Question

9．化简：$\frac{1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}{\sqrt{3}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}$．

Answer 1

分析 根据二次根式的数字特点，适当变形，分子分母同乘$\sqrt{2}$-1，进一步计算化简即可．

解答 解：∵（$\sqrt{2+\sqrt{2}}$+$\sqrt{2-\sqrt{2}}$）²=4+2$\sqrt{2}$，
∴$\sqrt{4+2\sqrt{2}}$=$\sqrt{2+\sqrt{2}}$+$\sqrt{2-\sqrt{2}}$，
另$\sqrt{2+\sqrt{3}}$=$\sqrt{\frac{8+4\sqrt{3}}{4}}$=$\frac{\sqrt{6}+\sqrt{2}}{2}$，
同理$\sqrt{2-\sqrt{3}}$=$\frac{\sqrt{6}-\sqrt{2}}{2}$，$\sqrt{4+2\sqrt{3}}$=$\sqrt{3}$+1，
∴原式=$\frac{（\sqrt{2}-1）（\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{（\sqrt{2}-1）（\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}）}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{\sqrt{6}+\sqrt{4+2\sqrt{2}}+\sqrt{4+2\sqrt{3}}-\sqrt{3}-\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\sqrt{6}-\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\sqrt{6}-\sqrt{2+\sqrt{3}}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\frac{\sqrt{6}-\sqrt{2}}{2}}$
=$\frac{（\sqrt{2}-1）（1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}）}{1+\sqrt{2-\sqrt{2}}+\sqrt{2-\sqrt{3}}}$
=$\sqrt{2}$-1．

点评 此题考查二次根式的化简，灵活利用二次根式的运算特点，适当变形解决问题．