历史上最早的皇帝:已知x，y,z互不相等，且xyz不等于0，x2＋yz＝z2，y2＋zx＝x2，求证：z2＋xy＝y2

来源：百度文库编辑：高考问答时间：2024/04/29 15:28:30

y2＋zx＝x2 => z=(x^2-y^2)/x
代入x2＋yz＝z2
=> x^4+xy(x^2-y^2)=(x^2-y^2)^2
=> x^4+x^3y-xy^3=x^4+y^4-2x^2y^2
=> x^3-xy^2-y^3+2x^2y=0

z2^＋xy-y^2=x^2＋yz＋xy-y^2
=x^2＋y(x^2-y^2)/x＋xy-y^2
=x^2-y^3/x＋2xy-y^2
=(x^3-xy^2-y^3+2x^2y)/x
=0/x(根据上式)
=0
=> z2^＋xy=y^2

试试反推法吧.我有时间再看看吧,挺麻烦的.

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