(3x+2)(x-1)-3(x+1)(x-2)=4
⇔3x2-3x+2x-2-3x2+6x-3x+6=4
⇔(3x2-3x2)+(-3x+2x+6x-3x)+(-2+6)=4
⇔2x+4=4
⇔2x=0
⇔x=0
Đúng 0
Bình luận (0)
Violympic toán 8
Các câu hỏi tương tự
Tìm x, biết:
(4x-1)2-(3x+2)(3x-2)= (7x-1)(x+2)+(2x+1)2-(4x2+7)
(5x-1)(x+1)-2(x-3)2= (x+2)(3x-1)-(x+4)2+(x2-x)
(-x+5)(x-2)+(x-7)(x+7)= (3x+1)2-(3x-2)(3x+2)
(2x+3)2-(5x-4)(5x+4)= (x+5)2-(3x-1)(7x+2)-(x2-1)
(1-3x)2-(x-2)(9x+1)= (3x-4)(3x+4)-9(x+3)2
1 tháng 12 2018 lúc 14:12
Tìm x biết:
1. (x-2)^2-(x-3)(x+3)=6
2. 4(x-3)^2-(2x-1)(2x+1)=10
3. (x-4)^2-(x-2)(x+2)=6
4.9(x+1)^2-(3x-2)(3x+2)=10
5. 3x +2(5-x)=0
6.x(2x-1)(x+5)-(2x^2+1)(x+4,5)=3,5
7, 3x^2-3x(x-2)=36
8. (3x^2-x+1)(x-1) +x^2(4-3x)=5/2
Tìm x
b) (x-5) (x-4) - (x+1)(x-2)=7
c) (3x-4)(x-2)=3x(x-9)-3
d)(x-3)(x^2+3x+9)+x(5-x^2)=6x
e) (3x-5)(x+1)-(3x-1)(x+1)=x-4
a) 5 - (x - 6) = 4 (3 - 2x)
b) x (x - 2) + 3x = (x + 3)2 + 11
c) (x + 2) (x - 2) + x = (x - 1)2 +1
d) (x + 1)2 - 4 (x - 3) = (x + 1) (x - 5)
f) (x - 3) (x + 4) - 2 (3x - 2) = (x - 4)2
Bài 1: giải các phương trình sau
a,(3x-1)(x+3)=(2-x)(5-3x) d,(3x+5)(2x+1)=(6x-2)(x-3)
b,(x+5)(2x-1)=(2x-3)(x+1) e,(x+2)2+2(x-4)=(x-4)(x-2)
c,(x+1)(x+9)=(x+3)(x+5) f,(x+1)(2x-3)-3(x-2)=2(x-1)2
a) (x+1/x-2)^2 + x+1/x-4 -3(2x-4/x-4)^2 = 0
b) 15x/x^2 +3x-4 - 1 = 12(1/x+4 + 1/3x-3)
c) x^2-4x+1/x+1 + 2 = - x^2-5x+1/2x+1
Giải phương trình:
a) (x^2+3x+1).((4x-3)/(3x+1)+2)=(4x+7).((4x-3)/(3x+1)+2)
b)(x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)
1) x ² + 2x -4 = -12 + 3x + x ²
2) x( 2x - 3 ) - x ² + 2 = x(x - 5) - 1
3) (x-1)(x+3) = x² -4
4) (6x + 2)(x-2) = 2x(3x-5)
5) (x-2) ² = (x-3)(x+2)
Phân tích các đa thức sau thành nhân tử :
a, x^2 + 4x + 3
b,16x - 5x^2 - 3
c, 2x^2 + 7x + 5
d, 2x^2 + 3x - 5
e,x^3 - 3x^2 + 1 - 3x
f, x^2 - 4x - 5
g, ( a^2 + 1 )^2 - 4a^2
h, x^3 - 3x^2 - 4x + 12
i, x^4 + x^3 + x + 1
k, x^4 - x^3 - x^2 + 1
l, ( 2x + 1 )^2 - ( x - 1 )^2
m,x^4 + 4x^2 - 5
Tìm x
a)(2x+1)2-4(x+2)2 =9
b)(3x-1)2 +2(x+3)2 +11(x+1)(1-x)=6
c)(x+1)3 - x2 (x+3)=2
d)(x-2)3 -x(x+1)(x-1)+6x2 =5
e)(x-3)(x2 +3x +9)-x(x+4)(x-4)=5
g)(x-2)3 -(x+5)(x2 -5x+25)+6x2 =11