Ta có
( 2 x 4 – 3 x 3 + x 2 ) : - 1 2 x 2 + 4 ( x – 1 ) 2 = 0 ⇔ 2 x 4 : ( - 1 2 x 2 ) - 3 x 3 : ( - 1 2 x 2 ) + x 2 : ( - 1 2 x 2 ) + 4 ( x 2 - 2 x + 1 ) = 0 ⇔ - 4 x 2 + 6 x – 2 + 4 x 2 – 8 x + 4 = 0
ó -2x + 2 = 0
ó x = 1
Đáp án cần chọn là: C
Ta có
( 2 x 4 – 3 x 3 + x 2 ) : - 1 2 x 2 + 4 ( x – 1 ) 2 = 0 ⇔ 2 x 4 : ( - 1 2 x 2 ) - 3 x 3 : ( - 1 2 x 2 ) + x 2 : ( - 1 2 x 2 ) + 4 ( x 2 - 2 x + 1 ) = 0 ⇔ - 4 x 2 + 6 x – 2 + 4 x 2 – 8 x + 4 = 0
ó -2x + 2 = 0
ó x = 1
Đáp án cần chọn là: C
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
Tìm x,biết
a) ( x+2)×(x+3)-(x -2)×(x+5)=0
b) (2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4)
c) (8-5x)×(x+2)+4(x-2)×(x+1)+2(x-2)×(x+2)=0
d) (8x-3)×(3x+2)-(4x+7)×(x+4)=(2x+1)×(5x-1)-33
Tìm x, biết.
a) x+ 5x2 = 0 b)(x+3)2+(4+x)(4-x)=10
c) 5x( x – 1) = x - 1 d) x2 -2x -3 = 0
tìm x biết
a)3x3-12x=0
b)(x-3)2-(x-3)(3-x)2=0
c)4(x+1)+(2x+1)2-8(x-1)(x+1)=11
d)(x+3)(x2-3x+9)-x(x-4)(x+4)
Tìm x , biết :
a, x mũ 2 - 2x + 1 = 25
b, 4 x mũ 2 - ( x + 4 ) mũ 2 = 0
c, 9 - 64 x mũ 2 = 0
d, 9 ( 4 x + 3 ) mũ 2 = 16 ( 3 x - 5 ) mũ 2
Tìm x biết
a) x(x-5)(x+5)-(x-2)(x^2+2x+4)=3
b) 25x^2-2=0
c) (x+2)^2=(2x-1)^2
d) (x+2)^2-x^2+4=0
e) (x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0
Tìm x,biết:
a) (x-3)^2-4=0
b) x^2-9=0
c) x(x-2x)-y^2-8=0
d) 2x(x-1)-2x^2+x-5=0
e) x(x-3)-(x+1)(x-2)=0
f) x(3x-1)-3x^2-7x=0
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0
b) 9.(2x+1)2-4.(x+1)2=0
c) x3-6x2+9x=0
d) x2.(x+1)-x.(x+1)+x.(x-1)=0
10 Phân tích các đa thức sau thành nhân tử:
a) 5xy(x-y)-2x+2y ; b) 6x-2y-x(y-3x)
c) x^2+4x-xy-4y ; d) 3xy+2z-6y-xz
11 Tìm x, biết: a) 4-9x^2=0 ; b) x^2+x+1/4=0 ; c) 2x(x-3)+(x-3)=0
d) 3x(x-4)-x+4=0 ; e) x^3-1/9x=0 ; f) (3x-y)^2-(x-y)^2=0