Question

10、（12a^2m+1b^m+3-20a^m+1b^2m+4+4a^m+1b^m+2）÷4a^mb^m+1=

3a^m+1b²-5ab^m+3+ab

．

Answer 1

分析：根據(jù)多項(xiàng)式除單項(xiàng)式，先用多項(xiàng)式的每一項(xiàng)除以單項(xiàng)式，再把所得的商相加進(jìn)行計(jì)算即可．

解答：解：（12a^2m+1b^m+3-20a^m+1b^2m+4+4a^m+1b^m+2）÷4a^mb^m+1
=12a^2m+1b^m+3÷4a^mb^m+1-20a^m+1b^2m+4÷4a^mb^m+1+4a^m+1b^m+2÷4a^mb^m+1
=3a^m+1b²-5ab^m+3+ab．

點(diǎn)評：本題考查了多項(xiàng)式除以單項(xiàng)式，關(guān)鍵就是將它轉(zhuǎn)化成單項(xiàng)式除以單項(xiàng)式，指數(shù)中含有字母，計(jì)算時容易出錯，要細(xì)心．