Question

用分解因式法解下列方程 [1]x(x+1)-5x-5=0 [2](2x+3)^2-(x-1)^2=0 [3](x+根号3)^2=4根号3x

Answer 1

1.
x(x+1)-5x-5=0
x(x+1)-5(x+1)=0
(x+1)(x-5)=0
x=-1或x=5
2.
(2x+3)²-(x-1)²=0
[(2x+3)+(x-1)][(2x+3)-(x-1)]=0
(3x+2)(x+4)=0
x=-2/3或x=-4
3.
(x+√3)²=4√3x
x²+2√3x+3=4√3x
x²-2√3x+3=0
(x-√3)²=0
x=√3

Answer 2

[1]x(x+1)-5x-5=0

x(x+1)-5(x+1)=0
(x+1)(x-5)=0
x+1=0 x-5=0
x=-1 x=5

[2](2x+3)^2-(x-1)^2=0
(2x+3-x+1)(2x+3+x-1)=0
(x+4)(3x+2)=0
x=-4 x=-2/3

[3](x+根号3)^2=4根号3x
x^2-2根号3x+3=0 (展开后,合并)
(x-根号3)^2=0
x=根号3