Question

(x+2分之x-1)÷(x平方-4分之x平方-2x+1)+(x-1分之1）

再帮忙做一道（x平方-xy分之x平方-2xy+y平方）÷[（y分之x)-(x分之y)]

Answer 1

=(x-1)/(x+2)÷[(x-1)^2/(x+2)(x-2)+1/(x-1)
=(x-2)/(x-1)+1/(x-1)
=1
（x平方-xy分之x平方-2xy+y平方）÷[（y分之x)-(x分之y)]
=(x-y)^2/x(x-y)÷(x/y-y/x)
=(x-y)/x÷(x+y)(x-y)/xy
=y/(x+y)

Answer 2

1）原式=(x-1)/(x+2)÷(x-1)^2/(x+2)(x-2)+1/(x-1)
=(x-2)/(x-1) * (x+2)(x-2) / (x-1)^2 + 1/(x-1)
=(x-2)/(x-1)+1/(x-1)
=(x-1/(x-1)
=1
2）原式==(x-y)^2/x(x-y)÷(x/y-y/x)
=(x-y)/x* xy /(x+y)(x-y)
=y/(x+y)