题目内容
3.已知$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$是二元一次方程组$\left\{\begin{array}{l}{mx+ny=8}\\{nx-my=1}\end{array}\right.$的解,则2mn的值为( )| A. | 6 | B. | 12 | C. | 18 | D. | 24 |
分析 根据方程组的解得定义将$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入二元一次方程组,得到关于m、n的方程组,解方程组求出m、n的值,再代入求解可得.
解答 解:根据题意,将$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入二元一次方程组$\left\{\begin{array}{l}{mx+ny=8}\\{nx-my=1}\end{array}\right.$,
得:$\left\{\begin{array}{l}{2m+n=8}\\{-m+2n=1}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{m=3}\\{n=2}\end{array}\right.$,
∴2mn=2×3×2=12,
故选:B.
点评 此题考查了二元一次方程组的解,方程组的解即为能使方程组中两方程成立的未知数的值.
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