50 de Hoc Sinh Gioi Toan 7

thi -lim -pic huyn

s 1:

thi hc sinh gii huyn

Mn Ton Lp 7

(Thi gian lm bi 120 pht)

Bi 1. Tm gi tr n nguyn dng: a)

EMBED Equation.DSMT4 ; b) 27 < 3n < 243

Bi 2. Thc hin php tnh:

Bi 3. a) Tm x bit:

b) Tm gi tr nh nht ca A = Khi x thay i

Bi 4. Hin nay hai kim ng h ch 10 gi. Sau t nht bao lu th 2 kim ng h nm i din nhau trn mt ng thng.

Bi 5. Cho tam gic vung ABC ( A = 1v), ng cao AH, trung tuyn AM. Trn tia i tia MA ly im D sao cho DM = MA. Trn tia i tia CD ly im I sao cho CI = CA, qua I v ng thng song song vi AC ct ng thng AH ti E. Chng minh: AE = BC

s 2:


Mn Ton Lp 7


Bi 1:(4 im)

a) Thc hin php tnh:

b) Chng minh rng : Vi mi s nguyn dng n th :

chia ht cho 10

Bi 2:(4 im)

Tm x bit:

a.

b.

Bi 3: (4 im)

a) S A c chia thnh 3 s t l theo . Bit rng tng cc bnh phng ca ba s bng 24309. Tm s A.

b) Cho . Chng minh rng:

Bi 4: (4 im)

Cho tam gic ABC, M l trung im ca BC. Trn tia i ca ca tia MA ly im E sao cho ME = MA. Chng minh rng:

a) AC = EB v AC // BE

b) Gi I l mt im trn AC ; K l mt im trn EB sao cho AI = EK . Chng minh ba im I , M , K thng hng

c) T E k . Bit = 50o ; =25o .

Tnh v

Bi 5: (4 im)

Cho tam gic ABC cn ti A c , v tam gic u DBC (D nm trong tam gic ABC). Tia phn gic ca gc ABD ct AC ti M. Chng minh:

a) Tia AD l phn gic ca gc BAC

b) AM = BC

Ht

p n 1ton 7

Bi 1. Tm gi tr n nguyn dng: (4 im mi cu 2 im) a)

EMBED Equation.DSMT4 ; => 24n-3 = 2n => 4n 3 = n => n = 1

b) 27 < 3n < 243 => 33 < 3n < 35 => n = 4Bi 2. Thc hin php tnh: (4 im)

=

=

Bi 3. (4 im mi cu 2 im) a) Tm x bit:

Ta c: x + 2 0 => x - 2.

+ Nu x - th => 2x + 3 = x + 2 => x = - 1 (Tho mn)

+ Nu - 2 x < - Th => - 2x - 3 = x + 2 => x = - (Tho mn)

+ Nu - 2 > x Khng c gi tr ca x tho mn b) Tm gi tr nh nht ca A = Khi x thay i + Nu x < 2006 th: A = - x + 2006 + 2007 x = - 2x + 4013

Khi : - x > -2006 => - 2x + 4013 > 4012 + 4013 = 1 => A > 1

+ Nu 2006 x 2007 th: A = x 2006 + 2007 x = 1

+ Nu x > 2007 th A = x - 2006 - 2007 + x = 2x 4013

Do x > 2007 => 2x 4013 > 4014 4013 = 1 => A > 1.

Vy A t gi tr nh nht l 1 khi 2006 x 2007

Bi 4. Hin nay hai kim ng h ch 10 gi. Sau t nht bao lu th 2 kim ng h nm i din nhau trn mt ng thng. (4 im mi) Gi x, y l s vng quay ca kim pht v kim gi khi 10gi n lc 2 kim i nhau trn mt ng thng, ta c:

x y = (ng vi t s 12 n s 4 trn ng h)

v x : y = 12 (Do kim pht quay nhanh gp 12 ln kim gi)

Do :

x = (gi) Vy thi gian t nht 2 kim ng h t khi 10 gi n lc nm i din nhau trn mt ng thng l giBi 5. Cho tam gic vung ABC ( A = 1v), ng cao AH, trung tuyn AM. Trn tia i tia MA ly im D sao cho DM = MA. Trn tia i tia CD ly im I sao cho CI = CA, qua I v ng thng song song vi AC ct ng thng AH ti E. Chng minh: AE = BC (4 im mi) ng thng AB ct EI ti F

ABM = DCM v:

AM = DM (gt), MB = MC (gt),

= DMC () => BAM = CDM

=>FB // ID => IDAC

V FAI = CIA (so le trong) (1) IE // AC (gt) => FIA = CAI (so le trong) (2)

T (1) v (2) => CAI = FIA (AI chung)

=> IC = AC = AF (3) v E FA = 1v (4)

Mt khc EAF = BAH (),

BAH = ACB ( cng ph ABC)

=> EAF = ACB (5)

T (3), (4) v (5) => AFE = CAB =>AE = BC

s 2:


Mn Ton Lp 7


Bi 1:(4 im)

a) Thc hin php tnh:


chia ht cho 10

Bi 2:(4 im)

Tm x bit:

a.

b.

Bi 3: (4 im)

c) S A c chia thnh 3 s t l theo . Bit rng tng cc bnh phng ca ba s bng 24309. Tm s A.

d) Cho . Chng minh rng:

Bi 4: (4 im)




c) T E k . Bit = 50o ; =25o .

Tnh v

Bi 5: (4 im)


c) Tia AD l phn gic ca gc BAC

d) AM = BC

Ht

p n 2 ton 7

Bi 1:(4 im):

a) (2 im)

b) (2 im)

=

=

=

= 10( 3n -2n)

Vy

EMBED Equation.DSMT4 10 vi mi n l s nguyn dng.

Bi 2:(4 im)

a) (2 im)

b) (2 im)

Bi 3: (4 im)

a) (2,5 im)

Gi a, b, c l ba s c chia ra t s A.

Theo bi ta c: a : b : c = (1)

v a2 +b2 +c2 = 24309 (2)

T (1)

EMBED Equation.DSMT4 = k

EMBED Equation.DSMT4 Do (2)

EMBED Equation.DSMT4

k = 180 v k =

+ Vi k =180, ta c: a = 72; b = 135; c = 30.

Khi ta c s A = a + b + c = 237.

+ Vi k =, ta c: a = ; b =; c =

Khi ta c s A =+( ) + () = .

b) (1,5 im)

T suy ra

khi

=

Bi 4: (4 im)

a/ (1im) Xt v c :

AM = EM (gt )

= (i nh )

BM = MC (gt )

Nn : = (c.g.c )0,5 im

AC = EB

V = =

(2 gc c v tr so le trong c to bi ng thng AC v EB ct ng thng AE )

Suy ra AC // BE . 0,5 im

b/ (1 im )

Xt v c :

AM = EM (gt )

= ( v )

AI = EK (gt )

Nn ( c.g.c )

Suy ra =

M + = 180o ( tnh cht hai gc k b )

+ = 180o

Ba im I;M;K thng hng

c/ (1,5 im )

Trong tam gic vung BHE ( = 90o ) c = 50o

= 90o - = 90o - 50o =40o

= - = 40o - 25o = 15o

l gc ngoi ti nh M ca

Nn = + = 15o + 90o = 105o

( nh l gc ngoi ca tam gic )

Bi 5: (4 im)

a) Chng minh ADB = ADC (c.c.c)

suy ra

Do

b) ABC cn ti A, m (gt) nn

ABC u nn

Tia BD nm gia hai tia BA v BC suy ra .

Tia BM l phn gic ca gc ABD

nn

Xt tam gic ABM v BAD c:

AB cnh chung ;

Vy: ABM = BAD (g.c.g)

suy ra AM = BD, m BD = BC (gt) nn AM = BC

s 3:

thi hc sinh gii

Mn Ton Lp 7


Cu 1: Tm tt c cc s nguyn a bit

Cu 2: Tm phn s c t l 7 bit n ln hn v nh hn

Cu 3. Cho 2 a thc P = x + 2mx + m v

Q = x + (2m+1)x + m

EMBED Equation.3 Tm m bit P (1) = Q (-1)

Cu 4: Tm cc cp s (x; y) bit:

Cu 5: Tm gi tr nh nht hoc ln nht ca cc biu thc sau :

A = +5

B =

Cu 6: Cho tam gic ABC c < 900. V ra pha ngoi tam gic hai on thng AD vung gc v bng AB; AE vung gc v bng AC.

a. Chng minh: DC = BE v DC BE

b. Gi N l trung im ca DE. Trn tia i ca tia NA ly M sao cho NA = NM. Chng minh: AB = ME v ABC = EMA

c. Chng minh: MA BC

p n 3 ton 7


0

=>= 0; 1; 2; 3 ; 4

* = 0 => a = 0

* = 1 => a = 1 hoc a = - 1

* = 2 => a = 2 hoc a = - 2

* = 3 => a = 3 hoc a = - 3

* = 4 => a = 4 hoc a = - 4


Gi mu phn s cn tm l x

Ta c:

=> => -77 < 9x < -70. V 9x 9 => 9x = -72

=> x = 8

Vy phn s cn tm l

Cu 3. Cho 2 a thc P = x + 2mx + m v

Q = x + (2m+1)x + m

EMBED Equation.3 Tm m bit P (1) = Q (-1)

P(1) = 12 + 2m.1 + m2 = m2 + 2m + 1

Q(-1) = 1 2m 1 +m2 = m2 2m

P(1) = Q(-1) th m2 + 2m + 1 = m2 2m 4m = -1 m = -1/4


=>

=> x2 = 4.49 = 196 => x = 14

=> y2 = 4.4 = 16 => x = 4

Do x,y cng du nn:

x = 6; y = 14

x = -6; y = -14

p dng tnh cht dy t s bng nhau ta c:

=>

=> -x = 5x -12

=> x = 2. Thay x = 2 vo trn ta c:

=>1+ 3y = -12y

=> 1 = -15y

=> y =

Vy x = 2, y = tho mn bi

Cu 5: Tm gi tr nh nht hoc ln nht ca cc biu thc sau :

A = +5

Ta c : 0. Du = xy ra x= -1.

