解方程:(1)(x+1) 2 -144=0;(2)3(x-2) 2 =x(x-2);(3)(3x-2) 2 =(2x-3) 2 ;(4)(x-3)(x-1)=2.
(1)(x+1) 2 -144=0 ∴(x+1) 2 =144 ∴x+1=±12, 解得:x 1 =11,x 2 =-13; (2)3(x-2) 2 =x(x-2) 3(x-2) 2 -x(x-2)=0 (x-2)[3(x-2)-x]=0, (x-2)(2x-6)=0, 解得:x 1 =2,x 2 =3; (3)(3x-2) 2 =(2x-3) 2 (3x-2) 2 -(2x-3) 2 =0 [(3x-2)+(2x-30)][(3x-2)-(2x-30)]=0 ∴(5x-32)(x+28)=0, 解得:x 1 =
(4)(x-3)(x-1)=2 x 2 -4x+3=2 ∴x 2 -4x+1=0 b 2 -4ac=16-4×1×1=12, ∴x=
∴x 1 =2+
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