE

E][JEJL

[,]DtJL
goEq,
c@luEI

r-Jar-lnr-lrjE]NlEl u ry]DTJENJE
clj9jEjgr-j.))) E EJCEI*JZDCGrlr-l"lrjfft.)w)e)g QJ
OAE]E
Bio
>ar
re)

\*J

*


a

Buoc 2: Gidicdc bditodn gdp phoitrOn dudng didA chgn

Xoc dinh vitri coc 6 trong bong ch0o coc bditoon tr6n duong di dO
v6, Trong hinh v6 tr6n, cqc bqi toon gdp phoi tr6n duong di do v6
theo th0 tu ld:
Bdi to6n 6 sd:

-

ti:HLLL:

Gi6icoc bqitoon trong coc 0 do x6c dinh r6ididn k6t quo voo bong
tro loi,

W dt1:

Bditoon 6 s6 l4:Tinh gio tri cOo bidu thric x3 + 3x2 + 3x + I khix = 9.
Bditoon 0 s6 75: Nggn
A, Chomo
Boi

toon

O

Lonzo

nOi
B,

noo duoc aoi ld "n6c nho cuo thd gioi"?

Soltoro

s6 l4 c6 dop sd lo

Kongri

.l000;

C, Urol

D. Everest

boi toon 6 s6 75 (dgng chon dop on

d0ng) c6 dop on d0ng lo D. To sd didn voo bdng tro loi nhu sou:

Bdi todn o sd:

2. Chon

.

cf,p bdng nhqu

Buoc /; Tinh nhdm hodc lom tr6n gidy nhop dd co k6t quo coc ph6p
tinh, boitoon d tdt co coc 6,

o Budc

2; Didn

coc cQp 6 ch0o phep tinh hodc boi toon co kdt quo

bdng nhou voo bong tro

loi,


Vi

d1t:D6ivoi bong ph6p tinh hodc bditoon

372+2.37.13+132

X5c dinh oi5 tri c0a

352+242-48.35

x2'- ziy * i2

tgix=75,y=l$

X5c dinh gi5 tri c0a

X5c dinh gi5 tri c0a
9x2 + 42x+ 49

x2-10x+26+
+y2 +2y

tqix=11,Y=7

thdy O s6 I vo O s6 7 chOo phep tinh vo boitoon cO kdt quo bdng nhou
(bdng 2500); 6 s6 9 vo 6 s6 l2 ch0o hoi boi toqn c6 kdt quo bdng nhou

To

(bdng

100);.,,

i-,1

E"rE

".(7

3. Siip xdp

o Budc

/; Tinh nhdm hodc lom tr6n gidy nhop dd

co kdt quo coc boi

toon o tdt co coc O.
o Bu6c 2; Di6n s6 xoc dinh vi tri O ch0o ph6p tinh hodc boi toon c6 kdt
qud tdng ddn voo bong tro ioi,
W dt1:D6iv6i bdng sou:
Tim nghiEm
c0a da thfc

x-4

g

-4

I
$
L

rrm gra tri

nn6 nhdt c0a

-

g:1'. r9

Kdt quo di6n vclo b ong tro loisO lo:

ffi

Tim gi5 tri c0a

thol mdn
0,5x+2=5

x

f n,it gon bidu thrlc
$ (, - 1)(x + 2)L-- _-,(,11l


wHrw

YEN

Vronca

sO tn* tu cta c6c 6 ch&a c6c s5, ph6p tinh, bii to6n trong
bdng sau theo thti tr,r gi6 tri c0a cic s6, ph6p tinh vi k6t qui bii
toin trong c6c 6 d6 ting din:

Hiy viiit

5+(-3)-8

crla da thrlc

l_.

&-r*:=::,::3i*j:+4rrr,*$

R0t ggn bi6u thr?c

(x-1Xx+2)-

I

-(x+1)x

t.."*,____"

..tr-€6 I

X6c clinh gii tri
cOa x th6a min:

-l)(x + 2) -x2 =5

X6c ilinh giA tri
crira bi6u thrlc

(a+b-c)(a-b+c)
t?i a=*1

*
nh6 nh6t c&a

-9+(x-1

n-*E

,b=3,

rfmtTaTi**"'""

c0a bi6u thuc

4x-2(2x-5)

tri
thrlc

Tim oiA
c0a biSu

,rlTfiiill"n
ro-t-4-(-e)I
Tim gi5 tr!
l0n nhAt c0a

17- (x -

10-[s+(-e)l

i,9*
,.

