Question

5．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2m-2n的算术平方根为$\sqrt{2}$．

Answer 1

分析 把x与y的值代入方程组求出m与n的值，即可求出所求算术平方根．

解答 解：把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程组得：$\left\{\begin{array}{l}{2m+n=8}\\{2n-m=1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{m=3}\\{n=2}\end{array}\right.$，
则$\sqrt{2m-2n}$=$\sqrt{2}$，
故答案为：$\sqrt{2}$

点评 此题考查了二元一次方程组的解，以及算术平方根，熟练掌握运算法则是解本题的关键．