thi HSGI. Trc nghim:Hy chn ch ci ng trc cu tr li ng trong cc cu sau:Cu 1: a thc f(x) = x4 + 2x3 + ax2 + 2x + b l bnh phng ca mt a thc th:A. a = 3; b = 1 B. a = 3; b = 0 C. a = 4; b = 1 D. a = 1; b = 1Cu 2: Cho phn thc 2x(x-1)2x. Gi tri ca phn thc bng 0 khi:A. x = 0 B. x = 0 hoc x = 1 C. x = 1 D. Khng c gi tr ca xCu 3: Kt qu ca php tnh (a6 - 1) : (a2 - 1) l:A. a4 + 1 B. a4 + a2 + 1C. a4 + 2a2 + 1 D. Khng thc hin cCu 4: Mt tam gic c di hai cnh bng 3cm v 8cm, gc xen gia bng 600. di cnh cn li l:A. 7cm B. 4cm C.55 D.63Cu5 Cho 12< x-1. Kt qu no sau y l ng?A. x = 0 B. x = 12C. 1 32 2x < < D. x = 4Cu 6 Bit5 4 x th (x - 5)2 bng:A. 2 B. 16 C. 32 D. 256Cu 7 Tng A = 3 - 32 + 33 - 34 + ... - 3100 c kt qu l:A. 1013 34B. 1013 32C. 3 - 3101D. 3101 - 3Cu 8 Mt tam gic c gc B - gc C = 300, tia phn gic ca gc A ct BC ti D. S o gc ADB l:A. 300B. 450 C. 600 D. 750 II. T lun:Cu 5: Gii cc phng trnh sau:a/ 2x3 + x2 - 5x + 2 = 0b/ 2x4 - 21x3 + 74x2 - 105x + 50 = 0c/ 2 1 2 1 4 x x + + Cu 6: Cho P = 228 71x xx ++. Tm gi tr nh nht v gi tr ln nht ca PCu 7: a/ Cho ba s chnh phng A, B, C. Chng minh rng: (A - B)(B - C)(C - A) chia ht cho 12.b/ Cho a3 + b3 + c3 = 3abc vi a, b, c khc 0. Tnh gi tr ca biu thc: P = 1 1 1a b cb c a| `| `| `+ + + . ,. ,. ,Cu 8: Cho tam gic ABC cn, AB = AC = 5cm; BC = 6cm. V cc -ng phn gic AD, BE, CF a/ Tnh di EFb/ Tnh din tch tam gic DEFCu 9: a/ Chng minh rng nu a + b + c3 th a4 + b4 + c4 a3 + b3 + c3

b/ Tm tt c cc tam gic vung c s o cc cnh l s nguyn v s o din tch bng s o chu vi. t hi hc sinh gii cp t r ng Bai1Phn tch a thc thnh nhn t : P(x) = 6x3 + 13x2 + 4x - 3 b) Tm GTNN ca biu thc A = (x - 1) (x+2)(x+3)(x+6) Bi 2 : a) Gii PT : x2 + 2x + 2 1 + x - 2 = 0 b) Gii BPT : x2 - x - 2 < 0 Bi 3 : a) Bit a - b = 7. Tnh GT ca biu thc a2(a+1) - b2(b-1) + ab - 3ab(a - b + 1) b) Chng minh rng : * Nu v x, y, n > 0 th * Nu a, b, c l 3 cnh ca tam gic th: Bi 4 : Ly im O trong ABC. Cc tia AO, BO, CO ct BC, AC, AB ln l t ti P, Q, R. Chng minh rng Bi 5 : Trn cnh AB ca hnh vung ABCD, ng i ta ly tu im E. Tia phn gic ca ct BC ti K. Chng minh AE + KC = DE 1 + > > . , . ,1im1im2a* Vi x1 (*) x - 1 0 1 1 x x ta c phng trnh x2 -3x + 2 + x-1 =0( )222 1 0 1 0 1 x x x x + ( Tho mn iu kin *)* Vi x< 1 (**) x - 1 0 1 1 x x ta c phng trnh x2 -3x + 2 + 1 - x=0( ) ( )24 3 0 1 3 0 x x x x + + x - 1 = 01 x ( Khng tha mn iu kin **) + x - 3 = 03 x ( Khng tho mn iu kin **)Vy nghim ca phng trnh l : x = 11im1imb * iu kin x 0 (1)* pt ( )2 222 22 21 1 1 18 4 4 x x x x xx x x x ]| ` | ` | ` | `+ + + + + + ] . , . , . , . , ] ]0.5im ( )222 2 22 2 21 1 1 18 2 4 4 x x x x xx x x x ]| ` | ` | ` | `+ + + + + + + ] . , . , . , . , ] ]