,Sa

\ \:J

Thti

tgr

3)2

cha cdc 6 sdp x€p theo y)u cdu ld:

I
I

|

6+JE
Tim oi6 tri
nn6 nn-at c0a
X5c dinh oi5 tri
cia bi6u thOc

n5+(n+1)

t?in=-1
ldn nhSt cria
-17 - (x -3)z


I

,.rlra:t\
$,,nr ]

Em

hiy chon d6p 6n dtng cho m6i ciu du&i diy bing cich t6 ddm

vio hinh trdn "O" tru0c m5i aap
Tich (2x

1.

6n d6.

- 3Xx - 2) bing:

-o
Cc.zx2-7x+6

Cn.

Voi mqix thi 6x

On. -z

3.

N6u x2 +x(6

-

Cn. 1
3

-2(3xOa. z

x5

-

1)

bing:

Cc

Co.

-+

=(x-1)(2-x) -2 thi x bing:
Ce. r
Cc -r

+

2x)

Tich (xa + x2y2 + y1(x

CR.

Cg. z*2 *o
Oo.zx2-7x-6

z*2

y5

-

Co

-4
3

y) bing:

CB. ,5 * yu

Cc. 15

Co.

lmot k6t qu6 khac)

5. N6u x2 - 3x = 0 thi tip cAc gi6 tri c[ra x ld:
On
Cc.

5.

to)
{o; g}

Cho hinh thang ABCD (AB

m
CR.

7.

= 9oo. Khi d6

rzso

oso

3xn-31xn

3*2n

A

CD). aict ADE = +so; EeD

i

bing:

Cc.

Oo.

roso

ooo

*yn*313xn-3 -

Cc.
yn

sso

Oo.

- t; beng'

oB.v'n
- y2n

= 600

rooo

= 1ooo, 0 -7oo, goc ngodi tqi dinh O ta soo tni 6

Ca.

*t y'*t)
-

CA.3*2n

Cc.

m

/

Ce. rso

Tu gi6cABCD co
bing.

Cn.

8.

Ca. {s}
Co. lmOt k6t qud khac)

CD.

3x2n

*

y2n

soo


9.

+7)-

Gi6tri c0ayth6a min (y-SXy

+sXy-

Cc.

Ce. -to

On. ro

(y

1)

+16 =0ld:

CD. mgi ye m

ro

10. Cho hinh thang vu6ng ABCD co: 6 = e = 9oo, AB = BC,

NEu AB = 5cm thi chu vi hinh thang bing:

CR.

Cg.

Cc.zo+5J7cm

Oi€n

XGt

qui thich hop vio ch5

zscm

Co.

zocm

eo

+zJ5cm

....... trong m6i cau sau:

1.

Voi x = -3, y =4 thi gia tri c0a bieu thuc 2x(3x- yXy

2.

Voi moi x thi (x + 3)(2x2

3.

Ba s5 tu nhi6n li6n ti6p md tich cia hai s6 Oau nh6 hcvn tich crla hai s6
cu6i t4 dcvn vi la ......... ; ......... ; ......... (vi6t theo thrir tL/ tu nh6 den lon).

4.

NOu lxl = 1 thi gi6 tri cOa bi6u thrlc A = (x + 3X3X
.......... hodc

5.

Tip c6c gia tri cia x thod m5n

6.

T?p cac gia tri cfra x th6a m6n (x - 1Xx -2)(x2 +2) =0 ld S =

7.

C5c goc cOa t(r gi6c ABCD co ti 16 A: B: C: D =
A=... .... ; 6=......... ; e=......... ; 6=........

8.

Cho hinh thang MNIQ (MN

-

-x) bing

5x + 1) bang

1x2

/

-1) + 2(x + 1) ldr

+ t)(x2 + 5) = O la S = .........

PQ). BiCt

1.2:3.

4

fr* F= 1590 va ft= 26.

NhuviYfr=...'....

g.

56tu nhi6n nth6a min 3xn-11rn+t *yn*t) -3*n-1rn+1=27 ld...-....

10. Trong trl giac ABCD thl AC + BD .... AB + CD

(dien ki hi6u so s6nh thich hcvp
', ., -).

vdr

AC + BD .... BC + AD


I

- a;.

i$,s

l

Em

hiy chon d6p in d6ng cho m6i cdu dudi diy bing c6ch t6 dim

vio hinh trdn "O" tru0c m6i Oap 6n d6.
11. Hi6u (15,6)2 - (4,4)2 bdng:

On.

