已知x^2+xy=12,xy+y^2=15,求(x+y)^2-(x+y)(x-y)的值

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已知x^2+xy=12,xy+y^2=15,求(x+y)^2-(x+y)(x-y)的值
已知x^2+xy=12,xy+y^2=15,求(x+y)^2-(x+y)(x-y)的值

已知x^2+xy=12,xy+y^2=15,求(x+y)^2-(x+y)(x-y)的值
前两个方程联立成方程组,二者相加,得x^2+2xy+y^2=27,即（x+y)^2=27,
二者相减,得y^2-x^2=3
而(x+y)^2-(x+y)(x-y)=(x+y)^2-（y^2-x^2）=27-（-3）=30

(x+y)²-(x+y)(x-y)
=x²+2xy+y²-x²+y²
=2xy+2y²
=30

x^2+xy=X(x+y)=12 ①
xy+y^2=y(x+y)=15 ②
∴①+②=X(x+y)+y(x+y)=(x+y)^2=27
①-②=X(x+y)-y(x+y)=(x+y)(x-y)=-3
∴(x+y)^2-(x+y)(x-y)=27-（-3）=30

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