A 5.Du = xy ra x= -1.

Vy: Min A = 5 x= -1.

B = = = 1 +

Ta c: x 0. Du = xy ra x = 0

x + 3 3 ( 2 v dng )

EMBED Equation.3 4 1+ 1+ 4

B 5

Du = xy ra x = 0

Vy : Max B = 5 x = 0.

Cu 6: a/

Xt ADC v BAF ta c:

DA = BA(gt)

AE = AC (gt)

DAC = BAE ( cng bng 900 + BAC )

=> DAC = BAE(c.g.c )

=> DC = BE

Xt AIE v TIC

I1 = I2 ( )

E1 = C1( do DAC = BAE)

=> EAI = CTI

=> CTI = 900 => DC BE

b/ Ta c: MNE = AND (c.g.c)

=> D1 = MEN, AD = ME

m AD = AB ( gt)

=> AB = ME (pcm) (1)

V D1 = MEN => DA//ME => DAE + AEM = 1800 ( trong cng pha )

m BAC + DAE = 1800=> BAC = AEM ( 2 )

Ta li c: AC = AE (gt) ( 3). T (1),(2) v (3) => ABC = EMA ( pcm)

c/ Ko di MA ct BC ti H. T E h EP MH

Xt AHC v EPA c:

CAH = AEP ( do cng ph vi gPAE )

AE = CA ( gt)

PAE = HCA ( do ABC = EMA cu b)

=> AHC = EPA

=> EPA = AHC

=> AHC = 900=> MA BC (pcm)

s 4:

thi hc sinh gii


Cu 1 ( 2 im)

Thc hin php tnh :

a-

b-

EMBED Equation.3Cu 2 ( 2 im)

a- Tm s nguyn a l s nguyn

b- Tm s nguyn x,y sao cho x - 2xy + y = 0

Cu 3 ( 2 im)

a- Chng minh rng nu a + c = 2b v 2bd = c (b+d) th vi b,d khc 0

b- Cn bao nhiu s hng ca tng S = 1+2+3+ c mt s c ba ch s ging nhau .

Cu 4 ( 3 im)

Cho tam gic ABC c gc B bng 450 , gc C bng 1200. Trn tia i ca tia CB ly im D sao cho CD = 2CB . Tnh gc ADECu 5 ( 1im)

Tm mi s nguyn t tho mn : x2 - 2y2 =1p n 4 CuHng dn chmim

1.aThc hin theo tng bc ng kt qu -2 cho im ti a1im

1.bThc hin theo tng bc ng kt qu 14,4 cho im ti a1im

2.aTa c : =

v a l s nguyn nn l s nguyn khi l s nguyn hay a+1 l c ca 3 do ta c bng sau :

a+1

-3

-1

1

3

a

-4

-2

0

2

Vy vi ath l s nguyn0,25

0,25

0,25

0,25

2.bT : x-2xy+y=0

Hay (1-2y)(2x-1) = -1

V x,y l cc s nguyn nn (1-2y)v (2x-1) l cc s nguyn do ta c cc trng hp sau :

Hoc

Vy c 2 cp s x, y nh trn tho mn iu kin u bi0,25

0,25

0,25

0,25

3.aV a+c=2b nn t 2bd = c (b+d) Ta c: (a+c)d=c(b+d)

Hay ad=bc Suy ra ( PCM)0,5

0,5

3.bGi s s c 3 ch s l =111.a ( a l ch s khc 0)

Gi s s hng ca tng l n , ta c :

Hay n(n+1) =2.3.37.a

Vy n(n+1) chia ht cho 37 , m 37 l s nguyn t v n+1 0 . Chng t rng: khng l s nguyn.

b) Cho a, b, c tho mn: a + b + c = 0. Chng minh rng: .

Cu 3: (2 im)

a) Tm hai s dng khc nhau x, y bit rng tng, hiu v tch ca chng ln lt t l nghch vi 35; 210 v 12.

b) Vn tc ca my bay, t v tu ho t l vi cc s 10; 2 v 1. Thi gian my bay bay t A n B t hn thi gian t chy t A n B l 16 gi. Hi tu ho chy t A n B mt bao lu ?

Cu 4: (3 im)

Cho cnh hnh vung ABCD c di l 1. Trn cc cnh AB, AD ly cc im P, Q sao cho chu vi (APQ bng 2. Chng minh rng gc PCQ bng 450.

Cu 5: (1 im)

Chng minh rng:

s 12:

thi hc sinh gii

(Thi gian lm bi 120 pht)Bi 1: (2 im)

a) Chng minh rng vi mi s n nguyn dng u c:

A=

b) Tm tt c cc s nguyn t P sao cho l s nguyn t.

Bi 2: ( 2 im)

a) Tm s nguyn n sao cho

b) Bit

Chng minh rng:

Bi 3: (2 im)

An v Bch c mt s bu nh, s bu nh ca mi ngi cha n 100. S bu nh hoa ca An bng s bu nh th rng ca Bch.

+ Bch ni vi An. Nu ti cho bn cc bu nh th rng ca ti th s bu nh ca bn gp 7 ln s bu nh ca ti.

+ An tr li: cn nu ti cho bn cc bu nh hoa ca ti th s bu nh ca ti gp bn ln s bu nh ca bn.

Tnh s bu nh ca mi ngi.

Bi 4: (3 im)

Cho (ABC c gc A bng 1200 . Cc ng phn gic AD, BE, CF .

a) Chng minh rng DE l phn gic ngoi ca (ADB.

b) Tnh s o gc EDF v gc BED.

Bi 5: (1 im)

Tm cc cp s nguyn t p, q tho mn:

s 13:

thi hc sinh gii


Bi 1: (2 im)

Tnh:

Bi 2: (3 im)

a) Chng minh rng: chia ht cho 77.

b) Tm cc s nguyn x t gi tr nh nht.

c) Chng minh rng: P(x) c gi tr nguyn vi mi x nguyn khi v ch khi 6a, 2b, a + b + c v d l s nguyn.

Bi 3: (2 im)

a) Cho t l thc . Chng minh rng:

v

b) Tm tt c cc s nguyn dng n sao cho: chia ht cho 7.

Bi 4: (2 im)


Bi 5: (1 im)

Chng minh rng: (a, b ( Z )

s 14:

thi hc sinh gii


Bi 1: (2 im)

a) Tm s nguyn dng a ln nht sao cho 2004! chia ht cho 7a.

b) Tnh

Bi 2: (2 im)

Cho chng minh rng biu thc sau c gi tr nguyn.

Bi 3: (2 im)

Hai xe my khi hnh cng mt lc t A v B, cch nhau 11 km i n C. Vn tc ca ngi i t A l 20 km/h. Vn tc ca ngi i t B l 24 km/h. Tnh qung ng mi ngi i. Bit h n C cng mt lc v A, B, C thng hng.

Bi 4: (3 im)

Cho tam gic nhn ABC. K AH ( BC (H ( BC). V AE ( AB v AE = AB (E v C khc pha i vi AC). K EM v FN cng vung gc vi ng thng AH (M, N ( AH). EF ct AH O. Chng minh rng O l trung im ca EF.

Bi 5: (1 im)

So snh: v

s 15:

thi hc sinh gii


Cu 1: (2 im)

Tnh : ;

Cu 2: (2 im)

a) Tm x, y nguyn bit: xy + 3x - y = 6

b) Tm x, y, z bit: (x, y, z )

Cu 3: (2 im)

a) Chng minh rng: Vi n nguyn dng ta c:

chia ht cho 10.

b) Tm s t nhin x, y bit:

Cu 4: (3 im)

Cho tam gic ABC, AK l trung tuyn. Trn na mt phng khng cha B, b l AC, k tia Ax vung gc vi AC; trn tia Ax ly im M sao cho AM = AC. Trn na mt phng khng cha C, b l AB, k tia Ay vung gc vi AB v ly im N thuc Ay sao cho AN = AB. Ly im P trn tia AK sao cho AK = KP. Chng minh:

a) AC // BP.

b) AK ( MN.

Cu 5: (1 im)

Cho a, b, c l s o 3 cnh ca mt tam gic vung vi c l s o cnh huyn. Chng minh rng: ; n l s t nhin ln hn 0.

s 16:

thi hc sinh gii


Cu 1: (2 im)

Tnh:

Cu 2: ( 2, 5 im)

1) Tm s nguyn m :

a) Gi tr ca biu thc m -1 chia ht cho gi tr ca biu thc 2m + 1.

b)

2) Chng minh rng: chia ht cho 30 vi mi n nguyn dng.

Cu 3: (2 im)

a) Tm x, y, z bit:

; v

b) Cho . Bit f(0), f(1), f(2) u l cc s nguyn. Chng minh f(x) lun nhn gi tr nguyn vi mi x nguyn.

Cu 4: (2,5 im)

Cho tam gic ABC c ba gc nhn, ng cao AH. min ngoi ca tam gic ABC ta v cc tam gic vung cn ABE v ACF u nhn A lm nh gc vung. K EM, FN cng vung gc vi AH (M, N thuc AH). a) Chng minh: EM + HC = NH.

b) Chng minh: EN // FM.

Cu 5: (1 im)

Cho l s nguyn t (n > 2). Chng minh l hp s.

s 17:

thi hc sinh gii


Cu 1: (2 im) Tnh nhanh:

Cu 2: (2 im)

a) Tnh gi tr ca biu thc vi

b) Tm x nguyn chia ht cho

Cu 3: ( 2 im)

a) Tm x, y, z bit v

b) Mt t phi i t A n B trong thi gian d nh. Sau khi i c na qung ng t tng vn tc ln 20 % do n B sm hn d nh 15 pht. Tnh thi gian t i t A n B.

