Question

初二因式分解题.

1.若x²-4x+y²-2y+5=0，则xy+2x²y²+x³y³的值为？
2.分解因式： -ax³y-axy³+2ax²y²（写出过程）

Answer 1

1.
x²-4x+y²-2y+5=0
(x-2)²+(y-1)²=0
则x-2=0且y-1=0,即x=2,y=1
xy+2x²y²+x³y³=xy(1+2xy+x²y²)=xy(1+xy)²=2(1+2)²=18

2.
ax³y-axy³+2ax²y²=axy(x²-y²+2xy)
=axy((x+y)²-2y²)
=axy(x+(1+根2)y)(x+(1-根2)y)