Ce.

za

12. R0t ggn bi6u thrlc 2x(3 + 8x)

Cn. s,

Cc.

so

-

(4x

-

az

Co.

ra

0,5)2 ta duoc:

CB. 3* - 0,5
()D. sx + o,s

--0,2s

Cc. tox _ o,2s

13. Gie tri c0a bi6u th0c (x - 10)2 - x(x + 80) tai x = 0,97 ld:

Cn. z
Ce. -z
Cc. e
Co. s
M. Gia tri c0a x th6a min x + (3x + 1)2 + 1= 9(x2 + 1) ld:
Cn. r
Ce. -r
Cc. z
Co. -z
15. ChoX*!=7vAxy =12thi lx-yl bing:
On. -r
Oe. r
Cc. z
Co. g
16. Cho X - Y = -10 vi xY = -21thi lx + yl bdng:
Cn. -+
Ca. +
Cc. s
Coo
17. Cho ABCD ld hinh thang cin,
Nhu vQy dO dai canh CD bdng:

CR.

socm

Cg. gscm

D- 60o, AD = 20cm, AB + CD = 40cm.

Cc. zoc,

Cp.

zscm

18. Cho hinh thang cin ABCD ( B ll CD) co AB = 13cm, CD = 25cm,
D = 45o. Nhu viy diQn tich hinh thang ABCD bing:

Cn.

zocm2 Ce.75cm2 Cc. a+"r2

Co.

r 14cm2


19. Tu gi5c trong hinh vE
Nhu viy x bing:

b,On

lir mQt hinh thang cin.

Cn.r+t-y Cc.z+t+y
Cg.=-t-y Co. =-t-y
20. VOi moi xe
min:

On.v.o

?

*.' -r.
u

.ari

R thi bi6u thfrc M = (x2 +2)2 - (x Cg.NI

,o

2)(x +2)(x + 4) tho6

Cc.M=o

CD.u=zo

Hiy tim nhirng cip O trong bdng sau chua ph6p tinh, bi6u th0c
ho{c bii to6n c6 ket qua bing nhau:
Tim gi6 tri crla x bi6t:

-

'

(2x + 3)2
(2x +
- 1) =22

1)(

352

+242

Tim gi6 tri lon nhdt

cia bi6u thuc:

B=48-y'-2y


\ Cdc cdp 6 tim duqc theo y1u cAu

tdr:

T,,T; f". I ; fl

I".fI

;

I"rf];

(-l ; [,. [l
I"uI ; [ua(-l; f],a(-l
[ur(-l
,u

chu6ng ngqi vit
bing c6ch giii c6c bdrito6n o cic chu6ng ngai vdt d6:
Em

hiy

OiCu Xhi€n xe vuo-t qua c6c

Chu6ng ngai vdt

OC

1:

W

Tip c6c gi5 tri cOa x th6a m6n 4(x - 1)' - g(* + 2)2 = g 1;

,W

Chu6ng ngai vit 2:
Caps6(x; y)th6a manx2

*y' -4x-2y+5=0

la

( .;

)

vC Oictr


W
W
W
W
W

Chu0ng ngai vit

3:

Cho hai s6. A = (3 + 1)(32 + 1)(3a + 1)(38 + 1)(316 + 1) ; B -g32
Khido:A =........ B.

Chu0ng ngqi vit

1.

4:

Gia tri nh6 nhSt cira bi6u

Chu6ng ngai vit

-

thfc M = (x - 1)(x + 5)(x2 + 4x + 5) la .........

5:

Gi6 tri lcvn nh6t cira bi6u thuc P = *x2

- 4* - y' + 2y lit

Chu6ng ngqi vQt 6:
Cho tam gi6c cin ABC cin taiA. K6 hai duong phAn giac BE vdt CD.
Ta co hai doan thing co d6 ddi bing DE la ........;
Chu0ng ngai vit

7:

Cho hinh thang vu6ng ABCD, A = 0 = 90o, AB = AD, BD vuong goc vcri
BC. NOu AB = 5cm thi di6n tich'hinh thang ABCD bing .... ..


v.-@
q
l'*,\

Em h6y girip Th6 tim duong trong m6 cung OC Oen duo. c O c6 cu

:nffiejffin @J n EJ
[J trJEDDD eiryJ H
r@

m DLE] DEJ rmM ffi

ifli ffi8iruffi:ryi ruEj n
H trJ DDtrJD EJ[I [J
[J [J CDCEJ Dtr [j

n EJ D@] DtrJ mrm

@J

.

