Question

19．解方程组$\left\{\begin{array}{l}{\sqrt{x+\frac{1}{y}}+\sqrt{x-y-2}=4}\\{2x-y+\frac{1}{y}=10}\end{array}\right.$．

Answer 1

分析 设$\sqrt{x+\frac{1}{y}}$=a≥0，$\sqrt{x-y-2}$=b≥0，则$2x-y+\frac{1}{y}$=10化为$x+\frac{1}{y}$+（x-y-2）=8，即a²+b²=8，可得原方程组变为$\left\{\begin{array}{l}{a+b=4}\\{{a}^{2}+{b}^{2}=8}\end{array}\right.$，解得a=b，进而解出．

解答 解：设$\sqrt{x+\frac{1}{y}}$=a≥0，$\sqrt{x-y-2}$=b≥0，
则$2x-y+\frac{1}{y}$=10化为$x+\frac{1}{y}$+（x-y-2）=8，即a²+b²=8，
∴原方程组变为$\left\{\begin{array}{l}{a+b=4}\\{{a}^{2}+{b}^{2}=8}\end{array}\right.$，解得a=b=2，
∴$\sqrt{x+\frac{1}{y}}$=2，$\sqrt{x-y-2}$=2，
化为$\left\{\begin{array}{l}{x+\frac{1}{y}=4}\\{x-y=6}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$，
经过检验满足原方程组．
∴原方程组的解为$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$．

点评 本题考查了根式的运算性质、换元法，考查了计算能力，属于中档题．