( ) ( )216 4 8 0 0 x x x x + + hoc x = -8So snh vi iu kin (1) , suy ra nghim ca phng trnh l x = - 81im0.5im3Ta c( ) ( ) ( )3 2 21 1 1 1 y y y y x y y + + + +v xy 0 x, y 0 x, y 0 y-10 v x-10

( ) ( ) ( )3 23 2 23 211 111 1 1 11 1xy y yyx x x x y x xx x x + + + + + +3 3 2 21 11 1 1 1x yy x y y x x + + + + + + ( ) ( )( ) ( )( ) ( ) ( )( )22 22 2 2 2 22 2 3 3 2 22 2 1 11 1 2 12 24 203 1 1 3x y xy x y x x y yx x y y xy x y xy xy x y xy x yxyxy x yxy y x xy| ` | `+ + + + + + + + + + + + + + + + + + + + +. , . , + + + 1im1im1im4Ta c: M = ( ) ( )2 210 16 10 24 16 x x x x + + + + +t a = x2 + 10x + 16 suy ra M = a( a+8) + 16 = a2 + 8a + 16 = ( a+ 4)2M = ( x2 + 10x + 20 )2 ( pcm)1im1im1im5a+ Hai tam gic ADC v BEC c: Gc C chung. CD CACE CB(Hai tam gic vung CDE v CABng dng)Do , chng dng dng (c.g.c). Suy ra: 0135 BEC ADC (v tam gic AHD vung cn ti H theo gi thit).Nn 045 AEB do tam gic ABE vung cn tiA. Suy ra: 2 2 BE AB m 1.5im1imbTa c: 1 12 2BM BE ADBC BC AC (doBEC ADC : )m 2 AD AH (tam gic AHD vung cn ti H)nn 1 1 22 2 2BM AD AH BH BHBC AC AC BE AB (doABH CBA : )Do BHM BEC :(c.g.c), suy ra: 0 0135 45 BHM BEC AHM 1.5im1imcTam gic ABE vung cn ti A, nn tia AM cn l phn gic gc BAC.Suy ra: GB ABGC AC , m( ) ( ) //AB ED AH HDABC DEC ED AHAC DC HC HC :Do : GB HD GB HD GB HDGC HC GB GC HD HC BC AH HC + + +1imUBND THNH PH Hu k thiCHN hc sinh gii tHNH PHPHNG Gio dc v o to lp 8 thCS - nm hc 2007 - 2008Mn :Ton chnh thc Thi gian lm bi: 120 phtBi 1: (2 im) Phn tch a thc sau y thnh nhn t:1.27 6 x x + +2.4 22008 2007 2008 x x x + + +Bi 2: (2im) Gii phng trnh: 1.23 2 1 0 x x x + + 2. ( )2 2 222 22 21 1 1 18 4 4 4 x x x x xx x x x| ` | ` | `| `+ + + + + +

. , . , . ,. ,Bi 3: (2im)1. Cn bc hai ca 64 c th vit di dng nh sau:64 6 4 +Hi c tn ti hay khng cc s c hai ch s c th vit cn bc hai ca chng di dng nhtrn v l mt s nguyn? Hy ch ra ton b cc s .2. Tm s dtrong php chia ca biu thc ( ) ( ) ( ) ( ) 2 4 6 8 2008 x x x x + + + + + cho a thc 210 21 x x + + .Bi 4: (4 im)Cho tam gic ABC vung ti A (AC > AB), ng cao AH (HBC). Trn tia HC ly im D sao cho HD = HA. ng vung gc vi BC ti D ct AC ti E.1. Chng minh rng hai tam gic BEC v ADC ng dng. Tnh di on BE theom AB .2. Gi M l trung im ca on BE. Chng minh rng hai tam gic BHM v BEC ng dng. Tnh s o ca gc AHM3. Tia AM ct BC ti G. Chng minh: GB HDBC AH HC+. biBi 1 (4 im)Cho biu thcA = 3 22 311:11x x xxxxx+

