Question

18．已知方程组$\left\{\begin{array}{l}{mx+n=5}\\{my-n=1}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$，则（m²-n²）的平方根是±$\sqrt{5}$．

Answer 1

分析 把$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{mx+n=5}\\{my-n=1}\end{array}\right.$，得出$\left\{\begin{array}{l}{m+n=5}\\{m-n=1}\end{array}\right.$，可得m²-n²=（m+n）（m-n）=5，即可求出（m²-n²）的平方根．

解答 解：∵方程组$\left\{\begin{array}{l}{mx+n=5}\\{my-n=1}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{m+n=5}\\{m-n=1}\end{array}\right.$，
∴m²-n²=（m+n）（m-n）=5，
∴（m²-n²）的平方根是±$\sqrt{5}$．

点评 本题主要考查了二元一次方程组的解和平方根，解题的关键是正确解方程组．