Question

已知x-3y=0,求[（2x+6)/(x^2-2xy+y^2)]*(x-y)的值.

Answer 1

[(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