若x/2=y/3,则分式7x^2-y^2/x^2-2xy+3y^2的值是 多少?

题目

若x/2=y/3,则分式7x^2-y^2/x^2-2xy+3y^2的值是 多少?
已知2x-y,求分式（1+3y^3/x^3-y^3）/（1+2y/x-y）的值？

答案

由已知得：x=2y/3,代入欲求式子,得
(7x^2-y^2)/(x^2-2xy+3y^2)
=[7*(2y/3)^2-y^2]/[(2y/3)^2-2*2y/3*y+3y^2]
=(28y^2/9-y^2)/(4y^2/9-4y^2/3+3y^2)
=(19y^2/9)/(19y^2/9)
=1