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.|vi x khc -1 v 1.a, Rt gn biu thc A.b, Tnh gi tr ca biu thc A ti x321 .c, Tm gi tr ca x A < 0.Bi 2 (3 im)Cho( ) ( ) ( ) ( ) bc ac ab c b a a c c b b a + + + + 2 2 2 2 2 2. 4 . Chng minh rng c b a .Bi 3 (3 im)Gii bi ton bng cch lp phng trnh. Mt phn s c t s b hn mu s l 11. Nu bt t s i 7 n v v tng mu ln 4 n v th s c phn s nghch o ca phn s cho. Tm phn s .Bi 4 (2 im)Tm gi tr nh nht ca biu thc A =5 4 3 22 3 4+ + a a a a .Bi 5 (3 im)Cho tam gic ABC vung ti A c gc ABC bng 600, phn gic BD. Gi M,N,I theo th t l trung im ca BD, BC, CD.a, T gic AMNI l hnh g? Chng minh.b, Cho AB = 4cm. Tnh cc cnh ca t gic AMNI.Bi 6 (5 im)Hnh thang ABCD (AB // CD) c hai ng cho ct nhau ti O. -ng thng qua O v song song vi y AB ct cc cnh bn AD, BC theo th t M v N.a, Chng minh rng OM = ON.b, Chng minh rng MN CD AB2 1 1 + .c, Bit SAOB= 20082 (n v din tch); SCOD= 20092 (n v din tch). Tnh SABCD. hng dn chm thi hc sinh gii cp Bi 1( 4 im ) a, ( 2 im )Vi x khc -1 v 1 th : A=) 1 ( ) 1 )( 1 () 1 )( 1 (:1122 3x x x x xx xxx x x+ + ++ + 0,5=) 2 1 )( 1 () 1 )( 1 (:1) 1 )( 1 (22x x xx xxx x x x+ ++ + + 0,5= ) 1 (1: ) 1 (2xx+0,5= ) 1 )( 1 (2x x +KL 0,5b, (1 im)Ti x = 321 = 35 th A = ]]]

]]]

+ )35( 1 )35( 12 0,25= )351 )(9251 ( + +0,25272102727238.934

KL0,5c, (1im)Vi x khc -1 v 1 th A + xvi mi x nn (1) xy ra khi v ch khi0 1 < x 1 > xKL0,50,25Bi 2 (3 im)Bin i ng thc c bc ac ab c b a ac a c bc c b ab b a 4 4 4 4 4 4 2 2 22 2 2 2 2 2 2 2 2 + + + + + + + +0,5Bin i c0 ) 2 ( ) 2 ( ) 2 (2 2 2 2 2 2 + + + + + ac c a bc c b ac b a0,5Bin i c0 ) ( ) ( ) (2 2 2 + + c a c b b a(*) 0,5V 0 ) (2 b a;0 ) (2 c b;0 ) (2 c a; vi mi a, b, cnn (*) xy ra khi v ch khi 0 ) (2 b a;0 ) (2 c b v 0 ) (2 c a;0,50,5T suy raa = b = c 0,5Bi 3 (3 im)Gi t s ca phn s cn tm l x th mu s ca phn s cn tm l x+11. Phn s cn tm l 11 + xx (x l s nguyn khc -11)0,5Khi bt t s i 7 n v v tng mu s 4 n v ta c phn s 157+xx (x khc -15)0,5Theo bi ra ta c phng trnh 11 + xx=715+xx0,5Gii phng trnh v tm c x= -5 (tho mn) 1T tm c phn s 65KL0,5Bi 4 (2 im)Bin i c A=3 ) 2 ( ) 2 ( 2 ) 2 (2 2 2 2+ + + + + a a a a a0,5=3 ) 1 )( 2 ( 3 ) 1 2 )( 2 (2 2 2 2+ + + + + a a a a a0,5V0 22> + a a v a a 0 ) 1 (2 nn a a a + 0 ) 1 )( 2 (2 2 do a a a + + 3 3 ) 1 )( 2 (2 20,5Du = xy ra khi v ch khi 0 1 a 1 a0,25KL 0,25Bi 5 (3 im)a,(1 im)Chng minh c t gic AMNI l hnh thang 0,5Chng minh c AN=MI, t suy ra t gic AMNI l hnh thang cn0,5b,(2im)Tnh c AD =cm33 4; BD = 2AD =cm33 8AM = BD21cm33 40,5Tnh c NI = AM =cm33 40,5DC = BC =cm33 8 ,MN = DC21cm33 40,5Tnh c AI =cm33 8 0,5Bi 6 (5 im)a, (1,5 im)Lp lun c BDODABOM, ACOCABON0,5Lp lun c ACOCDBOD 0,5NIMD CABONMDCB A ABONABOM OM = ON0,5b, (1,5 im)Xt ABD c ADDMABOM(1), xt ADC c ADAMDCOM(2)T (1) v (2) OM.(CD AB1 1+)1 +ADADADDM AM0,5Chng minh tng t ON.1 )1 1( +CD AB 0,5t c (OM + ON).2 )1 1( +CD AB MN CD AB2 1 1 +0,5b, (2 im)ODOBSSAODAOB, ODOBSSDOCBOC AODAOBSSDOCBOCSS AOD BOC DOC AOBS S S S . . 0,5Chng minh cBOC AODS S 0,5 2) ( .AOD DOC AOBS S S Thay s c 20082.20092 = (SAOD)2 SAOD = 2008.20090,5Do SABCD= 20082 + 2.2008.2009 + 20092 = (2008 + 2009)2 = 40172 (n v DT)0,5Bi 1 (4 im): Cho biu thc