DDC[JD Nitrj n

trl DE][J DEJ
Biito6n 6 si5 z: Tinh gi6 tric0a
Biri to6n 6 sti t 2: Tinh &2 '

@N)

bAJLN

(vd

nn ilt + d^6
xA

bi6u thucx3

+3x2+3x + 1 khix = 9.

1\ua * 1r' * lt

3"

A
f{//1-<\

3

9''
Bdri to6n 6 sti tg: ROt gon bi6u thrlc (a + b)3 - (a - b)3.
Biri to6n o s5 ze: Bietx3 -6x2 + 12x -8 = - B. Tim x.
Biri to6n 6 sti gZ: Tinh gi6 tri c0a bieu thuc *3 - 27x2 + 2tx - 27 tai X = 1.
.


&

Biri

toin 6 s6 gg: Cho tam gi6c ABC, trung

D

Bdri

to6n

4-)

353 1133

Bdrito6n o si5 sg: Tinh gi6 tri cia bi6u thuc
-3s.13.

tuytln AM. Goi I ld trung
diOm cria AM. Tia Bl cit AC tai D. Bi6t BD = 12cm, Tinh lD.

\Y/

O

s5 +g: ROt gon bi6u thrlc (x - 1)3 - (*'

*x + 1)(x -

1).

48

o

Bdi to6n 6 si5 Sg: Cho tam gi6rc ABC. BC = Bcm. Ggi D vd E IAn tucyt ld
trung di6m c6c canh AB, AC, M vd N lAn luqt ld trung di6m c6c canh
BD, CE. Tinh MN.

@

Bdri

N\\

\

toin 6 s6 e+: Tinh gi6 tri c0a bi6u thrlc

oa3

- 523 + 68.52.

16

toin 6 sri zo: Cho hinh thang ABCD. phin gi5c cia A va 0 cit
nhau tai l, ph6n gi6c cia 6 va e cit nhau tai K. Ta co AiD = 6Re
Biri

bing bao nhi6u do?

Biri to6n 6 s5 zs: Cho hinh thang vu6ng ABCD, A = 0 = 90o. Tu trung
di6m M c0a canh CD ke MH vu6ng goc voi AB, MH cit eO tai t. Cho
biCt AD = 16cff1, MH - lH = 1Ocm. Tinh dO dai iloan BC.

Duong di em chgn gi(tp Thd de.vd d bdng tr€n. Cdc bai toan gap phdi
tr1n dwong Thd di qua vd dap s6 la:

g

.sg

Biti todn 6 s6;

, E, hdy chon d6p 6n d0ng cho m6i ciu du6i day bing c5ch t6 dim

I hinh trdn "O" truoc m6l aap in d6.
1.

R0t gon bi6u thuc (x

()n.

-oo

- 3)(x2 + 3x + 9) - (33 * x3; ta ducyc:
Ce. x3
Cc. oo
Co. -r3


2.

Gia tri crla

on.

3.
4.

thrlc (x + 4)(x2

tr

NOu (x

Cn.

birOu

Ca.

+2)(x2 -

-s

2x +

s+s

Cc.

t

tg

Co.

gr

4) -x(x2 +2) =15 thi x bing:

Ce.

g

Tipc6cgiatri c0axth6a min

Cn. s = {-1; o; 1}

Cc.s={o;

- 4x + 16) tai x = 3 ld:

Cc.

g,s

(*'- t)u-(*o*

CD. -g,s

x2 +1)(x2

- 1)=0ld:

Ce. s = {-1; oi

1i

Co.s=t-1;1i

5. Bieu thuvc (x+2)(x2 - 2x+a) + (3 -x)(9 +3x +r'1 b5ng,
Ce.

Cn. -so

6.

Cho a + b =

Cn.

go

Ca.

roo

=Z

ll CD,
AM=ME=ED,BN=NF=FC.
Khi do d6 dai doan CD bdng:
o

Ce.

z

Oc.

e

Coa

-

CP.

zoo

la:

tsg

Cc.

ga

(x +y)(x2

-

xy + y2) = o la:

Cg.r=y=o

O hinh b6n, AB // MN ll EF

Cn.

ss

Co.

-ea

dogi6tri ciaa3-b3

=y=1

Cc. * =1;y
9.

Cc.

gg

Cap s6 (x; y) th6a m6n x2(x + 3) +y2(y + 5)

Cn.,

Co.

