初二分式难题！！！！

-10+10x^2/x^3+2x^2-x-2
=10(x^2-1)/[x^2(x+2)-(x+2)]
=10(x^2-1)/[(x+2)(x^2-1)]
=10/(x+2)

y/x=5/3
所以y/5=x/3
令y/5=x/3=k
y=5k
x=3k
(3x^2-5xy+2y^2)/(2x^2+3xy-5y^2)
=(3x-2y)(x-y)/(2x+5y)(x-y)
=(3x-2y)/(2x+5y)
=(9k-10k)/(6k+25k)
=-k/31k
=-1/31

对1.原式=10(x^2-1)/[(x+2)*(x^2-1)]=10/(x+2)
对2.上下除x*y，原式=[3*(x/y)-5+2*(y/x)]/[2*(x/y)+3-5*(y/x)]=-1/31

你唬我啊
哪有初二的题这么难