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.|+ ++ 2 2 2 2 2 221 1:y4xyAx xy y x y xa) Tm iu kin ca x, y gi tr ca A c xc nh.b) Rt gn A.c) Nu x; y l cc s thc tho mn: 3x2+ y2+ 2x 2y = 1, hy tm tt c cc gi tr nguyn dng ca A?Bi 2 (4 im):a) Gii phng trnh : 824493331042211511 ++++++ x x x xb) Tm cc s x, y, zbit :x2 + y2 + z2 = xy + yz + zx v2010 2009 2009 20093 + + z y xBi 3 (3 im): Chng minh rng vi mi n N th n5 v n lun c ch s tn cng ging nhau. Bi 4 (7 im): Cho tam gic ABC vung ti A. Ly mt im M bt k trn cnh AC. T C v mt ng thng vung gc vi tia BM, ng thng ny ct tia BM ti D, ct tia BA ti E.a) Chng minh:EA.EB = ED.EC v EAD ECB b) Cho 0120 BMC v 236AEDS cm . Tnh SEBC?c) Chng minh rng khi im M di chuyn trn cnh AC th tng BM.BD + CM.CA c gi tr khng i. d) K DH BC ( ) H BC . Gi P, Q ln lt l trung im ca cc on thng BH, DH. Chng minh CQ PD .Bi 5 (2 im): a) Chng minh bt ng thc sau:2 +xyyx (vi x v y cng du) b) Tm gi tr nh nht ca biu thc P = 2 22 23 5x y x yy x y x| `+ + + . , (vi x 0, y 0 )Phng Gio dc- o toTRC NINH*****p n v hng dn chm thi hc sinh giinm hc 2008 - 2009mn: Ton 8Bi 1 : (4 im)a) iu kin: x t y; y 0 (1 im)b) A = 2x(x+y) (2 im)c) Cn ch ra gi tr ln nht ca A, t tm c tt c cc gi tr nguyn dng ca A+ T (gt): 3x2 + y2 + 2x 2y = 1 2x2 + 2xy + x2 2xy + y2 + 2(x y) = 1 2x(x + y) + (x y)2 + 2(x y) + 1 = 2 A + (x y + 1)2 = 2 A = 2 (x y + 1)22 (do (x y + 1)0 (vi mi x ; y) A 2. (0,5)+ A = 2 khi( )x y 1 02x x y 2x y;y 0 + + ' t

