Question

9．已知單項(xiàng)式2x³y^m-1與$\frac{1}{5}$x^m-2ny⁴是同類項(xiàng)，求代數(shù)式：$\frac{2}{3}$（m-2n）-2（2n-m）+m-5的值．

Answer 1

分析 根據(jù)同類項(xiàng)的定義得到m-2n=3，m-1=4，然后解方程組$\left\{\begin{array}{l}{m-2n=3}\\{m-1=4}\end{array}\right.$求得m，n，再化簡$\frac{2}{3}$（m-2n）-2（2n-m）+m-5，把方程組的解代入進(jìn)行計(jì)算即可．

解答 解：∵單項(xiàng)式2x³y^m-1與$\frac{1}{5}$x^m-2ny⁴是同類項(xiàng)，
∴$\left\{\begin{array}{l}{m-2n=3}\\{m-1=4}\end{array}\right.$，
解得$\left\{\begin{array}{l}{m=5}\\{n=1}\end{array}\right.$，
∴$\frac{2}{3}$（m-2n）-2（2n-m）+m-5
=$\frac{2}{3}$m-$\frac{4}{3}$n-4n+2m+m-5
=$\frac{11}{3}$m-$\frac{16}{3}$n-5
=$\frac{11}{3}$×5-$\frac{16}{3}$-5
=8．

點(diǎn)評(píng) 本題考查了整式的加減-化簡求值，同類項(xiàng)的定義：所含字母相同，并且相同字母的次數(shù)也分別相同的項(xiàng)叫同類項(xiàng)．