Cu 4: (3 im) Cho tam gic ABC, trung tuyn AM. Trn na mt phng cha nh C b l ng thng AB dng on AE vung gc vi AB v AE = AB. Trn na mt phng cha nh B b l ng thng AC dng on AF vung gc vi AC v AF = AC. Chng minh rng:

a) FB = EC

b) EF = 2 AM

c) AM ( EF.

Cu 5: (1 im)

Chng t rng:

s 18:

thi hc sinh gii


Cu 1: (2 im)

a) Thc hin php tnh:

b) Tnh tng:

Cu 2: (2 im)

1) Tm x bit:

2) Trn qung ng Kp - Bc giang di 16,9 km, ngi th nht i t Kp n Bc Giang, ngi th hai i t Bc Giang n Kp. Vn tc ngi th nht so vi ngi th hai bng 3: 4. n lc gp nhau vn tc ngi th nht i so vi ngi th hai i l 2: 5. Hi khi gp nhau th h cch Bc Giang bao nhiu km ?

Cu 3: (2 im)

a) Cho a thc (a, b, c nguyn).

CMR nu f(x) chia ht cho 3 vi mi gi tr ca x th a, b, c u chia ht cho 3.

b) CMR: nu th (Gi s cc t s u c ngha).

Cu 4: (3 im)

Cho tam gic ABC c AB < AC. Gi M l trung im ca BC, t M k ng thng vung gc vi tia phn gic ca gc A, ct tia ny ti N, ct tia AB ti E v ct tia AC ti F. Chng minh rng:

a) AE = AF

b) BE = CFc)

Cu 5: (1 im)

i vn ngh khi 7 gm 10 bn trong c 4 bn nam, 6 bn n. cho mng ngy 30/4 cn 1 tit mc vn ngh c 2 bn nam, 2 bn n tham gia. Hi c nhiu nht bao nhiu cch la chn c 4 bn nh trn tham gia.

s 19:

thi hc sinh gii


Cu 1: (2 im)

a) Tnh gi tr ca biu thc:

b) Chng t rng:

Cu 2: (2 im)

Cho phn s: (x ( Z)

a) Tm x ( Z C t gi tr ln nht, tm gi tr ln nht .

b) Tm x ( Z C l s t nhin.

Cu 3: (2 im)

Cho . Chng minh rng:

Cu 4: (3 im)

Cho tam gic vung cn ABC (AB = AC), tia phn gic ca cc gc B v C ct AC v AB ln lt ti E v D.

a) Chng minh rng: BE = CD; AD = AE.

b) Gi I l giao im ca BE v CD. AI ct BC M, chng minh rng cc (MAB; MAC l tam gic vung cn.

c) T A v D v cc ng thng vung gc vi BE, cc ng thng ny ct BC ln lt K v H. Chng minh rng KH = KC.

Cu 5: (1 im)

Tm s nguyn t p sao cho:

; l cc s nguyn t.

s 20:

thi hc sinh gii


Cu 1: (2 im)

a) Thc hin php tnh:

;

b) Tm cc s nguyn t x, y sao cho: 51x + 26y = 2000.

Cu 2: ( 2 im)

a) Chng minh rng: 2a - 5b + 6c 17 nu a - 11b + 3c 17 (a, b, c ( Z).

b) Bit

Chng minh rng:

Cu 3: ( 2 im)

By gi l 4 gi 10 pht. Hi sau t nht bao lu th hai kim ng h nm i din nhau trn mt ng thng.

Cu 4: (2 im)

Cho (ABC vung cn ti A. Gi D l im trn cnh AC, BI l phn gic ca (ABD, ng cao IM ca (BID ct ng vung gc vi AC k t C ti N. Tnh gc IBN ?

Cu 5: (2 im)

S 2100 vit trong h thp phn to thnh mt s. Hi s c bao nhiu ch s ?

s 21:

thi hc sinh gii


Bi 1: (2 im)

a) Tnh gi tr ca biu thc

b) Chng minh rng:

Cu 2: (2 im)

a) Chng minh rng vi mi s nguyn dng n th:

chia ht cho 6.

b) Tm gi tr nh nht ca biu thc:

Cu 3: (2 im)

Mt t phi i t A n B trong thi gian d nh. Sau khi i c na qung ng t tng vn tc ln 20 % do n B sm hn d nh 10 pht. Tnh thi gian t i t A n B.

Cu 4: (3 im)

Cho tam gic ABC, M l trung im ca BC. Trn na mt phng khng cha C c b AB, v tia Ax vung gc vi AB, trn tia ly im D sao cho AD = AB. Trn na mt phng khng cha B c b AC v tia Ay vung gc vi AC. Trn tia ly im E sao cho AE = AC. Chng minh rng:

a) DE = 2 AM

b) AM ( DE.

Cu 5: (1 im)

Cho n s x1, x2, , xn mi s nhn gi tr 1 hoc -1. Chng minh rng nu x1. x2 + x2. x3 + + xn x1 = 0 th n chia ht cho 4.

s 22:

thi hc sinh gii



b) Chng minh rng tng:

Bi 2: (2 im)

a) Tm cc s nguyn x tho mn.

b) Cho p > 3. Chng minh rng nu cc s p, p + d , p + 2d l cc s nguyn t th d chia ht cho 6.

Bi 3: (2 im)

a) lm xong mt cng vic, mt s cng nhn cn lm trong mt s ngy. Mt bn hc sinh lp lun rng nu s cng nhn tng thm 1/3 th thi gian s gim i 1/3. iu ng hay sai ? v sao ?

b) Cho dy t s bng nhau:

Tnh

Bi 4: (3 im)

Cho tam gic nhn ABC, AB > AC phn gic BD v CE ct nhau ti I.

a) Tnh cc gc ca (DIE nu gc A = 600.

b) Gi giao im ca BD v CE vi ng cao AH ca (ABC ln lt l M v N. Chng minh BM > MN + NC.

Bi 5: (1 im) Cho z, y, z l cc s dng.

Chng minh rng:

s 23:

thi hc sinh gii


Bi 1: (2 im)

a) Tm x bit:

b) Tm tng cc h s ca a thc nhn c sau khi b du ngoc trong biu thc: A(x) =

Bi 2: (2 im)

Ba ng cao ca tam gic ABC c di bng 4; 12; x bit rng x l mt s t nhin. Tm x ?

Bi 3: (2 im)

Cho .

CMR biu thc sau c gi tr nguyn:

Bi 4: (3 im)

Cho tam gic ABC vung A c gc B =. Trn cnh AC ly im E sao cho gc EBA= . Trn tia i ca tia EB ly im D sao cho ED = BC. Chng minh tam gic CED l tam gic cn.

Bi 5: (1 im)

Tm cc s a, b, c nguyn dng tho mn :

v

s 24:

thi hc sinh gii


Bi 1: (2 im)

a) Tnh

b) Tm x bit

Bi 2: (2 im) Chng minh rng:

Nu

Th

Bi 3: (2 im)

Hai xe my khi hnh cng mt lc t A v B, cch nhau 11km i n C (ba a im A, B, C cng trn mt ng thng). Vn tc ca ngi i t A l 20 km/h. Vn tc ca ngi i t B l 24 km/h. Tnh qung ng mi ngi i. Bit h n C cng mt lc.

Bi 4: (3 im)

Cho tam gic ABC c gc A khc 900, gc B v C nhn, ng cao AH. V cc im D, E sao cho AB l trung trc ca HD, AC l trung trc ca HE. Gi I, K ln lt l giao im ca DE vi AB v AC.

Tnh s o cc gc AIC v AKB ?

Bi 5: (1 im)

Cho x = 2005. Tnh gi tr ca biu thc:

s 25:

thi hc sinh gii


Cu 1 . ( 2) Cho: .

Chng minh: .

Cu 2. (1). Tm A bit rng:

A = .

Cu 3. (2). Tm A( Z v tm gi tr .

a). A = . b). A = .

Cu 4. (2). Tm x:

a)

EMBED Equation.3 = 5 . b). ( x+ 2) 2 = 81. c). 5 x + 5 x+ 2 = 650

Cu 5. (3). Cho ( ABC vung cn ti A, trung tuyn AM . E ( BC,

BH,CK ( AE, (H,K ( AE). Chng minh ( MHK vung cn.

s 26:

thi hc sinh gii


Cu 1: (2)

Rt gn A=

Cu 2 (2)

Ba lp 7A,7B,7C c 94 hc sinh tham gia trng cy. Mi hc sinh lp 7A trng c 3 cy, Mi hc sinh lp 7B trng c 4 cy, Mi hc sinh lp 7C trng c 5 cy,. Hi mi lp c bao nhiu hc sinh. Bit rng s cy mi lp trng c u nh nhau.

Cu 3: (1,5)

Chng minh rng l mt s t nhin.

Cu 4 : (3)

Cho gc xAy = 600 v tia phn gic Az ca gc . T mt im B trn Ax v ng thng song song vi vi Ay ct Az ti C. v Bh ( Ay,CM (Ay, BK ( AC.Chng minh rng .

a, K l trung im ca AC.

b, BH =

c, u

Cu 5 (1,5 )

Trong mt k thi hc sinh gii cp Huyn, bn bn Nam, Bc, Ty, ng ot 4 gii 1,2,3,4 . Bit rng mi cu trong 3 cu di y ng mt na v sai 1 na:

a, ty t gii 1, Bc t gii 2.

b, Ty t gii 2, ng t gii 3.

c, Nam t gii 2, ng t gii 4.