1x23y2' + A = 1 khi( )2(x y 1) 12x x y 1x y;y 0 + + ' t T , ch cn ch ra c mt cp gi tr ca x v y, chng hn: 2 1x22 3y2'++ Vy A ch c th c 2 gi tr nguyn dng l: A = 1; A = 2 (0,5 im)Bi 2: (4 im) a) x 11 x 22 x 33 x 44115 104 93 82+ + + ++ +x 11 x 22 x 33 x 44( 1) ( 1) ( 1) ( 1)115 104 93 82+ + + + + + + + +x 126 x 126 x 126 x 126115 104 93 82+ + + + + +x 126 x 126 x 126 x 1260115 104 93 82+ + + + + ... x 126 0 + x 126 b) x2 + y2 + z2 = xy + yz + zx2x2 +2y2 + 2z2 2xy 2yz 2zx = 0(x-y)2 + (y-z)2 + (z-x)2 = 0x y 0y z 0z x 0 ' x y z x2009 = y2009 = z2009Thay vo iu kin (2) ta c 3.z2009 = 32010 z2009 = 32009 z = 3Vy x = y = z = 3Bi 3 (3 im)Cn chng minh:n5 n M10- Chng minh : n5 - n M2n5 n = n(n2 1)(n2 + 1) = n(n 1)(n + 1)(n2 + 1) M2 ( v n(n 1) l tch ca hai s nguyn lin tip)- Chng minh: n5 n M5 n5 - n = ... = n( n - 1 )( n + 1)( n2 4 + 5)= n( n 1 ) (n + 1)(n 2) ( n + 2 ) + 5n( n 1)( n + 1 )l lun dn n tng trn chia ht cho 5- V ( 2 ; 5 ) = 1nn n5 n M2.5 tc l n5 n M10Suy ra n5 v n c ch s tn cng ging nhau.Bi 4: 6 imI PQHEDAB CMCu a: 2 im* Chng minh EA.EB = ED.EC (1 im)- Chng minh EBD ng dng vi ECA (gg) 0,5 im- T suy ra. .EB EDEA EB ED ECEC EA 0,5 im* Chng minh EAD ECB (1 im)- Chng minh EAD ng dng vi ECB (cgc) 0,75 im- Suy ra EAD ECB 0,25 imCu b: 1,5 im- T BMC = 120o AMB = 60o ABM = 30o0,5 im- Xt EDB vung ti D c B= 30o ED = 12EB 12EDEB 0,5 im- L lun cho 2EADECBS EDS EB| `

