2x²-8=0 (x-2)²-x+2=0 (x-3)²+2x(x-3)=0 (x-2)²=2x-4 (2x-1)²=x²-4用一元二次方程的因式分解法解

问题描述:

2x²-8=0 (x-2)²-x+2=0 (x-3)²+2x(x-3)=0 (x-2)²=2x-4 (2x-1)²=x²-4
用一元二次方程的因式分解法解

1、2(X+2)(X-2)=0,X+2=0或X-2=0,X1=-2,X2=2.2、(X-2)^2-(X-2)=0(X-2)(X-3)=0X1=2,X2=3,3、(X-3)^2+2X(X-3)=.0(X-3)(X-3+2X)=0X-3=0或3X-3=0X1=3,X2=1,4、(X-2)^2-2(X-2)=0(X-2)(X-4)=0X1=2,X2=4.5、4X^2-4X+1-X^2+4=...