*2vd ab = -15. Khi do gia tri c0a a3 + b3 la:

7. Choa-b--4vdab =-21.Khi
On -raa Cs. taa
8.

Cc. -ss

Co.

(mqt ket qud kh6c)

10. O hinh b6n, tu giac ABCD
ldr hinh thang, AM = MD, BN = NC.
Khi do x bing:

On.

()c.

Ce.
Co.

+

o

s
z

e

* t...
ri. i.

OiCn k6t

qui thich hop vdo ch5
+c

= o thi a3 +b3

........ trong m5i cau sau:

*c3 bang ........

1.

NOu a + b

2.

NOu a2

3.

4.

Tip c6cgia tri c0axth6a mbn {2x_ 1)3 _ 4x212x-3) =5la S=........
Tam gi6c ABC vuong tai A. Hai canh goc vu6ng ld a, b, di6n tich
S = 10,625. N6u (b +c12 =85 thi 6 =........ ; e =........

ldt

5.

*:::n.

ld

6.

Cho x +y = 3 vax2 +y2 = 5.Khido

7.

Cho x

8.

Cho tam giac ABC c6n tai A, duong cao AH, phAn giac BD, HE ll BD.

.1

-

BietAH
9.

+b2 +c2

biOu

+3=2(a+b+c) thi a=

thuc (a + b -c)2

y = 5 vdx2 +y2

-G

b=........, e=........

-c)2 -2ab +2bcta duoc ket qud

*t *yt bing

........

= 15.Khi do *u - yt bdng ........

= 2 BD. Nhu viy A bing
^

........

Cho x +2y =5. Khi do gia tri crja birSu thrrc

bing ......

10. O hinh b6n, ABCD ld hinh thang
vuong, M ld trung di6m cia BC.
Khi do: MAE ......... MDC (dien
ki hi6u so s6nh thich hop >, <, =).

*)

*

4y2

-2x + 1o +

4xy

-

4y


t*-.&
I

Hiy viet so thrl tu cua c6c 6 ch&a c6c ph6p tinh, bi6u th0c, bii

"'. to6n trong bdng sau theo thr? tu gi6 tri c0a c6c ph6p tinh, bi6u thrlc
^ vi kiSt qua bii toin trong cic 6 d6 ting din.
X6c dinh gi6 tri crla bi6Ll

thuc.

xt*6x'+12x-8

':

- 12x2y +
+ 6x2y - y3

Bx3

a2

X6c dinh gi6 tri cOa mn,

X6c d!nh gi6 tri cia

x3-3xy1x-y)-y3
-x2 +2xy -y2

I

:

r::::::::

t_. .
I

litliiri,r,i:

+3) = 22

-l)

(27a3 + 5)

R[t

4(x + 2)2

+ (x

-

1)2

ggn

[*.')[','

uinn n,,

z r)
--x +3 e)
|

gi6 tri c0a x,
3(x

-

- 3x(x -

1)2

-

5) =21

X5c dinh gi6 tri c0a

,

-

x'

+Y.2

- yt) - 3(x + y)2
Oiltx - y =Z
z(xt

=i

[

I

[

+tl-axtt- x)-3 I
!

*0, nnn
t)t - (x - 1 )(x2 +xl tz, il(r-Z)t-x(x+1Xx-1) &* $

:

-

(x+8)2-2(x+8)(x

-

I ^r. . xt+y'
I bi6tx+Y=

.,,,o':1.',-,1*:1' *-.-" fl
..,.
I rm gra Ifl cua x, OreI:

$t

- y')

-gx3

:l::l:::::l:1:

ir:-l,iriririlr

+ 1)'

R0t ggn:

+b2 =52.

(x+1)3-(x-1)t- 6(x - 1)2 = -10

25m2+4n2=120A.

+ 6x(x

gi6 tri c0a x,
+y21

R0t ggn:

..t.

5m-2n=30va

$

y(xo

lrm gra tn cua x, btet:

bi6t:

','r.!.::::lai:

-

y2y1x2

(3a+1)(9a2-3a+1)

a+b='l 0vd

tqix=!,Y=-2

tt}lliIi ini"i
.:r'riliili;ili

-

X6c d!nh gi6 tri cia
ab, bi6t:

thuc.

rji.tiilliffi

y1x2

:=u

c dinh gi6 tri crla bi6u

:,'rilill',i'rl

R0t'g9n:

[,r1')' -(',r 1\'
\ 4) \ 4)

"

$F

Rut son:

+2. + 1) -8(x3+1)

1)(4x2


Thg tur cla cdc 6 duo. c sdp x6p theo y1u cAu ld:

1.
2.
3.
4.