. , t SECB = 144 cm20,5 imCu c: 1,5 im- Chng minhBHD ng dng viDHC (gg)0,5 im22BH BD BP BD BP BDDH DC DQ DC DQ DC 0,5 im- Chng minhDPB ng dng viCQD (cgc) ` 90oBDP DCQCQ PDma BDP PDC '+ 1 imCu d: 1 im- Chng minhBMI ng dng viBCD (gg) - Chng minh CM.CA = CI.BC 0,5 im- Chng minh BM.BD + CM.CA = BC2 c gi tr khng i 0,5 imCch 2: C th bin i BM.BD + CM.CA = AB2 + AC2 = BC2Bi 5: (2 im)a) v x, y cng du nn xy > 0, do 2 2x y2 x y 2xyy x+ + 2(x y) 0 bt ng thc ny lun ng, suy ra bt ban u ng (pcm)b)tx yty x+ 2 222 2x yt 2y x + Biu thc cho tr thnhP = t2 3t + 3 P = t2 2t t + 2 + 1 = t(t 2) (t 2) + 1 = (t 2)(t 1) + 1- Nu x; y cng du, theo c/m cu a) suy ra t 2.t 2 0 ;t 1 0 > ( ) ( ) t 2 t 1 0 P 1 . ng thc xy ra khi v ch khi t = 2 x = y (1)- Nu x; y tri du th x0y< v y0x 0 P > 1 (2)- T (1) v (2) suy ra: Vi mi x0 ; y0 th lun c P 1. ng thc xy ra khi v ch khi x = y. Vy gi tr nh nht ca biu thc P l Pmin= 1 (khi x = y) Bi 5: (2 im)- Gi R(x) l a thc d trong php chia f(x) : (x 2)(x2 x + 1), khi ta c: f(x) = (x 2).(x2 x + 1).P(x) + R(x)(1)- V a thc chia (x 2)(x2 x + 1) l a thc bc 3 nn a thc d R(x) c bc 2- T (1) d trong php chia f(x) : (x 2) chnh l d trong php chia R(x) : (x 2), m R(x) l a thc c bc 2, v f(x) : (x 2) d 4 (gt) R(x) = (x 2)(kx + p) +4- Lp lun tng t trn ng GD & T Nam Trc thi kho st cht lng hsg nm hc 2008-2009Mn: ton 8 Thi gian lm bi: 120 phtBi 1(4)Gii cc pt sau:a) 01312112 2 3++ ++ x x x x xb) 5200442003320022200112000++++++++x x x x xBi 2 (4) a)Tch ca 4 st nhin lin tip cng thm 1 l mt s chnh ph-ngb) 11...4131212 2 2 2< + + + +nBi 3 (3) Hai b nc cha y cng mt lng nc v mi b c1 vi x nc ra. Nu m vi b th nht th trong 20 pht b s ht nc. Nu m vi b th hai th trong 10 pht b s ht nc. Hi nu m hai vi cng mt lc thau bao lu s nc cn li trong b th nht nhiu hn s nc cn li trong b th hai l 3 ln, bit vn tc dng chy ca mi vi l khng i.Bi 4(3) Cho tam gic ABC cn ti A. Mt im D bt k ly trn cnh BC, k DEAB, DFac. Chng minh rng tng DE+DF khng i khi D di chuyn trn cnh BC.Bi 5 (4) Cho hnh vung ABCD c cnh di 20cm, Trn cnh CD ly im M. ng vung gc vi BM ct AD ti N.a) Tnh DN bit MC=5cmb) Tm v tr im M di DN ln nht.Bi 6 (2)Xc nh a phng trnh4x2+31y2=a + 6 - 17xyc nghim nguyn duy nht THI HC SINH GII MN TONLP 8QUN 1 TP H CH MINH NM HC 2002-2003( Thi gian lm bi : 90 pht)Bi 1: (3 im) Phn tch a thc thnh nhn t a) x2 +6x +5b) (x2-x +1) (x2 x+2) -12Bi 2: (4 im) a) Cho x+y+z = 0 .Chng minh x3 +y3 +z3 =3xyzab) Rt gn phn thc :3 3 32 2 23( ) ( ) ( )x y z xyzx y y z z x+ + + Bi 3 : (4 im) Cho x , y , z l di ba cnh ca tam gic A= 4x2y2 (x2 + y2 z2)2 .Chng minh A >0Bi 4 : (3 im)Tm s d trong php chia ca biu thc ( x+1)(x+3)(x+5)(x+7)+2002cho x2 +8x +12Bi 5: (6 im) Cho tam gic ABC vung ti A (AC >AB) ,ng cao AH .Trn tia HC ly HD= HA .ng vung gc vi BC ti D ct AC ti E a) Chng minh AE = AB b) Gi M l trung im ca BE .Tnh gc AHMPhng gd-t vnh tng kho st cht lng hsgMn:Ton8Thi gian lm bi 150 pht (khng k thi gian giao )I/ Trc nghim khch quan: Hy chn cu tr li ngtrong cc cu sau:Cu 1: Rt gn biu thc P=36 3 34 32 x xx vi x -1,6 x > 2 C.- 0,8x > -1,6 x< 2B.- 0,8x > -1,6 x > -2 D. - 0,8x > -1,6 x < -2Cu 3: Cho tam gic ABC cn A, AB = 32cm; BC = 24cm, -ng cao BK. Tnh di KC ta c:A.KC = 16 B. KC = 9 C.KC = 4 D. KC = 3Cu 4:Cho hnh thang ABCD ( AB// CD); AB = 3cm, CD = 5cm. Gi O l giao im ca cc ng thng AD v BC. Bit din tch tam gic OAB bng 27cm2. Tnh din tch hnh thang ta c:A.9 cm2B.25cm2C. 48cm2D. 75cm2II. T lun:Cu 1: Cho ba s t nhin:A = 444 ( c 2n ch s 4);B = 222 ( c n+1 ch s 2);C = 888 ( c n ch s 8);Chng minh rng A + B + C + 7 l s chnh phng.Cu 2: Chng minh rng tng cc bnh phng ca n s t nhin u tin :S = 12 + 22 +32 ++ (n-1)2 + n2 = 6) 1 2 )( 1 ( + + n n nCu 3: Gii v bin lun phng trnh n x18) 2 () 1 ( ) 1 2 ( 28) 2 (2 222 2++ + + + + x mm m xx mCu 4: Tm gi tr nh nht ca biu thc P = x2 + y2 xy x + y + 1Cu 5: Cho tam gic ABC c B v C l cc gc nhn, y BC di 20cm, ng cao AH di 10cm. Hnh ch nht MNPQ ni tip trong tam gic ABC sao cho M thuc AB, N thuc AC , P v Q thuc BC.a. t MQ = x; MN = y; Hy biu th y theo x.b. Tm gi tr ca x hnh ch nht MNPQ c din tch ln nht. Hng dn chm kho st cht lng hsg Mn: Ton 8I/Trc nghim khch quan ( 1 im)Cu 1 2 3 4p n ngD C B CCho im 0,25 0,25 0,25 0,25II/T lun Cu Ni dung im1(2 )Ta c: A +B+C+7=7 8 ... 88 2 ... 22 4 ... 441 2+ + ++ n n n=4* 7 1 ... 11 * 8 1 ... 11 * 2 1 ... 111 2+ + ++ n n n= 791 10* 891 10* 291 10* 41 2++++ n n n=21269 ... 6637 10 * 2

