[(2x+6)/(x^2-2xy+y^2)]*(x-y)是不是给错了,应该是
[(2x+6y)/(x^2-2xy+y^2)]*(x-y)吧?
x-3y=0得x=3y代入
[(2x+6y)/(x^2-2xy+y^2)]*(x-y)=[12y/(x-y)^2](x-y)
=12y/(x-y)
=12y/2y
=6