解方程:
(1)x2-5=0????????????????????(2)x2+2=3(x+2)
(3)x2+4x-1=0?????????????? ??(4)(x-2)2-3(x-2)=0.
网友回答
解:(1)x2-5=0,
x2=5,
x 1=,x 2=-;
(2)x2+2=3(x+2),
x2-3x-4=0,
(x-4)(x+1)=0,
x 1=4,x 2=-1;
(3)x2+4x-1=0,
(x+2)2=5,
x 1=-2+,x 2=-2-;
(4)(x-2)2-3(x-2)=0,
(x-2)(x-5)=0,
x 1=2,x 2=5;
解析分析:(1)利用直接开平方法求出一元二次方程的根即可;
(2)运用因式分解法将原式分解因式,得出(x-4)(x+1)=0,即可得出