Question

1．$\sqrt{2}$-1，$\frac{1}{\sqrt{2}-1}$的等差中项是$\sqrt{2}$．

Answer 1

分析 由等差中项可得2a=$\sqrt{2}$-1+$\frac{1}{\sqrt{2}-1}$，化简根式可得a值．

解答 解：设a为$\sqrt{2}$-1，$\frac{1}{\sqrt{2}-1}$的等差中项，
则$\frac{1}{\sqrt{2}-1}$-a=a-（$\sqrt{2}$-1），
∴2a=$\sqrt{2}$-1+$\frac{1}{\sqrt{2}-1}$
=$\sqrt{2}$-1+$\frac{\sqrt{2}+1}{（\sqrt{2}-1）（\sqrt{2}+1）}$
=$\sqrt{2}$-1+$\sqrt{2}$+1=2$\sqrt{2}$，
∴a=$\sqrt{2}$
故答案为：$\sqrt{2}$

点评 本题考查等差数列的通项公式，涉及根式的化简，属基础题．