已知:x+y=6,xy=7,求(3x+y)2+(x+3y)2的值.

问题描述:

已知:x+y=6,xy=7,求(3x+y)2+(x+3y)2的值.

原式=9x2+6xy+y2+x2+6xy+9y2
=10x2+12xy+10y2
=10(x2+y2)+12xy
=10(x+y)2-8xy,
当x+y=6,xy=7,原式=10×62-8×7=304.