已知x+y=6,xy=7,求(3x+y)的平方+(x+3y)的平方的值
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已知x+y=6,xy=7,求(3x+y)的平方+(x+3y)的平方的值
已知x+y=6,xy=7,求(3x+y)的平方+(x+3y)的平方的值
已知x+y=6,xy=7,求(3x+y)的平方+(x+3y)的平方的值
x+y=6
原式=9x²+6xy+y²+x²+6xy+9y²
=10x²+12xy+10y²
=10x²+20xy+10y²-8xy
=10(x+y)²-8xy
=10×6²-8×7
=304
解
x+y=6
两边平方
x²+2xy+y²=36
∵xy=7
∴x²+y²=36-14=22
(3x+y)²+(x+3y)²
=9x²+6xy+y²+x²+6xy+9y²
=10(x²+y²)+12xy
=10×22+12×7
=220+84
=304
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