Question

数学题！！！急

7、8两题，谢谢了！

Answer 1

解：(7)①原式=(x-1)+1/x-1/x=x-1[按乘法分配律展开]
②原式=(3-a)/[2(a-2)]÷[(a²-9)/(a-2)]=-(a-3)/[2(a-2)]×(a-2)/[(a+3)(a-3)]=-1/(2a+6)
(8)①原式=[(a-2)(a+2)/(a-2)²-1/(a-2)]×(a+2)/(a+1)=(a+1)/(a-2)×(a+2)/(a+1)=(a+2)/(a-2)=-5/3
②原式=[(x+y)²/(x-y)][(x-y)²/(x+y)]=(x+y)(x-y)=199
[先通分，分子分母分解因式，约分再代入数值]

追问

也谢谢你啦！

追答

不客气。

Answer 2

追问

第8题呢？

追答

Answer 3

x*[(x-1)/x]-[1/(1-x)]*[(x-1)/x]-(1/x)
=x-1+(1/x)-(1/x)
=x-1

[(3-a)/(2a-4)]÷[(a^2-4-5)/(a-2)]
={(3-a)/[2(a-2)]}*{(a-2)/[(a-3)(a+3)]}
=(-1/2)*[1/(a+3)]
=-1/(2a+6)

{[(a+2)/(a-2)]-[1/(a-2)]}*[(a+2)/(a+1)]
=(a+2)/(a-2)
=(5/2)/(-3/2)
=-5/3

[(x+y)^2/(x-y)]*[(x-y)^2/(x+y)]
=(x+y)(x-y)
=(100+99)(100-99)
=199

追问

谢谢！