VI x^+ xy =2 giup c6 diTdcb) Nhan ra rkng x +
a) Co the dSt phep tinh chia da thuTc x* cho da thtfc x^-x 1.
b) ?hai nghl den van dung cau a) de giai, tif do nhan ra rkng xf = Sx, + 5,
X2 = 8x2 + 5. Tat nhien phai nghl den bien đi A ve dcfn gian hdn vi chung ta da CO xf - Xi - 1 = 0 va X 2 - X2 - 1 = 0
Bai 4: V i X, y e Z nen : t ^ ;
5x2 + y2 _ 2xy + 2x - 6y + 1 < 0 » sx^ + y^ - 2xy + 2x - 6y + 2 < 0 o (x^ + y^ + 9 - 2xy + 6x - 6y) + (4x^ _ 4x + 1) - 8 < 0
o ( x - y + 3)' + ( 2 x - 1) ' < 8
Cty TNHH IVITV DWH Khang ViQt
Ta CO : (2x - 1)^ < 8 ; (2x - if so chinh phiTdng lẹ
Do vay chi c6 : (2x - 1)^ = 1^ <=> 2x - 1 = 1 hoac 2x - 1 = - 1 <r> x = 0 hoSc x = 1. V i (2x - 1)' = 1 nen (x - y + 3)^ < 7. Do d6 (x - y + 3)' = Ó; 1 ^ 2\
• x = 0 : T a c 6 : ( 3 - y ) ' = 0 ' ; 1 ^ 2' o y G { 3 ; 2 ; 4 ; 1 ; 5 } • x = l : T a c 6 : ( 4 - y ) ' = 0 ^ 1 ^ 2 ^ « y e { 4 ; 3 ; 5 ; 2 ; 6 }
Cac cap so nguyen (x ; y) can tim la : (0 ; 3 ) ; (0 ; 2 ) ; (0 ; 4 ) ; (0 ; 1 ) ; (0 ; 5 ) ; (1 ; 4 ) ; ( 1 ; 3 ) ; ( 1 ; 5 ) ; ( 1 ; 2 ) ; ( 1 ; 6 ) .
Nh§n xet : Neu a e Z thi a < 0 o a + 1 < 0. ?hat hien (2x - l)Ma so chinh phiTdng le m^ (2x - 1)^ < 8. Do vay chi c6 (2x - 1)^ = l ^ that doc daọ
hi 5: Cdch 1: U J V i
Gpi N la giao diem cua D M va diTdng tron (O). ; .(.Ih
A D N = 90" =^ A N la dif^ng kinh cua diTdng tron (O) Do do N la diem chinh giffa cua cung BC.
Qua M ve diTdng thing song song vdi AC Celt AB d S, ve diTdng thing song song vdi AB cat AC d K. Ta c6 tu" giac ASMK la hinh binh h^nh.
Ta c6 : SM // AC => BSM = B A C , SMB = ACB N Nen SMB = SBM (= A C B ) ^ ASBM can tai S ^ SB = SM
Mat khac B D M = B A N = - B A C = - B S M ' ' ' 2 2
S la giao diem cua diTdng trung triTc doan thing B M va cung chtfa g6c c6 só do bkng 2 B D M nen S la tam diTdng tr6n ngoai tiep tarn giac D B M => SD = SM. Chtfng minh tu-dng tif, ta c6 K D = K M .
Do do SK la dirdng trung triTc cua doan thing D M nen E s S, F s K. Tiir giac ASMK la hinh binh hanh do do tu" giac AEMF la hinh binh h l n h .