KOt

qud r0t gon cira (a + b)3 + (a

Rrjt gon bi6u thrlc (a + b)3

-

(a

-

b)3

-

-

6ab2 la ........

125

taix

=

-10 la ........

6a2b, ta clucvc ........

Trong tam gi6c ABC biet AX = XM = MB vd AY = YN = NC (hinh du0i)

-

........cffi, BC = ........cffi.

5.

Gi5tri coa bi6u,n*.

f .+.*

6.
7.

8.
9.
rLro{x t/l

b)3

Gi6 tri crla bi6u thrlc x3 - 15x2 +75x

N6u XY = 6ctrl, thi MN

r.

-

R0t gon bi6u thtlc (x

-

1)3

-

*{,r,X=-8, y=6Id ....

(x + 1)3 + 6(x + 1)(x

-

1) ta ducvc ........

Chox +y= 1. Khi do gia tri c&a bi6u thucx3 +y3 +3xy bing... . .
Cho x - y = 1. Khi do gi5 tri crla bi6u thrlc x3 - y3 - 3xy bing ........
N6ux2 +y2

-2x'4y+5=0thix=

y-

......


10. Cho hai s6 x, y kh6c nhau vd x2 - y = y2 - x. Khi do gi6 tri cOa bi6u
thuc x2 + 2xy + y2 - 3x * 3y la .......

?

4 a."':
.i*r

ri,l

hiy chgn tl6p 6n dfing cho m6i ciu du6i day bing cich t6 d6m

Em

'' vao vdng trdn "O" truoc m5i oap 6n d6.
1.

NOu 2(a2 + b2; = (a

*

b)2 thi ta co:

CA.a=b

Cc.a=-b

2.

()B.ab=1

Co.

lmOt k6t qud khac)

Giatri cira bi6uthuc(x +2)(x2 *2x+4) -x(x -3)(x +3) tai x=2td.
Cn. zo
Oe zs
Qc. zq
Oo. zs

3. N6u a, b, c ld dO dai ba canh crla tam gi6c ABC md
a2

+b2 * c' =ab + bc + ca thi tam gi6c ABC la

Cn.

Ce. fam

tam gi5c vuong

Co.

CC. tam giac d6u

4.

Neu (2x

Cn.

s.

-

1)3

-

- 3) = 5 thi x bdng:
Ce. -r
Cco

tn*c

Cn. zrs

6.

Gia

Cn.

tricia

lmOt k6t qud kh6c)

4x212x

r

Gi6 tri cta bi6u

gi6c cin

tI
{84627.t- 4

Ca. -zrs
bi6u thuc

-tzo

Co.

,u, X = 6, v =

Cc. zro

-e

s
td:

Co. -zro

lx6 -?*ov- *)*'y' -1vt taix = -3 vd y =ztd:
2 ', B',
27 3

Cs:

rzo

()c.

7. DUng hinh thang ABCD, biet AD ll

-rzs

Co. rzs

BC, AD = 6cm, AB = 3crt,
CD = 2,5cm vd A = 35o. VOi c6c d0 ki6n tr6n ta dung ducyc:

On. r hinh thang

Cc.

s hinh thang

Ce. z hinh thang
Co. xnOng dyng duoc


8. 0 hinh b6n, AM = MD = DB,
AN = NE = EC, NQ ll AB.
Bi6t MN = Scm thi d6
doan QC bing:

dai

CR. tocm

Ce.

Cc.
Co.

9.

gcm
ecm
zcm

Gi5 tri

35e78 +2.9111

cia bi6u thuc

15(816

Cn. -r

Ca.

r

10. O hinh b6n, ABCD li hinh thang
cin (AB ll CD), AC vu6ng goc
voi BD, MN la ducrng trung binh.
AiCt cnieu cao c[ra hinh thang ld
1Ocm. Khi d6 do dai dudng trung
binh MN nhQn gi5 tri ndo duoi

div:

CR.

ecm

Cg.

Cc.

tocm

Co. ttcm

gcm

td:

-e.g1e)

Ccz


v*-@
c

it^\
Em hdy OiCn t6t

qui thich hop vdro ch6 ....... trong c6c cdu sau:

1.

Phin tich da thr?c 8x(x - y) - 6y(y - x) thdnh nhin trl ta duoc .......

