Question

16．化简$\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+2\sqrt{2}+1}$．

Answer 1

分析 可以先求出所求式子的倒数的值，然后再求出所求式子的值，本题得以解决．

解答 解：∵$\frac{\sqrt{3}+2\sqrt{2}+1}{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}$
=$\frac{\sqrt{3}+\sqrt{2}+\sqrt{2}+1}{\sqrt{3}（\sqrt{2}+1）+\sqrt{2}（\sqrt{2}+1）}$
=$\frac{（\sqrt{3}+\sqrt{2}）+（\sqrt{2}+1）}{（\sqrt{2}+1）（\sqrt{3}+\sqrt{2}）}$
=$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}$
=$\sqrt{2}-1+\sqrt{3}-\sqrt{2}$
=$\sqrt{3}-1$，
∴$\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+2\sqrt{2}+1}$=$\frac{1}{\sqrt{3}-1}$=$\frac{\sqrt{3}+1}{2}$．

点评 本题考查实数的运算，解题的关键是明确题意，可以发现先求所求式子的倒数的值，再通过转化的思想解答．