Cdch 2 : Goi N , I , K Ian lifdt la giao diem cua EF vdi BD, A M , D M . Ta CO : A D 1 D M , N F1 D M
=> A D // NF => A D N + D N F = I8O"
Ma A D N + ACB = ISO" (ADBC noi tiep) Suy ra DNF = ACB => Ttf gidc BNFC noi tiep. A M A D CO NF // A D , K D = K M l A = I M Ta CO : A N M - END (doi xtfng true),
LuySn giai dg iruOc k1 thi vao I6p 10 b a m j g n Bac, Trung, Nam mOn Toan _ Nguyjn Pile iSn
> Tỉ gi^c BNEM noi tiep M E // AC
ENM = EBM DNE - EMB
Nen EMB = ACB
Xet A I E M va A I F A c6 E I M = F I A (doi dinh), I M = l A , I M E = I A F (ME / / AC) Do d6 A I E M = A I F A (g.c.g) => M E = A F
Tỉ giac A E M F c6 M E // A F va ME = A F . Vay tuT gidc A E M F Ik hmh blnh hknh. Nhan xet : V i phat hien day la hhi toan "ngtfcJc" cua mot bai toan quen thuoc, do vay dilng ky thuat " H I N H D U Y N H A T " de c6 diTdc cdch giai 1. Bai 6: Ve ES 1 B C tai S. Goi J Ik giao diem cua D S va M H .
Ve EN 1 A C tai N , D T 1 A B tai T , D I 1 BC tai I .
Ta c6 : M L // DT, M K // EN, ES // M H // D I . Ma BD, CE la cac diTcJng phan giac cua tarn giac A B C nen DT = D I , EN = ES.
M K D M ADEN c6 M K // EN : ADEN c6 M K // EN : ADES CO MJ // ES =: M K M J ^ Do vay EN ES EN DE MJ _ D M ES ~ DE • D M ^ ' D E , EN = ES ^ M K = MJ ADET CO M L // DT ASDI CO JH // D I => M L ^ EM DT ~ ED SJ JH ; ADES c6 MJ // ES E M SJ ED SD . Nen CO M L _ JH DT ~ D I , DT = D I => M L = JH. SD D I Do vay : M H = MJ + JH = M K + M L
Nhfin x e t: Day la mot bki toin quen thupc doi vdi hpc sinh gioi todn Idp 8. B a i 7 : V i a b + bc + ca = l . T a c 6 :
â + 1 = â + ab + be + ca = ăa + b) + c(a + b) = (a + b)(a + c) TiTdng tir, ta CO : b H 1 = (a + b)(b + c ) ; c^ + 1 = (a + c)(b + c) Dodo: "'^ ^ p = Vâ + l.Vb^ + 1 ^/?" l.Vc^ + 1 /â + 1 4 = V(a + b)^ + 7(b + c)^ + x/(c + a)^ = a + b + b + c + = 2(a + b + c) = 2\/(a + b + c)^ c + a = 2, | ^ L ( a - b r + ( b - c r + ( c - a r J + 3 > 2 V 3
Dau " = " xay ra<=>a = b = c = V3
VSy gia tri nho nhát ciia bieu tMc P Ik 2 V3 .
Nh§n x e t: De nhan ra tuf ab + be + ca = 1 giup c6 diTPc P = 2(a + b + c) Tiif a, b, c diTdng vk ab + be + ca = 1, khong kho khkn l^m de c6 dmc P > 2 Vs .
* D ] g s 6 5 9
D E THI T U Y E N SINH VAO L(3P 10 C H U Y § N , TRUflNG THPT C H U Y E N DAI HQC S I / PHAM HA NOI TRUflNG THPT C H U Y E N DAI HQC S I / PHAM HA NOI
NAM HQC 2011 - 2012
Bai 1: (2 diem) Cho bieu thiJc:
A = X - y x ' • + — + + y-2) ^ 4x'* + 4x^y + y^ - 4
X + y + xy + X
^ 2 y - x 2 y ^ + x y - x ^ vdi X > 0, y > 0, X ;^ 2y, y ;^ 2 - 2x1 a) Rut gon bieu thtfc Ạ
2 b) Cho y = 1, hay tim x sao cho A = - . b) Cho y = 1, hay tim x sao cho A = - .
