Question

分式计算。。。谢谢各位了。

Answer 1

解14题：
原式=[(a+1)/(a-1)]-[(a²+a)/(a²-1)]
=[(a+1)/(a-1)]-[a(a+1)]/[(a+1)(a-1)]
=(a+1)/(a-1)-a/(a-1)
=(a+1-a)/(a-1)
=1/(a-1)

解15题：
原式={[(a+b)(a-b)]/[ab(a-b)]}÷[(2ab+a²+b²)/(2ab)]
=[(a+b)/(ab)]÷[(a+b)²/(2ab)]
=[(a+b)/(ab)]×[2ab/(a+b)²]
=2/(a+b)

解16题：
原式=[(x+y)(x-y)/(x+y)]-[(4x²-4xy+y²)/(2x-y)]
=x-y-[(2x-y)²/(2x-y)]
=x-y-(2x-y)
=x-y-2x+y
= -x

Answer 2

14、（a+1)/(a-1)-(a²+a)/(a²-1)
=（a+1)/(a-1)-a(a+1)/(a-1)(a+1)
=（a+1)/(a-1)-a/(a-1)
=1/(a-1)
15、(a²-b²)/(a²b-ab²)÷[1+(a²+b²)/2ab]
=(a-b)(a+b)/ab(a-b)×2ab/(a²+b²+2ab)
=(a+b)/ab×2ab/(a+b)²
=2/(a+b)
16、(x²-y²)/(x+y)-[4x(x-y)+y²]/(2x-y)
=(x-y)(x+y)/(x+y)-(4x²-4xy+y²)/(2x-y)
=(x-y)-(2x-y)²/(2x-y)
=(x-y)-(2x-y)
=x-y-2x+y
=-x