Question

9．如果$\sqrt{x+\sqrt{2}}$+|y-1|=0，则|$\frac{1}{x+y}$|=（　　）

• A． 1-$\sqrt{2}$
• B． 1+$\sqrt{2}$
• C． $\sqrt{2}$-1
• D． -$\sqrt{2}$-1

Answer 1

分析 由$\sqrt{x+\sqrt{2}}$+|y-1|=0，可得$\left\{\begin{array}{l}{x+\sqrt{2}=0}\\{y-1=0}\end{array}\right.$，解得x，y．即可得出．

解答 解：∵$\sqrt{x+\sqrt{2}}$+|y-1|=0，
∴$\left\{\begin{array}{l}{x+\sqrt{2}=0}\\{y-1=0}\end{array}\right.$，解得x=-$\sqrt{2}$，y=1．
则|$\frac{1}{x+y}$|=$\frac{1}{|-\sqrt{2}+1|}$=$\frac{1}{\sqrt{2}-1}$=$\sqrt{2}$+1．
故选：B．

点评 本题考查了函数解析式的求法，考查了计算能力，属于基础题．