Question

19．化简$\frac{1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}{\sqrt{3}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}$，可得（　　）

• A． $\sqrt{3}$-$\sqrt{2}$
• B． $\frac{\sqrt{2}+1}{2}$
• C． 2-$\sqrt{2}$
• D． $\sqrt{2}$-1

Answer 1

分析 先利用完全平方公式得到$\sqrt{4+2\sqrt{2}}$=$\sqrt{2+\sqrt{2}}$+$\sqrt{2-\sqrt{2}}$，$\sqrt{2-\sqrt{3}}$=$\frac{\sqrt{6}-\sqrt{2}}{2}$，$\sqrt{2+\sqrt{3}}$=$\frac{\sqrt{6}+\sqrt{2}}{2}$，再把原式分子分母都乘以（$\sqrt{2}$-1），使分母中出现4+2$\sqrt{2}$，然后进行代换把分母化为1+$\sqrt{2-\sqrt{2}}$+$\sqrt{2-\sqrt{3}}$，则利用约分即可得到原式的值．

解答 解：∵（$\sqrt{2+\sqrt{2}}$+$\sqrt{2-\sqrt{2}}$）²=2+$\sqrt{2}$+2$\sqrt{（2+\sqrt{2}）（2-\sqrt{2}}）$+2-$\sqrt{2}$=4+2$\sqrt{2}$，
∴$\sqrt{4+2\sqrt{2}}$=$\sqrt{2+\sqrt{2}}$+$\sqrt{2-\sqrt{2}}$，
∵$\sqrt{2-\sqrt{3}}$=$\sqrt{\frac{4-2\sqrt{3}}{2}}$=$\frac{\sqrt{（\sqrt{3}-1）^{2}}}{\sqrt{2}}$=$\frac{\sqrt{3}-1}{\sqrt{2}}$=$\frac{\sqrt{6}-\sqrt{2}}{2}$，$\sqrt{2+\sqrt{3}}$=$\frac{\sqrt{6}+\sqrt{2}}{2}$，
∴原式=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{（\sqrt{2}-1）（\sqrt{3}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}）}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{\sqrt{6}+\sqrt{4+2\sqrt{2}}+\sqrt{4+2\sqrt{3}}-\sqrt{3}-\sqrt{2+\sqrt{2}}-\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{\sqrt{6}+\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}}+\sqrt{3}+1-\sqrt{3}-\sqrt{2+\sqrt{2}}-\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\sqrt{6}-\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\sqrt{6}-\frac{\sqrt{6}+\sqrt{2}}{2}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\frac{\sqrt{6}-\sqrt{2}}{2}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}+\sqrt{2-\sqrt{3}}}}$
=$\sqrt{2}$-1．
故选D．

点评 本题考查了二次根式的化简求值：次根式的化简求值，一定要先化简再代入求值．二次根式运算的最后，注意结果要化到最简二次根式，二次根式的乘除运算要与加减运算区分，避免互相干扰．