题中x-1是在分母上的吧
关键是由x的范围确定绝对值内的符号,
然后去掉绝对值号,最后与分母进行约分
当x>1时,x+1>0,x-1>0 故有
(|x+1||x-1|)/(x-1)=(x+1)(x-1)/(x-1)=x+1
当-1≤ x≤1时,x+1≥0,x-1≤0 故有
(|x+1||x-1|)/(x-1)=(x+1)[-(x-1)]/(x-1)=-(x+1)
当x<-1时,x+1<0,x-1<0 故有
(|x+1||x-1|)/(x-1)=[-(x+1)][-(x-1)]/(x-1)=x+1