题目

1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.（x+y分之x²-x+y分之y²）·x-y分之xy
4.（1-1-x分之1)÷1-x分之1+x
分式的加减法

提问时间：2021-03-21

答案

解1题：
原式=[1/(a+1)]-[2/(1-a)]
=[1/(a+1)]+[2/(a-1)]
={ (a-1)/[(a+1)(a-1)] }+{2(a+1)/[(a+1)(a-1)] }
=[(a-1)+2(a+1)]/[(a+1)(a-1)]
=(a-1+2a+2)/(a²-1)
=(3a+1)/(a²-1)
解2题：
原式=[x/(x-y)]-[y/(x+y)]
={ x(x+y)/[(x+y)(x-y)] }-{ y(x-y)/[(x+y)(x-y)] }
=[x(x+y)-y(x-y)]/[(x+y)(x-y)]
=(x²+xy-xy+y²)/(x²-y²)
=(x²+y²)/(x²-y²)
解3题：
原式=[x²/(x+y)-y²/(x+y)]×[xy/(x-y)]
=[(x²-y²)/(x+y)]×[xy/(x-y)]
=[(x+y)(x-y)/(x+y)]×[xy/(x-y)]
=xy
解4题：
原式=[1-1/(1-x)]÷[(1+x)/(1-x)]
={ [(1-x)-1]/(1-x) }×[(1-x)/(1+x)]
=[-x/(1-x)]×[(1-x)/(1+x)]
=-x/(1+x)