Em hy xc nh th t ng ca gii cho cc bn.

s 27:

thi hc sinh gii


Bi 1: (3 im): Tnh

Bi 2: (4 im): Cho chng minh rng:a)

b)

Bi 3:(4 im) Tm bit:

a)

b)

Bi 4: (3 im) Mt vt chuyn ng trn cc cnh hnh vung. Trn hai cnh u vt chuyn ng vi vn tc 5m/s, trn cnh th ba vi vn tc 4m/s, trn cnh th t vi vn tc 3m/s. Hi di cnh hnh vung bit rng tng thi gian vt chuyn ng trn bn cnh l 59 giy

Bi 5: (4 im) Cho tam gic ABC cn ti A c , v tam gic u DBC (D nm trong tam gic ABC). Tia phn gic ca gc ABD ct AC ti M. Chng minh:

e) Tia AD l phn gic ca gc BAC

f) AM = BC

Bi 6: (2 im): Tm bit:

---------------------------------------------------------

s 28:

thi hc sinh gii


Bi 1. Tnh

Bi 2. Tm gi tr nguyn dng ca x v y, sao cho:

Bi 3. Tm hai s dng bit: tng, hiu v tch ca chng t l nghch vi cc s 20, 140 v 7

Bi 4. Tm x, y tho mn: = 3 Bi 5. Cho tam gic ABC c gc ABC = 500 ; gc BAC = 700 . Phn gic trong gc ACB ct AB ti M. Trn MC ly im N sao cho gc MBN = 400. Chng minh: BN = MC.

s 29:

thi hc sinh gii




Cu 3: Trong 3 s x, y, z c 1 s dng , mt s m v mt s 0. Hi mi s thuc loi no bit:


Cu 5: Tnh tng:

Cu 6: Cho tam gic ABC c < 900. V ra pha ngi tam gic hai on thng AD vung gc v bng AB; AE vung gc v bng AC.

d. Chng minh: DC = BE v DC BE

e. Gi N l trung im ca DE. Trn tia i ca tia NA ly M sao cho NA = NM. Chng minh: AB = ME v

f. Chng minh: MA BC

s 30:

thi hc sinh gii


Cu 1: So snh cc s:

a.

B =251

b.2300 v 3200Cu 2: Tm ba s a, b, c bit a t l thun vi 7 v 11; b v c t l nghch vi 3 v 8 v 5a - 3b + 2c = 164

Cu 3: Tnh nhanh:

Cu 4. Cho tam gic ACE u sao cho B v E hai na mt phng i nhau c b AC.

a. Chng minh tam gic AED cn.

b. Tnh s o gc ACD?

Tuyn tp cc thi hc sinh gii lp 7

Mt s kinh nghim nh v tm ch s tn cng v ng dng vo cc bi ton chng minh chia ht ca cc lp 6,7

I. phn m u : Tm ch s tn cng ca mt lu tha

y l nhng bi ton tng i phc tp ca hc sinh cc lp 6,7 nhng li l nhng bi ton ht sc l th , n to cho hc sinh lng say m khm ph t cc em ngy cng yeu mn ton hn . c nhng bi c s m rt ln tng nh l mnh khng th gii c . Nhng nh pht hin v nm bt c qui lut , vn dungj qui lut cc em t gii c v t nhin thy mnh lm c mt vic v cng ln lao . t gieo vo tr tu cc em kh nng khm ph , kh nng t nghin cu

Tuy l kh nhng chng ta hng dn cc em mt cch t t c h thng ,l rch v cht ch th cc em vn tip fhu tt . y l mt kinh nghim nh m ti mun trnh by v trao i cng cc bn

II. Ni dung c th : 1. L thuyt v tm ch s tn cng : phn ny rt quan trng , cn l gii cho hc sinh mt cch k lng ,y

n = mt s c tn cng l 0 khi lu tha bc n c tn cng vn l 0




*a = vi a chn : mt s c tn cng l 5 khi nhn vi mmt s chn s c ch s tn cng l 0

*a = vi a l : mt s c tn cng l 5 khi nhn vi mt s l s c tn cng l 5

Qua cc cng thc trn ta c quy tc sau : Mt s tn nhin c ch s tn cng l : (0,1,5,6) khi nng ln lu tha vi s m t nhin th c ch s t nhin khng thay i

Kt lun trn l cha kho gi cc bi ton v tm ch s tn cng ca mt lu tha

2. Cc bi ton c bn .

Bi ton 1 : Tm ch s tn cng ca cc lu tha sau

a) 2100 ; b) 3100 ; c) 4100 d) 5100 ; e) 6100 ; f) 7100 g) 8100 ; 9100

Ta nhn thy cc lu tha 5100 , 6100 thuc v dng c bn trnh by trn

nay cn li cc lu tha m c s l 2, 3 , 4 , 7 , 8 , 9

Mun gii cc bi ton ny th ta phai a chng v mt trong 4 dng c bn trn . thc cht ch c a v hai dng c bn l : n = , n =

gii bi ton 1

a) 2100 = 24*25 = (4)25 = (16)25 =

b) 3100 = 34*25 = (4)25 = (81)25 =

c) 4100 = 44*50 =(2)50 = (16)50 =

d) 7100 = 74*25 =(4)25 = 240125 =

e) 8100 = 84*25 = (4)25 = 409625 =

f) 9100 = 92*50 = (2)50 = 8150 =

Bi ton 2 : tm ch s tn cng ca cc s sau :

a) 2101 ; b) 3101 ; c) 41o1 , d) 7101 ; e) 8101 ; f) 9101

Gii bi ton 2

_ nhn xt u tin .

s m ( 101 khng chia ht cho 2 v 4 )

_ Ta vit 101 = 4.25 +1

101 = 2 .50 +1

_ p dng cng thc am+n = am.an

ta c : a) 2101 = 24.25+1 = 2100 . 2 = .2 =

b) 3101 = 3100+1 = 3100 . 3 = .3 =

c) 41o1 = 4100 +1 = 4100 . 4 = . 4 =

d) 7101 = 7100+1 = 7100 . 7 = .7 =

e) 8101 = 8100+1 = 8100 . 8 = .8 =

f) 9101 = 9100 +1 = 9100 . 9 = . 9 =

3. Mt s bi ton phc tp hn

Bi ton 3: Tm ch s tn cng ca cc lu tha sau :

a) 12921997 ; b) 33331997 ; c) 12341997 ; d) 12371997 ; e) 12381997 ;

f) 25691997 Bi gii

Nhn xt quan trng : Thc cht ch s tn cng ca lu tha bc n ca mts t nhin ch ph thuc vo ch s tn cng ca s t nhin m thi (c s) . Nh vy bi to 3 thc cht l bi ton 2

a) 12921997 = 12924. 499 +1= (12924)499 .1292 =

b) 33331997 = 33334. 499 +1 =(33334)499 +1 . 3333 = 499 .3333 =

c) 12341997 = 12344 .499 +1 = (12344)499 . 1234 = ()499 . 1234 =

d) 12371997 = 12374 .499 +1 = (12374) 499. 1237 = 499 .1237 =

4. vn dng vo cc bi ton chng minh chia ht p dng du hiu chia ht

Ta d dng nhn thy : Nu hai s c ch s tn cng ging nhau th khi thc hin php tr s c ch s tn cng l 0 ta s c cc bi ton chng minh chia ht cho { 2,5,10 } . Nu mt s c tn cng l 1 v mt s c tn cng l 3 chng hn ta s c bi ton chng minh tng hai s chia ht cho 2 (v ch s tn cng ca tng l 4)

Cc bi ton c th : Hy chng minh

a) 12921997 + 33331997 5

Theo bi ton trn ta c

12921997 =

33331997 =

nh vy tng ca hai s ny s c tn cng l 5 12921997 + 33331997 5

b) Chng minh 16281997 + 12921997 10

Ap dng qui tc tm ch s tn cng ta c

16281997 s c tn cng l

12921997 S C tn cng l

Nh vy 16281997 + 12921997 10 (v ch s tn cng ca tng ny s l 0)

Ta cng c th vn dung hiu ca hai s hoc tch ca hai s ra cc bi ton chng minh tng t

III. Kt lun: Trn y ti trnh by phn c bn ca vn tm ch s tn cng ca mt lu tha v nhng ng dng ca n trong bi ton chng minh chia ht trong tp hp s t nhin

Trong nhng nm hc qua ti trc tip hng dn cho mt s hc sinh cc em t ra rt thch th v xem nh l nhng khm ph mi ca chnh cc em vi cch t vn nh trn cc em t ra c v c nhiu bi rt hay ...

Cch t vn cung nh trnh by ni chc s khng trnh khi phn sai st mong cc ng nghip gp chn thnh

thi -lim -pic huyn

Mn Ton Lp 7

Nm hc 2006-2007


Bi 1. Tm gi tr n nguyn dng:

a)

EMBED Equation.DSMT4 ; b) 27 < 3n < 243

Bi 2. Thc hin php tnh:

Bi 3. a) Tm x bit:


Bi 4. Hin nay hai kim ng h ch 10 gi. Sau t nht bao lu th 2 kim ng h nm i din nhau trn mt ng thng.

Bi 5. Cho tam gic vung ABC ( A = 1v), ng cao AH, trung tuyn AM. Trn tia i tia MA ly im D sao cho DM = MA. Trn tia i tia CD ly im I sao cho CI = CA, qua I v ng thng song song vi AC ct ng thng AH ti E. Chng minh: AE = BC

p n ton 7

Bi 1. Tm gi tr n nguyn dng: (4 im mi cu 2 im)

a)

EMBED Equation.DSMT4 ; => 24n-3 = 2n => 4n 3 = n => n = 1

b) 27 < 3n < 243 => 33 < 3n < 35 => n = 4

Bi 2. Thc hin php tnh: (4 im)

=

=

Bi 3. (4 im mi cu 2 im) a) Tm x bit:

Ta c: x + 2 0 => x - 2.

+ Nu x - th => 2x + 3 = x + 2 => x = - 1 (Tho mn)

+ Nu - 2 x < - Th => - 2x - 3 = x + 2 => x = - (Tho mn)

+ Nu - 2 > x Khng c gi tr ca x tho mn


+ Nu x < 2006 th: A = - x + 2006 + 2007 x = - 2x + 4013

Khi : - x > -2006 => - 2x + 4013 > 4012 + 4013 = 1 => A > 1

+ Nu 2006 x 2007 th: A = x 2006 + 2007 x = 1

+ Nu x > 2007 th A = x - 2006 - 2007 + x = 2x 4013

Do x > 2007 => 2x 4013 > 4014 4013 = 1 => A > 1.