5
B^i 2: (2 diem) Mot nhom cong nhan dkt ke hoach san xuat 200 san pham. Trong 4 ngay dau hp thiTc hien diing miJc de ra, nhiirng ngay c6n lai hp da 1km viTdt mdrc moi ngky 10 san pham, nen da hoan thanh ke hoach sdm 2 ngkỵ Hoi theo ke hoach moi ngay nhom cong nhan can san xuat bao nhieu san pham?
Bai 3: (2 diem) Cho parabol (P): y = x^ vk diTdng thing (d): y = mx - + 3, m la tham sọ Tim taft ca cdc gia tri cua m de diTcfng thing (d) ck't parabol (P) tai hai diem phan biet c6 hoanh dp x,, X j . Vdi gia tri nao cua m thi x,, X2 la dp dki cac canh goc vuong cua mot tam gidc vuong c6 dp dki canh huyen
B^i 4: (2 diem) Cho diTdng tr6n (O) diTdng kinh AB = 10. Day cung CD cua dirdng tron (O) vuong goc vdi AB tai diem E sao cho AE = 1. Cac tiep tuyen tai B va C cua difdng tron (O) c i t nhau tai K, A K va CE cil nhau tai M . ^) Chu-ng minh AAEC ^ AOBK. Tinh BK.
Bai 5: (1 diem) Cho hinh thoi ABCD c6 BAD= 120". C^c diem M N chay tren cdc canh BC CD tu"dng ufng sao cho MAN = 30". Chiang minh r^ng tam dircJng tron ngoai tiep tam giac MAN thuoc mot du"dng thing co dinh.
Bai 6: (1 diem) ChuTng minh bat ding thufc:
1 ^ 1 ^ 1 1 +... + > 4 +... + > 4 HI/6NG DAN GIAI Bai 1: a) A = X - y _^ X " + y + y - 2 2y - X 4x + 4x^y + y^ - 4 ,2 X ' + y + xy + X x2 2y + xy - X _ ( x - y ) ( x + y) + x^ +y^ + y - 2 (2x^ + y)^ - 4 ( 2 y - x ) ( x + y) ' x ( x + y) + x + y - y^ + x^ + y ' + y - 2 . (2x^ + y + 2)(2x' + y - 2) ( 2 y - x ) ( x + y) ' (x + y)(x + l) (x + y)(x + l) 2x^ + y - 2 (2y - x)(x + y)'(2x2 + y + 2)(2x^ + y - 2) X + 1 ( 2 y - x ) ( 2 x 2 + y + 2) X + 1
b) Vdi y = 1, ta co: A = 7 — r , dieu kien x 2. (2 - x)(2x2 + 3) (2 - x)(2x2 + 3) , A = - o ^V^i . = | o 5 ( x + l ) = 2(2-x)(2x^ + 3) 5 (2 - x)(2x2 + 3) 5 o 5x + 5 = 8x^ + 12 - 4x^ - 6x o 4x^ - 8x^ + 1 Ix - 7 = 0 <z> 4x3 _ _ + 4x + 7x - 7 = 0 o (X - ^^(^^^2 _ 4x + 7 ) = 0 o ( x - l ) [ ( 2 x - l ) S 6 ] = 0 o x - 1 = 0 o x = 1.
J'U i i K i g limi ^>';it. nv<
Nhfin xet:
' a) Day Ih bai loan dẹ
b) Neu phiTdng trinh ax^ + bx^ + cx + d = 0 co a + b + c + d = 0 thi c6 mpt
^ nghiem la 1.
