Question

这六道题用因式分解法来解 谢谢

Answer 1

(1) x^2+x=0
x(x+1)=0
x1=0, x2=-1.
(2) x^2-(2根号3)x=0
x(x-2根号3)=0
x1=0, x2=2根号3.
(3) 3x^2-6x=-3
3x^2-6x+3=0
3(x^2-2x+1)=0
3(x-1)^2=0
x1=x2=1.
(4) 4x^2-121=0
(2x+11)(2x-11)=0
x1=-11/2, x2=11/2.
(5) 3x(2x+1)=4x+2
6x^2+3x-4x-2=0
6x^2-x-2=0
(2x+1)(3x-2)=0
x1=-1/2, x2=2/3.
(6) (x-4)^2=(5-2x)^2
(x-4)^2-(5-2x)^2=0
[(x-4)+(5-2x)][(x-4)+(5-2x)]=0
(x-4+5-2x)(x-4-5+2x)=0
(-x+1)(3x-9)=0
-3(x-1)(x-3)=0
x1=1, x2=3.

Answer 2

Answer 3

追答