Vy A t gi tr nh nht l 1 khi 2006 x 2007

Bi 4. Hin nay hai kim ng h ch 10 gi. Sau t nht bao lu th 2 kim ng h nm i din nhau trn mt ng thng. (4 im mi) Gi x, y l s vng quay ca kim pht v kim gi khi 10gi n lc 2 kim i nhau trn mt ng thng, ta c:

x y = (ng vi t s 12 n s 4 trn ng h)

v x : y = 12 (Do kim pht quay nhanh gp 12 ln kim gi)

Do :

=> x = (gi)

Vy thi gian t nht 2 kim ng h t khi 10 gi n lc nm i din nhau trn mt ng thng l gi

Bi 5. Cho tam gic vung ABC ( A = 1v), ng cao AH, trung tuyn AM. Trn tia i tia MA ly im D sao cho DM = MA. Trn tia i tia CD ly im I sao cho CI = CA, qua I v ng thng song song vi AC ct ng thng AH ti E. Chng minh: AE = BC (4 im mi) ng thng AB ct EI ti F

ABM = DCM v:

AM = DM (gt), MB = MC (gt),

= DMC () => BAM = CDM

=>FB // ID => IDAC

V FAI = CIA (so le trong) (1) IE // AC (gt) => FIA = CAI (so le trong) (2)

T (1) v (2) => CAI = FIA (AI chung)

=> IC = AC = AF (3)

v E FA = 1v (4)

Mt khc EAF = BAH (),

BAH = ACB ( cng ph ABC)

=> EAF = ACB (5)

T (3), (4) v (5) => AFE = CAB

=>AE = BC

BI TP V CC I LNG T L

1. Ba n v kinh doanh gp vn theo t l 2 : 3 : 5. Hi mi n v c chia bao nhiu tin nu tng s tin li l 350 000 000 v tin li c chia theo t l thun vi s vn ng gp.

2. Hai nn nh hnh ch nht c chiu di bng nhau. Nn nh th nht c chiu rng l 4 mt, nn nh th hai c chiu rng l 3,5 mt. lt ht nn nh th nhtngi ta dng 600 vin gch hoa hnh vung. Hi phi dng bao nhiu vin gch cng loi lt ht nn nh th hai?

3. Khi tng kt cui nm hc ngi ta thy s hc sinh gii ca trng phn b cc khi 6,7,8,9theo t l 1,5 : 1,1 : 1,3 : 1,2. Hi s hc sinh gii ca mi khi lp, bit rng khi 8 nhiu hn khi 9 l 3 hc sinh gii.

4. Ba i my san t lm 3 khi lng cng vic nh nhau. i th nht, th hai, th ba hon thnh cng vic ln lt trong 4 ngy, 6 ngy, 8 ngy. Hi mi i c my my, bit rng i th nht c nhiu hn i th hai l 2 my v nng sut cc my nh nhau.

5. Vi thi gian mt ngi th lnh ngh lm c 11 sn phm th ngi th hc ngh ch lm c 7 sn phm. Hi ngi th hc vic phi dng bao nhiu thi gian hon thnh mt khi lng cng vic m ngi th lnh ngh lm trong 56 gi?

6. Mt vt chuyn ng trn cc cnh ca mt hnh vung. Trn hai cnh u vt chuyn ng vi vn tc 5m/s, trn cnh th ba vi vn tc 4m/s, trn cnh th t vi vn tc 3m/s. Hi di ca cnh hnh vung bit rng tng s thi gian vt chuyn ng trn 4 cnh l 59s.

BI TP HNH HC

1. Cho 2 gc v k b. Ot v Ot ln lt l phn gic ca hai gc v t im M bt k trn Ot h MH Ox ( HOx ). Trn tia Oz ly im N sao cho ON = MH. ng vung gc k t N ct tia Ot ti K. Tnh s o gc KM^O ?

2. Cho tam gic ABC c B^ = 300 , C^ = 200.ng trung trc ca AC ct BC ti E ct BA ti F.Chng minh rng : FA = FE.

3. Cho tam gic ABC tia phn gic ca gc B v gc C ct nhau ti O. Qua O k ng thng song song vi BC ct AB D v AC E. Chng minh rng : DE = BD + EC.

4. Cho tam gic ABD c =. K AH vung gc vi BD (H BD ) trn tia i ca tia BA ly BE = BH, ng thng EH ct AD ti F. Chng minh rng : FH = FA = FD.

5. Cho tam gic cn ABC (AB = AC) trn tia i ca tia CA ly im D bt k .

a) Chng minh rng : = 2 + .

b) Gi s = 300, = 900, hy tnh gc CBD.

MT S BI TON KH

1. Tm x, y, bit :

a) (x 1)2 + (y + 2)2 = 0

b) + = 0

2. Trong mt cuc chy ua tip sc 4 100m ( Mi i tham gia gm 4 vn ng vin, mi VV chy xong 100m s truyn gy tip sc cho VV tip theo. Tng s thi gian chy ca 4 VV l thnh tch ca c i, thi gian chy ca i no cng t th thnh tch cng cao ). Gi s i tuyn gm : ch, mo, g, vt c vn tc t l vi 10, 8, 4, 1. Hi thi gian chy ca i tuyn l ? giy. Bit rng vt chy ht 80 giy?

3. Tm cc s nguyn x, y tha mn :

s 31:

thi hc sinh gii


Bi 1 (3):

1, Tnh: P =

2, Bit: 13 + 23 + . . . . . . .+ 103 = 3025.

Tnh: S = 23 + 43 + 63 + . . . .+ 203

3, Cho: A =

Tnh gi tr ca A bit l s nguyn m ln nht.

Bi 2 (1):

Tm x bit:

3x + 3x + 1 + 3x + 2 = 117

Bi 3 (1):

Mt con th chy trn mt con ng m hai phn ba con ng bng qua ng c v on ng cn li i qua m ly. Thi gian con th chy trn ng c bng na thi gian chy qua m ly.

Hi vn tc ca con th trn on ng no ln hn ? Tnh t s vn tc ca con th trn hai on ng ?

Bi 4 (2):

Cho ABC nhn. V v pha ngoi ABC cc u ABD v ACE. Gi M l giao im ca BE v CD. Chng minh rng:

1, ABE = ADC

2,

Bi 5 (3):

Cho ba im B, H, C thng hng, BC = 13 cm, BH = 4 cm, HC = 9 cm. T H v tia Hx vung gc vi ng thng BC. Ly A thuc tia Hx sao cho HA = 6 cm.

1, ABC l g ? Chng minh iu .

2, Trn tia HC ly im D sao cho HD = HA. T D v ng thng song song vi AH ct AC ti E.

Chng minh: AE = AB

s 32 thi hc sinh gii


Bi 1 (4):

Cho cc a thc:

A(x) = 2x5 4x3 + x2 2x + 2

B(x) = x5 2x4 + x2 5x + 3

C(x) = x4 + 4x3 + 3x2 8x +

1, Tnh M(x) = A(x) 2B(x) + C(x)

2, Tnh gi tr ca M(x) khi x =

3, C gi tr no ca x M(x) = 0 khng ?

Bi 2 (4):

1, Tm ba s a, b, c bit:

3a = 2b; 5b = 7c v 3a + 5b 7c = 60

2, Tm x bit:

Bi 3 (4):

Tm gi tr nguyn ca m v n biu thc

1, P = c gi tr ln nht

2, Q = c gi tr nguyn nh nht

Bi 4 (5):

Cho tam gic ABC c AB < AC; AB = c, AC = b. Qua M l trung im ca BC k ng vung gc vi ng phn gic trong ca gc A, ct cc ng thng AB, AC ln lt ti D, E.

1, Chng minh BD = CE.

2, Tnh AD v BD theo b, c

Bi 5 (3):

Cho ABC cn ti A, . D l im thuc min trong ca ABC sao cho .

Tnh gc ADB ?

s 33:

thi hc sinh gii


Bi 1 (3): Tnh:

1,

2, (63 + 3. 62 + 33) : 13

3,

Bi 2 (3):

1, Cho v a + b + c 0; a = 2005.

Tnh b, c.

2, Chng minh rng t h thc ta c h thc:

Bi 3 (4): di ba cnh ca tam gic t l vi 2; 3; 4. Ba chiu cao tng ng vi ba cnh t l vi ba s no ?

Bi 4 (3):

V th hm s:

y =

Bi 5 (3):

Chng t rng:

A = 75. (42004 + 42003 + . . . . . + 42 + 4 + 1) + 25 l s chia ht cho 100

Bi 6 (4):

Cho tam gic ABC c gc A = 600. Tia phn gic ca gc B ct AC ti D, tia phn gic ca gc C ct AB ti E. Cc tia phn gic ct nhau ti I.

Chng minh: ID = IE

s 34:

thi hc sinh gii


Bi 1 (5):

1, Tm n N bit (33 : 9)3n = 729 2, Tnh :

A = +

Bi 2 (3):

Cho a,b,c R v a,b,c 0 tho mn b2 = ac. Chng minh rng:

=

Bi 3 (4):

Ba i cng nhn lm 3 cng vic c khi lng nh nhau. Thi gian hon thnh cng vic ca i , , ln lt l 3, 5, 6 ngy. Bit i nhiu hn i l 2 ngi v nng sut ca mi cng nhn l bng nhau. Hi mi i c bao nhiu cng nhn ?

Cu 4 (6):

Cho ABC nhn. V v pha ngoi ABC cc u ABD v ACE.

1, Chng minh: BE = DC.

2, Gi H l giao im ca BE v CD. Tnh s o gc BHC.

Bi 5 (2):

Cho m, n N v p l s nguyn t tho mn: = .

Chng minh rng : p2 = n + 2.

s 35:

thi hc sinh gii


Bi 1: (2 im)

a, Cho

Trong hai s A v B s no ln hn v ln hn bao nhiu ln ?

b) S c chia ht cho 3 khng ? C chia ht cho 9 khng ?

