已知x+y=2,xy=2分之1,求x³y-2²y²+xy³的值
问题描述：
已知x+y=2,xy=2分之1,求x³y-2²y²+xy³的值
答
x³y-2x²y²+xy³
=xy(x²-2xy+y²)
=xy[(x²+2xy+y²)-4xy]
=xy[(x+y)²-4xy]
=2分之1×(2²-4×2分之1)
=2分之1×2
=1