问题 解答题 已知x2-y2=xy,且xy≠0,求代数式x2y-2+x-2y2的值. 答案 ∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2-y2)2+2x2y2x2y2=3x2y2x2y2=3.