Question

5．計(jì)算：（-a²bc+2ab²-$\frac{3}{5}$ac）•（-$\frac{2}{3}$ac）²．

Answer 1

分析 根據(jù)單項(xiàng)式與多項(xiàng)式相乘，先用單項(xiàng)式乘多項(xiàng)式的每一項(xiàng)，再把所得的積相加計(jì)算即可．

解答 解：（-a²bc+2ab²-$\frac{3}{5}$ac）•（-$\frac{2}{3}$ac）²=（-a²bc+2ab²-$\frac{3}{5}$ac）•$\frac{4}{9}$a²c²=-$\frac{4}{9}$a⁴bc³+$\frac{8}{9}$a³b²c²-$\frac{4}{5}$a³c³．

點(diǎn)評(píng) 本題考查了單項(xiàng)式與多項(xiàng)式相乘，熟練掌握運(yùn)算法則是解題的關(guān)鍵，計(jì)算時(shí)要注意符號(hào)的處理．