Question

19．化簡(jiǎn)$\frac{1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}{\sqrt{3}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}$，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），使分母中出現(xiàn)4+2$\sqrt{2}$，然后進(jìn)行代換把分母化為1+$\sqrt{2-\sqrt{2}}$+$\sqrt{2-\sqrt{3}}$，則利用約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．
故選D．

點(diǎn)評(píng) 本題考查了二次根式的化簡(jiǎn)求值：次根式的化簡(jiǎn)求值，一定要先化簡(jiǎn)再代入求值．二次根式運(yùn)算的最后，注意結(jié)果要化到最簡(jiǎn)二次根式，二次根式的乘除運(yùn)算要與加減運(yùn)算區(qū)分，避免互相干擾．