Đề tuyển sinh lớp 10 môn Toán năm 2021 - 2022 trường THPT chuyên Hà Tĩnh

SO GIAO DVC VA DAO T(O HA TiNH

DE THI CHiNH THUC

Cau 1. (1,5 diim)

Cho a, b, c la cac s6 thgc doi m9t phan bimt, rut g<;m binu thuc:

KY THI TUYEN v Ao LOP 10 TRUONG THPT CHUYEN HA TiNH NAM HQC: 2021 - 2022 MON: TOAN CHUYEN Thai gian lam bai: 15 0 phut

A=-----'-(a_- _b8) 3 _+9( _b-_c:) 3 _+_(;c _-<a) _ 3 _

a2 (b-c)+b2 (c- a)+c2 (a- b) ·

Cau 2. (2,5 diim)

a) Giai phucmg trinh: 9x2 = (x2 +x-s)(.J3x+l-lr.

{ x3 +xy2 +2y3 = 0 b) Giai hm phucmg trinh: ✓----- 2x3 -x +8y2 +3y = 4

Cau 3. (2,5 diim)

a)Timcacs6nguyen m ,n thoa man m(m+l)(m+2) =n 2 •

b) Cho x, y la cac s6 thgc ducmg thoa man X + y + xy = 3. Tim gia tri l&n nhdt cua binu thuc:

Cau 4. (2,5 diim).

Cho nua duong trcm tam O duong kinh AB. G9i I la dinm chinh giua cung AB. Tren cung l&n AB cua duong trcm tam I ban kinh IA ldy dinm C sao cho tam giac ABC nh9n. G9i M, N lful lugt la giao dinm cua CA, CB

v6i nua duong trcm duong kinh AB ( M khac A, N khac B); J la giao dinm cua AN v6i BM.

a) Chung minh b,,.MBC va b,,.NAC la cac tam giac can.

b) Chung minh I la trgc tam cua tam giac CMN.

) G . K 1' d"" ' L'J 'nh ' ;.. CJ c 91 a trung 1em cua , t1 ti so OK.

Cau 5. (1,0 diim)

Cho top hgp X = {l; 2; 3; 4; 5; 6; 7; 8; 9}, chia tpp hgp X thanh hai top hgp khac rfing va khong c6 phful tu

chung. Chung minh rfulg v6i m9i each chia thi luon tfin tl;li 3 s6 a, b, C trong m9t top hgp thoa man a+ C = 2b.

----HET----

H9 va ten thi sinh: .......................................... S6 bao danh: ......................................... .

LOI GIA.I CHI TIET

Cau 1. (1,5 diim)

Cho a, b, c la cac s6 thàc doi m<)t phan biTt, rut gQn bi€u thuc:

Loi giai

Ta biUt rfulg nUu x+ y+z = 0 thi x3 + y3 +z3 -3.xyz = (x+ y+z)(x2 + y2 +z2 -xy-yz-zx) = 0 do do:

Taco:

x3 + y3 +z3 = 3xyz.

a2 ( b - c) + b2 ( c - a)+ c2 ( a - b) = ( a - b) c2 -( a2 - b2 ) c + ab ( a - b)

= (a-b)[c2 -(a+b)c+ab] = (a-b)(c-a)(c-b) = -(a-b)(b-c)(c-a).

DVt x = a-b, y = b-c, z = c-a khi do ta co:

VWy A=-3.

Cau 2. (2,5 diim)

x3 + y3 +z3 A=-----=-3. xyz

a) Giai phucmg trinh: 9x2 = (x2 +x-s)(.J3x+1-1f.

{ x3 +xy2 +2y3 = 0 b) Giai hT phucmg trinh: ✓ - 2x3 -x +8y2 +3y = 4

Loi giai

a) DiXu kiTn: x 2:-½- Phucmg trinh tucmg ducmg:

Taco:

9x2 ( .J3x + 1 + 1 r = ( x2 + X - 5 )(-J3x + l -1 r ( .J3x + 1 + 1 r

¢? 9x2 ( .J3x+ 1 +if= (x2 +x-5)ã9x2

x=O

{:} x2 +x-5 = (-J3x+ 1 + 1 r (1)"

(1)$ x 2 -2x-7 = 2✓3x+l $ 2x 2 -4x-l4 = 4✓3x+l

$ 2x 2 -7x-l5+✓3x+l (.J3x+l-4) = 0

3(x-5)✓3x+l $ (x-5)(2x+3)+ Fx+l = 0 3x+l +4

$(x-5)(2x+3+ J!!ff1 ]=o 3x+l +4

x= 5 $ 2 3 3✓3x+l 0. x+ + = -= ✓3x+l +4

Do x2'.-_!_=?2x+3>0=?2x+3+ J!!ff1 =0 vilnghi . 3 3x+l +4

V<y phucmg trinh da cho c6 nghi=m x = 0, x = 5.

1 --<x<0 b)ri.ã::. ki> 2 3 >0 ✓2 - - v1eu en: x - x ? . - 1 x>- -✓2

Phucmg trinh thu nhdt cua h= tucmg ducmg:

x3 +xy2 +2y3 = 0 ? (x3 + l)+(xy 2 + l)= 0 ( X + y) ( X2 - xy + y 2 ) + y 2 ( X + y) = 0

$ (x+ y)(x 2 -xy+2y 2) = 0

X=-y $ x 2 -xy+2y 2 = 0

Ven X = -y, thay vao phucmg trinh thu hai ta dugc: ✓2x 3 -x +8x 2 -3x-4 = 0.

{

x-0 ;. 3 -X 2: 0 X::::; 0 - Neu ✓2x -x+x=O${ 3 2 ( )( ) l- 2x -x = x x 2x + 1 x -l = 0 x = --2

V 6i x = 0, ta thdy phucmg trinh vo nghi=m va phucmg trinh nh@ m<)t nghi=m la x = - .

Do d6 xet x :;t= _ _!_, phucmg trinh tucmg ducmg: 2