Bai 3: PhiTdng trinh hoanh dp giao diem cua (?) (d) ML:
x^ = mx - m^ + 3 <=> x^ - mx + m^ - 3 = 0 (*)
(d) Celt (?) tai hai diem phan biet o (*) c6 hai nghiem phan biet <=> A = m^ - 4(m^ - 3) > 0 o -3m^ + 1 2 > 0 < o m ^ < 4 o - 2 < m < 2
Cty TNHH MTV DWH Khano ViOt
(d) c^t (?) tai hai diem phan biet co hoanh dp x,, x,, X j la dp dai cac canh cua goc vuong cua mpt tam giac vuong co dp d^i canh huy^n bang ^
A > 0 X , > 0 X , > 0 X2 > 0 X| + ^^2 = ^(dinhlỉy-ta-go) -2 < m < 2 X| + X j = m > 0 ' ' ' ' x.x, = m^ - 3 > 0 ( X , + X 2) ' - 2 X, X 2 = ^ ^ ^ ^ . ^ . ^ , -2 < m < 2 <•(•.; '•• m > 0 m > v/3hoacm < - N/ 3 <=> m = m 2- 2 ( m^ - 3) = : l ^ 2 • Nhanxet: i ,
• (d) c^t (?) tai hai diem phan biet x,, X2 o ?hirdng trinh hoanh dp giao diem cua hai nghiem phan biet.
• X|, X2 la dp dai cac canh cua goc vuong cua mpt tam giac vuong cp dp dai canh huyen bang ^ o x, > 0, Xj > 0 va xf + x^ = | (dinh li ?y-ta-go)
Bai 2: Gpi so san pham theo ke hoach moi ngay nhom cong nhan can san xuat la x (san pham) (Dieu kien: x e N*)
ThcJi gian hoan thanh theo ké hoach la (ngay)
X
So san pham lam trong 4 ngay dau la 4x (san pham)
Sp san pham nhffng ngay con lai phai lam la: 200 - 4x (san pham)
So san pham moi ngay nhom cong nhan can san xuát trong nhCTng ngay con lai la x + 1 0 (san pham)
Thdi gian hoan thanh sp san pham con lai la: (san pham) '! fi ^
= 3U
X + 10 Theo de bai ta cp phúdng trinh: 200 ^200-4x ^ Theo de bai ta cp phúdng trinh: 200 ^200-4x ^
x + 10 + 4 = 2 ^ 200 2 0 0 - 4 x ^ 100 100 - 2x w = t) <=> . _ 3 X x + 10 X x + 10 . , ^ lOOx + 1000 - lOOx + 2x' = 3x' + 30x o x' + 30x - 1000 = 0 A '= 225 + 1000 = 1225, V A' = 35
Luy^n giJi dS tru6c kl thi vAo Bp 10 ba mign B^c, Trung, N.m mOn Tpan _ Nguyin BOrc Jin
X, = = 20 (thich hdp) ; X j = = -50 (khong thich hdp)
Vay theo ke hoach moi ngay nh6m cong nhan can san xuat 20 san pham.
Nhan xet: Bai toan lap phiTdng trinh dang nang suaft, b ^ i todn quen thupc.
PhiTdng trinh thiTcJng lap \h:
Bai 4: a) K C , K B la cdc tiep tuyen cua diTctng tr5n (O) ^, ^
=> O K la tia phan gidc cua g6c COB
1 1
=> KOB = - C O B . M a CAE = - C O B (h? qua g6c npi tiep) 2 2
X 6 t A A E C va A 0 B K c 6 : .-i
AEC - OBK (= 90"). CAE = KOB Do do AAEC ^ AOBK (g.g). Do do AAEC ^ AOBK (g.g). Ma A E = 1, A B = 10 nen OE = 4.
Ta CO AECO vuong tai E
CÊ + OÊ = OC^ (dinh l i Py-ta-go) C E = VOC^ - OÊ = Vs^ - 4^ = 3 A E CE Ta c6: OB B K 3.5 (AAEC ^ AOBK) => B K = V a y B K = — = 15 (dvđ)
b) Xet A A B K c6 M E // B K (vi M E 1 A B . B K 1 AB) M E A E