一题关于初二因式分解的题目x-y=1,xy=3,求x³y-2x²y²+xy³的值~
问题描述：
一题关于初二因式分解的题目
x-y=1,xy=3,求x³y-2x²y²+xy³的值~
答
原式=xy(x²-2xy+y²)=xy(x-y)2=3
答
x^3y-2x^2y^2+xy^3
=xy(x^2-2xy+y^2)
=xy(x-y)^2
=3*1^2
=3
答
x³y-2x²y²+xy³
=xy(x^2 -2xy+y^2)
=xy(x-y)^2
=3×1^2
=3