Đề tuyển sinh 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