Question

初二数学：已知x^2-xy-2y^2=0,且x≠0,y≠0,求代数式x^2-2xy-5y^2/x^2+2xy+5y^2的值.

^2表示平方. 欢迎高手指点！

Answer 1

x^2-xy-2y^2=0
(x-2y)(x+y)=0
x=2y或x=-y

(x^2-2xy-5y^2)/(x^2+2xy+5y^2)
=(x^2-xy-2y^2-xy-3y^2)/(x^2-xy-2y^2+3xy+7y^2)
=(-xy-3y^2)/(3xy+7y^2)
因为y≠0
原式=(-x-3y)/(3x+7y)
当x=2y时,
原式=(-2y-3y)/(6y+7y)= -5/13
当x= - y时,
原式=(y-3y)/(-3y+7y)= -1/2