题目内容
已知:x2+xy=12,xy+y2=15,求(x+y)2-(x+y)(x-y)的值.
分析:根据已知x2+xy=12,xy+y2=15,将两式分别相加和相减即可求出(x+y) 2=27,(x+y)(x-y)=-3,即可得出答案.
解答:解:∵x2+xy=12,xy+y2=15,
∴x2+xy+xy+y2=12+15,
∴(x+y) 2=27,
x2+xy-(xy+y2)=12-15,
∴(x+y)(x-y)=-3,
∴原式=27-(-3)=30.
∴x2+xy+xy+y2=12+15,
∴(x+y) 2=27,
x2+xy-(xy+y2)=12-15,
∴(x+y)(x-y)=-3,
∴原式=27-(-3)=30.
点评:此题主要考查了整式的化简求值和因式分解的应用,根据已知得出(x+y) 2=27,(x+y)(x-y)=-3是解题关键.
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