2.

KOt

3.

Gi6 tri c0a bi6u thOc 20,09.45 + 20,09.47 +20,Og.8 la .......

4.

TQp c6c gi6 tri crla x Oe S1x + 3)

- 2x(3 + x) = 0 la .......

5.

Gi5 tri cira bi6u thOc 15,75.175

-

6.

KEt qu6 ph6n tich da thuc (7x

4)2

7.

Tap hep c6cs6xth6a min (x -5)2

8.

Cho tam gi6c ABC, A = 50o. M ld mQt di6m tr6n canh BC (M kh6c B, M
khac C). Ggi D vd E lAn lucvt ld di6m O6i xung cria M qua AB vd AC.

qui phin tich da thrjrc ,u

*lI

-

thinh

nhAn tr? ta .......

15.75.55

-

-

15,75.2O la .......

(2x + t12 tnann nhAn trl la .......

-50

= 1 ld S

={.......;

i

Khi d6 oAE =..... . (d9).
9.

Cho tam gi6c d6u ABC cqnh 6cm vd m6t duong thing d nim ngodi

tam gi6c. DUng tam gi6c A'B'C' O6i xung voi tam gi6c ABC qua duong
thing d rdi vC Clucrng cao A'H' crla tam gi6c A'B'C'
Khi d6 A'H'= ........... cm

10. TAp hqp c6c

s6 x th6a m6n (x + 3)3

-

125

=o

la S =

{.......;

}


qi'rl

r- a

ri: iii

i

Em

hiy chgn d6p 6n d0ng cho m6i ciu sau bing c6ch tO dim

hinh tron

1.

"O" truoc
thfc

m6i

GI6p

6n d6.

- b)x + (b - a)y - a + b thdnh nhAn t0, ta ducvc.
Ce. 1a-bXx*y-1)
Cn. 1a-b)(x+y+1)
Co. (mQt k6t qud khac)
OC. 1a-b)(x-y+1)

2.

Phin tich

CIa

(a

Cap s6 nguy6n duong (x; y) thda m6n x(y + 1)

Ca. (z; t)

Cn. (r; z)
Cc. (r;r)

3.

Co.

Ket qud phin tich da thuc (a + b)2

Cn. 1a+b+c)(a+b-c)
CC. 1r+b-c+ab)c

4.

a

-

s+g

-.'thanh nhAn trl ld:
Ce. 1a-c+b)(a-c-b)

Co.(a+b-c-ab)c

Ce.

-s+s

o-c. sqz

Oo. -zqz

Cg.

a2

-s

Cc.a2-ga Co.a2+6a+9

+,scm

CB.

4,8cm

Cc.

scm

Co.

s,scm

Giatri cirabi6uthuc(2x+5y)(4x2-1Oxy +ZSy2)tai x=

Cn.

8.

kh6c)

Cho tam giac ABC vu6ng tai A, AB = 3crl, AC = 4cm. Kd AH vu6ng g6c
vcvi BC. Goi D vd E lAn luqt ldr diem d6i xung c[ra H qua AB vir AC.
Nhu viy dO dai doan DE bing:

OR.

7.

qui

3 ld mot nhAn trir cia.

CR.a+g

5.

(mQt ket

li

Cho x + 2y = -7. Khi do gi6 tri c0a bieu thrlc x3 + 6x2y +12xy2 + 8y3 la:

Cn.

5.

Y=3

rog

Oe. -tas

Gi6tri c0a bi6uthuc(x-!+5)2

Cn.

roo

Ca.

rot

Cc.

rag

-2,y=-1
Oo. -tog

ld:

-2(x-y+5)+1taix=5, Y=-1 ld:

Cc.

toz

Qo. ros


9.

Cho hinh thang vu6ng ABCD,
AD. CB' c5t RO tqi M. Nhu v6y:

Cn. elttD ,

A= 6=

AfrE

Cc ffi=AfrE
10. Cho tam gi6c ABC. Ve tia phdn
gi6c ngodi Ax crla goc A (hinh
b6n). Goi M ld di6m tiy f tr6n
tia Ax (M kh5c A). Khi do ta co:

90o. VO B'O6i xung voi B qua

Ce. eMD . ArraB
Co. lmOt k6t qud khac)
x

Cn.rua+MC>AB+AC
Ce.NIa+MCCc.vn+MC=AB+AC

Co.

lmOt k6t qud khac).