Cu 2: (2 im)

Trn qung ng AB di 31,5 km. An i t A n B, Bnh i t B n A. Vn tc An so vi Bnh l 2: 3. n lc gp nhau, thi gian An i so vi Bnh i l 3: 4. Tnh qung ng mi ngi i ti lc gp nhau ?

Cu 3:

a) Cho vi a, b, c l cc s hu t.

Chng t rng: . Bit rng

b) Tm gi tr nguyn ca x biu thc c gi tr ln nht.

Cu 4: (3 im)

Cho (ABC dng tam gic vung cn BAE; BAE = 900, B v E nm hai na mt phng khc nhau b AC. Dng tam gic vung cn FAC, FAC = 900. F v C nm hai na mt phng khc nhau b AB.

a) Chng minh rng: (ABF = (ACE

b) FB ( EC.

Cu 5: (1 im)

Tm ch s tn cng ca

s 36:

thi hc sinh gii

(Thi gian lm bi 120 phtCu 1: (2 im)

a) Tnh

b) Cho

Chng minh rng .

Cu 2: (2 im)

a) Chng minh rng nu th (gi thit cc t s u c ngha).

b) Tm x bit:

Cu 3: (2im)

a) Cho a thc vi a, b, c l cc s thc. Bit rng f(0); f(1); f(2) c gi tr nguyn.

Chng minh rng 2a, 2b c gi tr nguyn.

b) di 3 cnh ca tam gic t l vi 2; 3; 4. Ba ng cao tng ng vi ba cnh t l vi ba s no ?Cu 4: (3 im) Cho tam gic cn ABC (AB = AC0. Trn cnh BC ly im D, trn tia i ca tia CB ly im E sao cho BD = CE. Cc ng thng vung gc vi BC k t D v E ct AB, AC ln lt M, N. Chng minh rng:

a) DM = EN

b) ng thng BC ct MN ti trung im I ca MN.

c) ng thng vung gc vi MN ti I lun i qua mt im c nh khi D thay i trn cnh BC.

Cu 5: (1 im)

Tm s t nhin n phn s c gi tr ln nht.

s 37:

thi hc sinh gii

(Thi gian lm bi 120 phtCu 1: (2 im)

a) Tnh:

A =

B =

b) Tm cc gi tr ca x :

Cu 2: (2 im)

a) Cho a, b, c > 0 . Chng t rng: khng l s nguyn.

b) Cho a, b, c tho mn: a + b + c = 0. Chng minh rng: .

Cu 3: (2 im)

a) Tm hai s dng khc nhau x, y bit rng tng, hiu v tch ca chng ln lt t l nghch vi 35; 210 v 12.

b) Vn tc ca my bay, t v tu ho t l vi cc s 10; 2 v 1. Thi gian my bay bay t A n B t hn thi gian t chy t A n B l 16 gi. Hi tu ho chy t A n B mt bao lu ?

Cu 4: (3 im)


Cu 5: (1 im)

Chng minh rng:

s 38:

thi hc sinh gii

(Thi gian lm bi 120 phtBi 1: (2 im)

a) Chng minh rng vi mi s n nguyn dng u c:

A=

b) Tm tt c cc s nguyn t P sao cho l s nguyn t.

Bi 2: ( 2 im)

a) Tm s nguyn n sao cho

b) Bit

Chng minh rng:

Bi 3: (2 im)

An v Bch c mt s bu nh, s bu nh ca mi ngi cha n 100. S bu nh hoa ca An bng s bu nh th rng ca Bch.

+ Bch ni vi An. Nu ti cho bn cc bu nh th rng ca ti th s bu nh ca bn gp 7 ln s bu nh ca ti.

+ An tr li: cn nu ti cho bn cc bu nh hoa ca ti th s bu nh ca ti gp bn ln s bu nh ca bn.

Tnh s bu nh ca mi ngi.

Bi 4: (3 im)

Cho (ABC c gc A bng 1200 . Cc ng phn gic AD, BE, CF .

a) Chng minh rng DE l phn gic ngoi ca (ADB.

b) Tnh s o gc EDF v gc BED.

Bi 5: (1 im)

Tm cc cp s nguyn t p, q tho mn:

s 39:

thi hc sinh gii

(Thi gian lm bi 120 phtBi 1: (2 im)

Tnh:

Bi 2: (3 im)

a) Chng minh rng: chia ht cho 77.

b) Tm cc s nguyn x t gi tr nh nht.

c) Chng minh rng: P(x) c gi tr nguyn vi mi x nguyn khi v ch khi 6a, 2b, a + b + c v d l s nguyn.

Bi 3: (2 im)

a) Cho t l thc . Chng minh rng:

v

b) Tm tt c cc s nguyn dng n sao cho: chia ht cho 7.

Bi 4: (2 im)


Bi 5: (1 im)

Chng minh rng: (a, b ( Z )

s 40:

thi hc sinh gii


a) Tm s nguyn dng a ln nht sao cho 2004! chia ht cho 7a.

b) Tnh

Bi 2: (2 im)

Cho chng minh rng biu thc sau c gi tr nguyn.

Bi 3: (2 im)

Hai xe my khi hnh cng mt lc t A v B, cch nhau 11 km i n C. Vn tc ca ngi i t A l 20 km/h. Vn tc ca ngi i t B l 24 km/h. Tnh qung ng mi ngi i. Bit h n C cng mt lc v A, B, C thng hng.

Bi 4: (3 im)

Cho tam gic nhn ABC. K AH ( BC (H ( BC). V AE ( AB v AE = AB (E v C khc pha i vi AC). K EM v FN cng vung gc vi ng thng AH (M, N ( AH). EF ct AH O. Chng minh rng O l trung im ca EF.

Bi 5: (1 im)

So snh: v

s 41:

thi hc sinh gii

(Thi gian lm bi 120 pht)Cu 1: (2 im)

Tnh : ;

Cu 2: (2 im)

a) Tm x, y nguyn bit: xy + 3x - y = 6

b) Tm x, y, z bit: (x, y, z )

Cu 3: (2 im)

a) Chng minh rng: Vi n nguyn dng ta c:

chia ht cho 10.

b) Tm s t nhin x, y bit:

Cu 4: (3 im)

Cho tam gic ABC, AK l trung tuyn. Trn na mt phng khng cha B, b l AC, k tia Ax vung gc vi AC; trn tia Ax ly im M sao cho AM = AC. Trn na mt phng khng cha C, b l AB, k tia Ay vung gc vi AB v ly im N thuc Ay sao cho AN = AB. Ly im P trn tia AK sao cho AK = KP. Chng minh:

a) AC // BP.

b) AK ( MN.

Cu 5: (1 im)

Cho a, b, c l s o 3 cnh ca mt tam gic vung vi c l s o cnh huyn. Chng minh rng: ; n l s t nhin ln hn 0.

s 42:

thi hc sinh gii


Tnh:

Cu 2: ( 2, 5 im)

1) Tm s nguyn m :

a) Gi tr ca biu thc m -1 chia ht cho gi tr ca biu thc 2m + 1.

b)

2) Chng minh rng: chia ht cho 30 vi mi n nguyn dng.

Cu 3: (2 im)

a) Tm x, y, z bit:

; v

b) Cho . Bit f(0), f(1), f(2) u l cc s nguyn. Chng minh f(x) lun nhn gi tr nguyn vi mi x nguyn.

Cu 4: (2,5 im)

Cho tam gic ABC c ba gc nhn, ng cao AH. min ngoi ca tam gic ABC ta v cc tam gic vung cn ABE v ACF u nhn A lm nh gc vung. K EM, FN cng vung gc vi AH (M, N thuc AH). a) Chng minh: EM + HC = NH.

b) Chng minh: EN // FM.

Cu 5: (1 im)

Cho l s nguyn t (n > 2). Chng minh l hp s.

s 43:

thi hc sinh gii

(Thi gian lm bi 120 pht)Cu 1: (2 im) Tnh nhanh:

Cu 2: (2 im)

a) Tnh gi tr ca biu thc vi

b) Tm x nguyn chia ht cho

Cu 3: ( 2 im)

a) Tm x, y, z bit v

b) Mt t phi i t A n B trong thi gian d nh. Sau khi i c na qung ng t tng vn tc ln 20 % do n B sm hn d nh 15 pht. Tnh thi gian t i t A n B.

Cu 4: (3 im) Cho tam gic ABC, trung tuyn AM. Trn na mt phng cha nh C b l ng thng AB dng on AE vung gc vi AB v AE = AB. Trn na mt phng cha nh B b l ng thng AC dng on AF vung gc vi AC v AF = AC. Chng minh rng:

a) FB = EC

b) EF = 2 AM

c) AM ( EF.

Cu 5: (1 im)

Chng t rng:

s 44:

thi hc sinh gii


a) Thc hin php tnh:

b) Tnh tng:

Cu 2: (2 im)

1) Tm x bit:

2) Trn qung ng Kp - Bc giang di 16,9 km, ngi th nht i t Kp n Bc Giang, ngi th hai i t Bc Giang n Kp. Vn tc ngi th nht so vi ngi th hai bng 3: 4. n lc gp nhau vn tc ngi th nht i so vi ngi th hai i l 2: 5. Hi khi gp nhau th h cch Bc Giang bao nhiu km ?

Cu 3: (2 im)

a) Cho a thc (a, b, c nguyn).

CMR nu f(x) chia ht cho 3 vi mi gi tr ca x th a, b, c u chia ht cho 3.

b) CMR: nu th (Gi s cc t s u c ngha).

