若 $x^{2}-2 x-2=\left(x^{2}-4 x+3\right)^{0}$, 则 $x$ 的值为 $(\quad)$
【点拨】根据题意x^2-2x-2=1,x^2-4x+3neq0,然后利用因式分解法求解即可.【解析】解:becausex^2-2x-2=left(x^2-4x+3right)^0,thereforex^2-2x-2=1,x^2-4x+3neq0,由x^2-2x-2=1整理得x^2-2x-3=0,therefore(x-3)(x+1)=0,thereforex=3或x=-1,x^2-4x+3neq0可知(x-3)(x-1)neq0,thereforexneq3或xneq1,therefore若x^2-2x-2=left(x^2-4x+3right)^0,则x的值为-1,故选A.【点睛】本题考查了解一元二次方程一因式分解法,零指数幂的意义,正确分解因式是解题关键.