题目内容
填上适当的式子,使以下等式成立:
(1)xy•(
(2)(
(1)xy•(
2y+x-1
2y+x-1
)=2xy2+x2y-xy(2)(
5a2n-4a2n+2+3a3n-1
5a2n-4a2n+2+3a3n-1
)÷an=5an-4an+2+3a2n-1.分析:(1)已知一个因式和积,求另一个因式,用除法.只需用(2xy2+x2y-xy)÷xy即可得出结果.
(2)已知商和除数,求被除数,只要用商乘除数即可解答.
(2)已知商和除数,求被除数,只要用商乘除数即可解答.
解答:解:(1)(2xy2+x2y-xy)÷xy
=2y+x-1.
(2)(5an-4an+2+3a2n-1)×an
=5a2n-4a2n+2+3a3n-1.
故答案为2y+x-1,5a2n-4a2n+2+3a3n-1.
=2y+x-1.
(2)(5an-4an+2+3a2n-1)×an
=5a2n-4a2n+2+3a3n-1.
故答案为2y+x-1,5a2n-4a2n+2+3a3n-1.
点评:本题实际上考查了多项式除以单项式的法则:多项式除以单项式,先用多项式的每一项除以这个单项式,再把所得的商相加.还考查了单项式与多项式相乘的运算法则:单项式与多项式相乘,就是用单项式去乘多项式的每一项,再把所得的积相加.
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