Em lay girip Tho tim duong trong m6 cung Oe OCn dugc O c6 c0
r6t vi giai c6c bii to6n in trong c6c 6 tr6n duong di di chon.

ci

OA
f,4-\

W
r]DDtrJDEJDDD dfu

EJ

DF]DN@]DE]M$

Hl.[]E]nHmm .n
npjfllmffinrumiE]
NEJCE] EtrtrE

@nnffi@JnnNl t3
nEJAUEln,Eiffi c
DDDE]Dt6]DtJ i[J

+$NE]D@]DEJDC im


-4.

EI

Bai

toin

O

s5 6: Tinh gia tri crla bi,3u thwc 27y3

-

27y2x + 9yx2

- x3 voi

1

v- -x.
3
Biri to6n o sii tg: X6c dinh ba chir s6 tan cung c0a s6 ggga +

Bii to6n

999.

tam gi€rc ABC c6 AB = 5cn, AC = 12cm, BC = 13cm
vi duong thing d nim ngoii tam gi6c. V6 tam gi6c A'B'C' d6i xung vcri
tam gi6c ABC qua dudng thing d. Dien tich tam gi6c A'B'C' ld s5 ndo
trong c6c s6 duoi dAy?

A.

O si5 ZS: Cho

35cm2

B.

7ocm2

c.

3ocm2

D. Gocm2

Biri to6n 6 sti zg: Cho hai s6 a, b > 0 vd a * b.
Bi6u thrlc M = (a2 *b\2

a\
,a,

A.M>0 B.M=0
Bdri

- 4a2b2 co tinh ch6t ndo ducri d6y?

C.M<0

to6n 6 s5 gz: Cho xa(x

-

1) + x3(x

D.(m6tk6tqu6khac)

-

1) = 0. X6c dinh tQp hcrp c6c

gi6 tri c0a x.

6

Bii to6n O sti ge: Cho tam gi6c dOu ABC canh Scm. Qua A k6 duong
thing d ll BC. V6 di6m D d6i xung v0i C qua d. Khi d6 d0 dai doan CD
bdng.......

D

Bii to6n

O sO +O:

Cho ducrng thing xy vd hai di6m
A, B (hinh b6n). VE di6m A' O6i
xung voi diOm A qua xy. A'B cdt
xy tai C. TrGn duong thing xy
l6y di6m M kh6c di6m C. Ki hiQu
chu vi tam gi6c XYZ lit CVxvz.
Khing dinh ndo sau diy lir d0ng?
A. CVa6s'CVnMe
B. CVa6s'CVnMe
C. CVass = CVRMs
D. (m6t k6t qud kh6c)
Bdri

tai

to6n

O

s6 +z: X6c dinh gia tri c0a bi6u tht?c (2x

x=3,y=2.

- y)3 + 3(2x - y) + 1

ffi;

.

llri.i:I,


Bii to6n 0 sii +e: BiCt (3x +y)3 -

3(3x +y)2+3(3x +y)

-1 =-27.Tinh

3x+y.
Biri toin O si5 SO: Cho tam gi6c d6u ABC, hai duong cao BD vd CE.
Gqi H ld giao di6m cOa BD vd CE. Khi do AH ld truc O6i xung c0a hinh
thang

.\

cin ndo?

Bii to6n 6 s5 sg: Hinh bon td mqt
hinh thang. Truc dOi x0ng c0a

\

hinh thang do

ld:

M

A. Duong thdng MN

thing
C. Duong thing
B. Duong

UT
PR

D. Ducrng thang SQ

t

E. (khong phdi cdc ducrng thing tr6n)
Bdri

to6n

O

sti SZ: Cho x + y = 3.

Xdc dinh gi6 tri cira bi6u thr?c x2 + 2xy + y2
Bdri

to6n

O

-

4x

-

4y + 1.

s5 eg: Cho a + b +c = 0.

Gia tri c0a bi6u thuc a3 + b3 + a2c + b2c

A.0

8.

1

- abc bing s6 ndo duoi dAy:

c. -1

Bii toin 6 sti zs: Cho x 3 + 27 + (x + 3Xx -

D.2
3) = 0. X6c dinh

tip c6c gi5

tri c0a x.

e,6

.(.,

t \*J

Duong di em chgn gi1p Thd de v€ d bdng tr€n. Cac
tr€n duong Thd di qua vd dap s6ld:

krs:i*

L

bdti

todn gAp phdi

L*t-*L- L-l-- L- L*L-L- Leic".: l--i-*i--*L-L*L- L--L--L*I--