Cu 4: (3 im)

Cho tam gic ABC c AB < AC. Gi M l trung im ca BC, t M k ng thng vung gc vi tia phn gic ca gc A, ct tia ny ti N, ct tia AB ti E v ct tia AC ti F. Chng minh rng:

a) AE = AF

b) BE = CFc)

Cu 5: (1 im)

i vn ngh khi 7 gm 10 bn trong c 4 bn nam, 6 bn n. cho mng ngy 30/4 cn 1 tit mc vn ngh c 2 bn nam, 2 bn n tham gia. Hi c nhiu nht bao nhiu cch la chn c 4 bn nh trn tham gia.

s 45:

thi hc sinh gii



b) Chng t rng:

Cu 2: (2 im)

Cho phn s: (x ( Z)

a) Tm x ( Z C t gi tr ln nht, tm gi tr ln nht .

b) Tm x ( Z C l s t nhin.

Cu 3: (2 im)

Cho . Chng minh rng:

Cu 4: (3 im)

Cho tam gic vung cn ABC (AB = AC), tia phn gic ca cc gc B v C ct AC v AB ln lt ti E v D.

a) Chng minh rng: BE = CD; AD = AE.

b) Gi I l giao im ca BE v CD. AI ct BC M, chng minh rng cc (MAB; MAC l tam gic vung cn.

c) T A v D v cc ng thng vung gc vi BE, cc ng thng ny ct BC ln lt K v H. Chng minh rng KH = KC.

Cu 5: (1 im)

Tm s nguyn t p sao cho:

; l cc s nguyn t.

s 46:

thi hc sinh gii


a) Thc hin php tnh:

;

b) Tm cc s nguyn t x, y sao cho: 51x + 26y = 2000.

Cu 2: ( 2 im)

a) Chng minh rng: 2a - 5b + 6c 17 nu a - 11b + 3c 17 (a, b, c ( Z).

b) Bit

Chng minh rng:

Cu 3: ( 2 im)

By gi l 4 gi 10 pht. Hi sau t nht bao lu th hai kim ng h nm i din nhau trn mt ng thng.

Cu 4: (2 im)

Cho (ABC vung cn ti A. Gi D l im trn cnh AC, BI l phn gic ca (ABD, ng cao IM ca (BID ct ng vung gc vi AC k t C ti N. Tnh gc IBN ?

Cu 5: (2 im)

S 2100 vit trong h thp phn to thnh mt s. Hi s c bao nhiu ch s ?

s 47:

thi hc sinh gii


a) Tnh gi tr ca biu thc

b) Chng minh rng:

Cu 2: (2 im)

a) Chng minh rng vi mi s nguyn dng n th:

chia ht cho 6.

b) Tm gi tr nh nht ca biu thc:

Cu 3: (2 im)

Mt t phi i t A n B trong thi gian d nh. Sau khi i c na qung ng t tng vn tc ln 20 % do n B sm hn d nh 10 pht. Tnh thi gian t i t A n B.

Cu 4: (3 im)

Cho tam gic ABC, M l trung im ca BC. Trn na mt phng khng cha C c b AB, v tia Ax vung gc vi AB, trn tia ly im D sao cho AD = AB. Trn na mt phng khng cha B c b AC v tia Ay vung gc vi AC. Trn tia ly im E sao cho AE = AC. Chng minh rng:

a) DE = 2 AM

b) AM ( DE.

Cu 5: (1 im)

Cho n s x1, x2, , xn mi s nhn gi tr 1 hoc -1. Chng minh rng nu x1. x2 + x2. x3 + + xn x1 = 0 th n chia ht cho 4.

s 48:

thi hc sinh gii



b) Chng minh rng tng:

Bi 2: (2 im)

a) Tm cc s nguyn x tho mn.

b) Cho p > 3. Chng minh rng nu cc s p, p + d , p + 2d l cc s nguyn t th d chia ht cho 6.

Bi 3: (2 im)

a) lm xong mt cng vic, mt s cng nhn cn lm trong mt s ngy. Mt bn hc sinh lp lun rng nu s cng nhn tng thm 1/3 th thi gian s gim i 1/3. iu ng hay sai ? v sao ?

b) Cho dy t s bng nhau:

Tnh

Bi 4: (3 im)

Cho tam gic nhn ABC, AB > AC phn gic BD v CE ct nhau ti I.

a) Tnh cc gc ca (DIE nu gc A = 600.

b) Gi giao im ca BD v CE vi ng cao AH ca (ABC ln lt l M v N. Chng minh BM > MN + NC.

Bi 5: (1 im) Cho z, y, z l cc s dng.

Chng minh rng:

s 49:

thi hc sinh gii


a) Tm x bit:

b) Tm tng cc h s ca a thc nhn c sau khi b du ngoc trong biu thc: A(x) =

Bi 2: (2 im)

Ba ng cao ca tam gic ABC c di bng 4; 12; x bit rng x l mt s t nhin. Tm x ?

Bi 3: (2 im)

Cho .

CMR biu thc sau c gi tr nguyn:

Bi 4: (3 im)

Cho tam gic ABC vung A c gc B =. Trn cnh AC ly im E sao cho gc EBA= . Trn tia i ca tia EB ly im D sao cho ED = BC. Chng minh tam gic CED l tam gic cn.

Bi 5: (1 im)

Tm cc s a, b, c nguyn dng tho mn :

v

s 40:

thi hc sinh gii


a) Tnh

b) Tm x bit

Bi 2: (2 im) Chng minh rng:

Nu

Th

Bi 3: (2 im)

Hai xe my khi hnh cng mt lc t A v B, cch nhau 11km i n C (ba a im A, B, C cng trn mt ng thng). Vn tc ca ngi i t A l 20 km/h. Vn tc ca ngi i t B l 24 km/h. Tnh qung ng mi ngi i. Bit h n C cng mt lc.

Bi 4: (3 im)

Cho tam gic ABC c gc A khc 900, gc B v C nhn, ng cao AH. V cc im D, E sao cho AB l trung trc ca HD, AC l trung trc ca HE. Gi I, K ln lt l giao im ca DE vi AB v AC.

Tnh s o cc gc AIC v AKB ?

Bi 5: (1 im)

Cho x = 2005. Tnh gi tr ca biu thc:

s 50:

thi hc sinh gii

(Thi gian lm bi 120 pht)Cu 1 . ( 2) Cho: .

Chng minh: .

Cu 2. (1). Tm A bit rng:

A = .

Cu 3. (2). Tm A( Z v tm gi tr .

a). A = . b). A = .

Cu 4. (2). Tm x:

a)

EMBED Equation.3 = 5 . b). ( x+ 2) 2 = 81. c). 5 x + 5 x+ 2 = 650

Cu 5. (3). Cho ( ABC vung cn ti A, trung tuyn AM . E ( BC,

BH,CK ( AE, (H,K ( AE). Chng minh ( MHK vung cn.

thi hc sinh gii ton lp 7

Cu 1: (2)

Rt gn A=

Cu 2 (2)

Ba lp 7A,7B,7C c 94 hc sinh tham gia trng cy. Mi hc sinh lp 7A trng c 3 cy, Mi hc sinh lp 7B trng c 4 cy, Mi hc sinh lp 7C trng c 5 cy,. Hi mi lp c bao nhiu hc sinh. Bit rng s cy mi lp trng c u nh nhau.

Cu 3: (1,5)

Chng minh rng l mt s t nhin.

Cu 4 : (3)

Cho gc xAy = 600 v tia phn gic Az ca gc . T mt im B trn Ax v ng thng song song vi vi Ay ct Az ti C. v Bh ( Ay,CM (Ay, BK ( AC.Chng minh rng .

a, K l trung im ca AC.

b, BH =

c, u

Cu 5 (1,5 )

Trong mt k thi hc sinh gii cp Huyn, bn bn Nam, Bc, Ty, ng ot 4 gii 1,2,3,4 . Bit rng mi cu trong 3 cu di y ng mt na v sai 1 na:

a, ty t gii 1, Bc t gii 2.

b, Ty t gii 2, ng t gii 3.

c, Nam t gii 2, ng t gii 4.

Em hy xc nh th t ng ca gii cho cc bn.

s 51:

thi hc sinh gii

(Thi gian lm bi 120 pht)Bi 1: (3 im): Tnh

Bi 2: (4 im): Cho chng minh rng:a)

b)

Bi 3:(4 im) Tm bit:

a)

b)

Bi 4: (3 im) Mt vt chuyn ng trn cc cnh hnh vung. Trn hai cnh u vt chuyn ng vi vn tc 5m/s, trn cnh th ba vi vn tc 4m/s, trn cnh th t vi vn tc 3m/s. Hi di cnh hnh vung bit rng tng thi gian vt chuyn ng trn bn cnh l 59 giy

Bi 5: (4 im) Cho tam gic ABC cn ti A c , v tam gic u DBC (D nm trong tam gic ABC). Tia phn gic ca gc ABD ct AC ti M. Chng minh:

g) Tia AD l phn gic ca gc BAC

h) AM = BC

Bi 6: (2 im): Tm bit:

---------------------------------------------------------

P N THI

Bi 1: 3 im

=

= 0.5

= 1

=

0.5

==

0.5

=

0.5

Bi 2:

a) T suy ra

0.5

khi 0.5

=

0.5

b) Theo cu a) ta c:

0.5

t 1

hay 0.5

vy

0.5

Bi 3:

a)

0.5

hoc 1

Vi hay

0.25

Vi hay

0.25

b)

0.5

0.5

0.5

0.5

Bi 4:

Cng mt on ng, cn tc v thi gian l hai i lng t l nghch 0.5

Gi x, y, z l thi gian chuyn ng ln lt vi cc vn tc 5m/s ; 4m/s ; 3m/s

Ta c: v

1

hay: 0.5

Do :

; ; 0.5

Vy cnh hnh vung l: 5.12 = 60 (m) 0.5

Bi 5:

-V hnh, ghi GT, KL ng

0.5

a) Chng minh ADB = ADC (c.c.c) 1

suy ra

Do


ABC u nn

Tia BD nm gia hai tia BA v BC suy ra . Tia BM l phn gic ca gc ABD

nn


AB cnh chung ;

Vy: ABM = BAD (g.c.g) suy ra AM = BD, m BD = BC (gt) nn AM = BC

Bi 6:

Ta c 8(x-2009)2 = 25- y2 8(x-2009)2 + y2 =25 (*) 0.5

V y2 0 nn (x-2009)2 , suy ra (x-2009)2 = 0 hoc (x-2009)2 =1 0.5

Vi (x -2009)2 =1 thay vo (*) ta c y2 = 17 (loi)

Vi (x- 2009)2 = 0 thay vo (*) ta c y2 =25 suy ra y = 5 (do ) 0.5

T tm c (x=2009; y=5)

0.5

s 52:

thi hc sinh gii

(Thi gian lm bi 120 pht) Bi 1. Tnh

Bi 2. Tm gi tr nguyn dng ca x v y, sao cho:

Bi 3. Tm hai s dng bit: tng, hiu v tch ca chng t l nghch vi cc s 20, 140 v 7

Bi 4. Tm x, y tho mn: = 3 Bi 5. Cho tam gic ABC c gc ABC = 500 ; gc BAC = 700 . Phn gic trong gc ACB ct AB ti M. Trn MC ly im N sao cho gc MBN = 400. Chng minh: BN = MC.

s 52:

thi hc sinh gii

(Thi gian lm bi 120 pht)Bi 1:(4 im)

a) Thc hin php tnh:


chia ht cho 10

Bi 2:(4 im)

Tm x bit:

a.

b.

