已知x+2y=0,求分式2xy+y^2/x^2-xy
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已知x+2y=0,求分式2xy+y^2/x^2-xy
![已知x+2y=0,求分式2xy+y^2/x^2-xy](/uploads/image/z/2252372-68-2.jpg?t=%E5%B7%B2%E7%9F%A5x%2B2y%3D0%2C%E6%B1%82%E5%88%86%E5%BC%8F2xy%2By%5E2%2Fx%5E2-xy)
x=-2y
(2xy+y^2)/(x^2-xy)
=(-4y^2+y^2)/(4y^2+2y^2)
=-3y^2/(6y^2)
=-1/2
再问: 我算出来也是-1/2,可答案说是-15/28
(2xy+y^2)/(x^2-xy)
=(-4y^2+y^2)/(4y^2+2y^2)
=-3y^2/(6y^2)
=-1/2
再问: 我算出来也是-1/2,可答案说是-15/28
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