已知x2+xy=2,y2+xy=5,则12x2+xy+12y2=___.
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已知x2+xy=2,y2+xy=5,则
x
1 |
2 |
![已知x2+xy=2,y2+xy=5,则12x2+xy+12y2=___.](/uploads/image/z/5874609-57-9.jpg?t=%E5%B7%B2%E7%9F%A5x2%2Bxy%3D2%EF%BC%8Cy2%2Bxy%3D5%EF%BC%8C%E5%88%9912x2%2Bxy%2B12y2%3D___%EF%BC%8E)
∵x2+xy=2,y2+xy=5,
∴x2+2xy+y2=7,
则原式=
1
2(x2+2xy+y2)=
7
2,
故答案为:
7
2
∴x2+2xy+y2=7,
则原式=
1
2(x2+2xy+y2)=
7
2,
故答案为:
7
2
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