Bi 3: (4 im)

e) S A c chia thnh 3 s t l theo . Bit rng tng cc bnh phng ca ba s bng 24309. Tm s A.

f) Cho . Chng minh rng:

Bi 4: (4 im)




c) T E k . Bit = 50o ; =25o .

Tnh v

Bi 5: (4 im)


i) Tia AD l phn gic ca gc BAC

j) AM = BC

Ht

P N V HNG DN CHM MN TON 7

Bi 1:(4 im):

p nThang im

a) (2 im)

b) (2 im)

3 n + 2 - Vi mi s nguyn dng n ta c:

=

=

=

= 10( 3n -2n)

Vy

EMBED Equation.DSMT4 10 vi mi n l s nguyn dng.

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

1 im

0,5 im

Bi 2:(4 im)

p nThang im

a) (2 im)

b) (2 im)

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

Bi 3: (4 im)

p nThang im

a) (2,5 im)

Gi a, b, c l ba s c chia ra t s A.

Theo bi ta c: a : b : c = (1)

v a2 +b2 +c2 = 24309 (2)

T (1)

EMBED Equation.DSMT4 = k

EMBED Equation.DSMT4 Do (2)

EMBED Equation.DSMT4

k = 180 v k =

+ Vi k =180, ta c: a = 72; b = 135; c = 30.

Khi ta c s A = a + b + c = 237.

+ Vi k =, ta c: a = ; b =; c =

Khi ta c s A =+( ) + () = .

b) (1,5 im)

T suy ra

khi

=

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

Bi 4: (4 im)

p nThang im

V hnh

0,5 im

a/ (1im) Xt v c :

AM = EM (gt )

= (i nh )

BM = MC (gt )

Nn : = (c.g.c )0,5 im

AC = EB

V = =

(2 gc c v tr so le trong c to bi ng thng AC v EB ct ng thng AE )

Suy ra AC // BE . 0,5 im

b/ (1 im )

Xt v c :

AM = EM (gt )

= ( v )

AI = EK (gt )

Nn ( c.g.c ) 0,5 im Suy ra =

M + = 180o ( tnh cht hai gc k b )

+ = 180o

Ba im I;M;K thng hng 0,5 im

c/ (1,5 im )

Trong tam gic vung BHE ( = 90o ) c = 50o

= 90o - = 90o - 50o =40o

0,5 im

= - = 40o - 25o = 15o

0,5 im

l gc ngoi ti nh M ca

Nn = + = 15o + 90o = 105o

( nh l gc ngoi ca tam gic ) 0,5 im

Bi 5: (4 im)

-V hnh

a) Chng minh ADB = ADC (c.c.c) 1 im

suy ra

0,5 im

Do

0,5 im


ABC u nn

0,5 im

Tia BD nm gia hai tia BA v BC suy ra .

Tia BM l phn gic ca gc ABD

nn

0,5 im


AB cnh chung ;

Vy: ABM = BAD (g.c.g)

suy ra AM = BD, m BD = BC (gt) nn AM = BC0,5 im

Lu : Nu hc sinh lm theo cch khc ng vn t im ti a.

s 53:

thi hc sinh gii

(Thi gian lm bi 120 pht)Cu 1 ( 2 im)

Thc hin php tnh :

a.

b.

EMBED Equation.3Cu 2 ( 2 im)

a. Tm s nguyn a l s nguyn

b. Tm s nguyn x, y sao cho x- 2xy + y = 0

Cu 3 ( 2 im)

a. Chng minh rng nu a + c = 2b v 2bd = c(b + d) th vi b, d khc 0

b. Cn bao nhiu s hng ca tng S = 1 + 2 + 3 + c mt s c ba ch

s ging nhau .

Cu 4 ( 3 im)

Cho tam gic ABC c gc B bng 450 , gc C bng 1200. Trn tia i ca tia CB ly im D sao cho CD = 2CB . Tnh gc ADECu 5 ( 1im)

Tm mi s nguyn t tho mn : x2- 2y2 = 1p n chm Ton 7CuHng dn chmim

1.aThc hin theo tng bc ng kt qu -2 cho im ti a1im

1.bThc hin theo tng bc ng kt qu 14,4 cho im ti a1im

2.aTa c : =

v a l s nguyn nn l s nguyn khi l s nguyn hay a+1 l c ca 3 do ta c bng sau :

a+1

-3

-1

1

3

a

-4

-2

0

2

Vy vi ath l s nguyn0,25

0,25

0,25

0,25

2.bT : x- 2xy + y = 0

Hay (1- 2y)(2x - 1) = -1

V x,y l cc s nguyn nn (1 - 2y)v (2x - 1) l cc s nguyn do ta c cc trng hp sau :

Hoc

Vy c 2 cp s x, y nh trn tho mn iu kin u bi0,25

0,25

0,25

0,25

3.aV a + c = 2b nn t 2bd = c(b + d) Ta c: (a + c)d =c(b + d)

Hay ad = bc Suy ra ( PCM)0,5

0,5

3.bGi s s c 3 ch s l =111.a ( a l ch s khc 0)

Gi s s hng ca tng l n , ta c :

Hay n(n + 1) =2.3.37.a

Vy n(n+1) chia ht cho 37 , m 37 l s nguyn t v n + 1 < 74

( Nu n = 74 khng tho mn )

Do n=37 hoc n + 1 = 37

Nu n =37 th n + 1 = 38 lc khng tho mn

Nu n + 1=37 th n = 36 lc tho mn

Vy s s hng ca tng l 360,25

0,25

0,5

4K DH Vung gc vi AC v ACD =600 do CDH = 300Nn CH = CH = BC

Tam gic BCH cn ti C

EMBED Equation.3CBH = 300 ABH = 150

M BAH = 150 nn tam gic AHB cn ti H

Do tam gic AHD vung cn ti H Vy ADB = 450 + 300 =7500,5

0,5

1,0

1,0

5T : x2- 2y2 =1suy ra x2- 1 = 2y2Nu x chia ht cho 3 v x nguyn t nn x = 3 lc y = 2 nguyn t tho mn

Nu x khng chia ht cho 3 th x2-1 chia ht cho 3 do 2y2 chia ht cho 3 M(2;3) =1 nn y chia ht cho 3 khi x2 =19 khng tho mn

Vy cp s (x,y) duy nht tm c tho mn iu kin u bi l (2;3)

0,25

0,25

0,25

0,25

s 54:

thi hc sinh gii

(Thi gian lm bi 120 pht)Bi 1 (4) -

Rt gn biu thc

a- A = a - 2 + 3 - 2a - 5 + a

b- vi n N

Bi 2 (4 ) .

Chng minh rng : nu a,b,c l cc s khng m tho mn cc iu kin sau : a + 3 c = 8 v a + 2 b = 9 th N = a + b - c - l s khng dng . Tm a,b,c N = 0

Bi 3 (4 ) .

Cho biu thc A =

Biu thc A c gi tr ln nht hay nh nht ? Tm gi tr

Cu 4 (4 )

Cho tam gic cn ABC c ACB = 100 0 . Phn gic trong ca CAB ct CB ti D . Chng minh rng AD + DC = AB

Bi 5 ( 4 )

Cho tam gic ABC c AB = AC . Trn ng thng vung gc vi AC ti C ly im D sao cho hai im B , D nm khc pha i vi ng thng AC . Gi K l giao im ca ng thng qua B vung gc vi AB v ng thng qua trung im M ca CD v vung gc vi AD .

Chng minh KB = KD

-------------------------*****-------------------------

s 55:

thi hc sinh gii

(Thi gian lm bi 120 pht)Bi 1: Thc hin php tnh (2 im)a/

b/

Bi 2: So snh (2 im)a/ vi

b/ vi 6Bi 3: Tm x, y, z bit (4,5 im)a/ 3(x-2) 4(2x+1) 5(2x+3) = 50

b/

c/ v 10x - 3y - 2z = -4

Bi 4: (6 im)Cho hm s . Bit th hm s i qua im A(-1; -1)

a/ Tm m

b/ V th hm s vi m tm cc/ im no sau y khng thuc th hm s trn.B(-2; -2)C(5; 1)D(2; 10)

d/ Tnh din tch tam gic OBC

Bi 5: (5,5 im)Cho ABC, gc B = 600, AB = 7cm, BC = 14cm. Trn BC ly im D sao cho gc BAD = 600. Gi H l trung im ca BD

a/ Tnh di HD

b/ Chng minh rng DAC cn

c/ ABC l tam gic g?

d/ Chng minh rng AB2 + CH2 = AC2 + BH2 =======(((=======

(Cn b coi thi khng gii thch g thm)

M

D

B

A

H

C

I

F

E

D

B

A

H

I

F

E

M

PAGE 1

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50 de Hoc Sinh Gioi Toan 7