√[（x+1）（-x²+2x-1）]
=√[-(x+1)(x²-2x+1)]
=√[-(x+1)(x-1)²]
若二次根式有意义需
-(x+1)(x-1)²≥0
∵x0,
∴-(x+1)>0成立
∴原式=|x-1|*√[-(x+1)]
=(1-x)*√(-x-1)
已经最简了