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.|

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.| + nn (pcm)0,50,50,50,52(2 )- Ta c ng thc sau:23= (1+1)3 = 13 +3.12.1+3.1.12 +1333= (2+1)3 = 23 +3.22.1+3.2.12 +13(n+1)3= n3 +3.n2.1+3.n.12 +13Cng tng v ri rt gn ta c:(n+1)3=1+3(12+22++n2) +3(1+2++n)+nThay 1+2++(n-1)+n=2) 1 ( + n nta c:3(12+22++n2)=(n+1)3-(n+1)-32) 1 ( + n n=1/2(n+1)(2n2+n)=1/2n(n+1)(2n+1)Vy S= 12+22++n2= 6) 1 2 )( 1 ( + + n n n (pcm)0,250,250,250,250,250,250,250,25BHCNAM K3(2 )Phng trnh cho 8m2x-32x=8m2+32m+32m2x-4x=m2+4m+4 (m-2)(m+2)x=(m+2)2 (*)- Nu m 2 t th pt c nghim duy nht x=22+mm- Nu m=2 th pt (*) 0x=4, pt v nghim- Nu m=-2 th pt (*) 0x=0, pt v s nghim0,250,250,50,50,250,254Ta c P=x2-x(y+1) + (y2+y+1)= (x-21 + y)2+(y2+y+1)-221

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.| + y= 43 2 32122+ ++ ,`

.| +y y yx=) 132(432122+ + + ,`

.| + y yyx=32)31(432122+ + + ,`

.| + yyxDo : P32 vi mi x, y. Du = xy ra khi x-21 + y=0 v y+1/3=0 x=2/3 v y=-1/3.Vy GTNN ca P=2/30,250,250,250,255( 2 )Gi K l giao im ca AH v MNa/ AMN ng dngABC nn 101020x yAHAKBCMN suy ra y=20-2xb/SMNPQ = (20 - 2x)x0,50,50,250,5QP= 20x- 2x2=-2(x2-10x+25)+50=-2(x-5)2+50suy ra SMNPQ 50 nn SMNPQln nht l 50m2 khi v ch khi x=5m ( khi MN l ng trung bnh ca tam gic ABC)0,25 thi hc sinh gii ton 8Bi 1: C/m rngA=75( + +...+ +4+1)+25 l s chia ht cho 100Bi 2: Cho a+b+c=1 vChng minh Bi 3: Tnh gi tr ca a thcP(x)= ti x=11Bi 4:An v Bnh cng lc t lng sang lng B cng mt b sng ri quay v A ngay. An i b, Bnh i thuyn vi vn tc ring ca thuyn bng vn tc i b ca An. Hi ai quay v sm hn?Bi 5:Cho tam gic ABC. Gi M l trung im ca BC. C/m rng AM 0Cu2:a. Gii phng trnh: 7911819846873 +++ x x x x = 4b. Cho x-2y = 1. Tm gi tr nh nht caA = x2+y2+4c. Tm s d ca php chia a thc x2008 x3 + 5 cho a thc x2 1Cu3:Cho AD l ng phn gic ca tam gic nhn ABC(AB