Question

一元二次方程——分解因式法用分解因式法解方程：（要有具体过程）1）（2x-1）^2=3x（2x-1）2)4(x-3)^2

Answer 1

(2x-1)^2=3x(2x-1)

(2x-1)(2x-1-3x)=0

(2x-1)(-x-1)=0

-2(x-1/2)(x+1)=0

x=1/2,x=-1

4(x-3)^2=25(x-2)^2

(2x-6)^2-(5x-10)^2=0

(2x-6+5x-10)(2x-6-5x+10)=0

(7x-16)(4-3x)=0

-21(x-16/7)(x-4/3)=0

x=16/7,x=4/3

(2x+1)^2+4(2x+1)=-4

(2x+1)^2+4(2x+1)+4=0

(2x+1+2)^2=0

(2x+3)^2=0

x=-3/2

(x+3)(x-1)=4x-4

(x+3)(x-1)-4(x-1)=0

(x-1)(x+3-4)=0

(x-1)(x-1)=0

x=1