Question

14．若a³（3aⁿ-2a^m+4a^k）=3a⁹-2a⁶+4a⁴，求-3k²（n³mk+2km²）的值．

Answer 1

分析 先根據(jù)單項(xiàng)式乘以多項(xiàng)式展開，得到$\left\{\begin{array}{l}{3+n=9}\\{3+m=6}\\{3+k=4}\end{array}\right.$，求出$\left\{\begin{array}{l}{n=6}\\{m=3}\\{k=1}\end{array}\right.$，代入代數(shù)式，即可解答．

解答 解：∵a³（3aⁿ-2a^m+4a^k）=3a³⁺ⁿ-2a^3+m+4a^3+k=3a⁹-2a⁶+4a⁴，
∴$\left\{\begin{array}{l}{3+n=9}\\{3+m=6}\\{3+k=4}\end{array}\right.$
∴$\left\{\begin{array}{l}{n=6}\\{m=3}\\{k=1}\end{array}\right.$
∴-3k²（n³mk+2km²）=-3×1²×（6³×3×1+2×1×3²）=-1998．

點(diǎn)評(píng) 本題考查了單項(xiàng)式乘以多項(xiàng)式，解決本題的關(guān)鍵是把多項(xiàng)式展開，求出m，n，k的值．