解
(10a³bc)/(-5a²b³c²)
=(5a²bc)(2a)/(5a²bc)(-b²c)
=-(2a)/(b²c)如果分式x+1分之x的2次方-1的为0，那么x为？(x²-1)/(x+1)=0(x-1)(x+1)/(x+1)=0x-1=0x=1约分m的2次方-m分之m的2次方-2m+1解(m²-2m+1)/(m²-m)=(m-1)²/[m(m-1)]=(m-1)/m