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22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
a) Ta có: \(\frac{7\sqrt{2}+2\sqrt{7}}{\sqrt{14}}-\frac{5}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\sqrt{14}\left(\sqrt{7}+\sqrt{2}\right)}{\sqrt{14}}-\frac{5\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{2}\right)-5\left(\sqrt{7}-\sqrt{5}\right)}{2}\)
\(=\frac{2\sqrt{7}+2\sqrt{2}-5\sqrt{7}+5\sqrt{5}}{2}\)
\(=\frac{2\sqrt{2}-3\sqrt{7}+5\sqrt{5}}{2}\)
b) Ta có: \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{\sqrt{2}\left(6+2\sqrt{5}\right)}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(6-2\sqrt{5}\right)}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\left|\sqrt{5}+1\right|}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left|\sqrt{5}-1\right|}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\left(\sqrt{5}+1\right)}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left(\sqrt{5}-1\right)}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{10}+\sqrt{2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{5\sqrt{2}+\sqrt{10}}+\frac{6\sqrt{2}-2\sqrt{10}}{5\sqrt{2}-\sqrt{10}}\)
\(=\frac{6+2\sqrt{5}}{5+\sqrt{5}}+\frac{6-2\sqrt{5}}{5-\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\sqrt{5}\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}}=2\)
c) Đặt \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
Ta có: \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
\(\Leftrightarrow A^3=32-12\cdot\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(=32-12A\)
\(\Leftrightarrow A^3+12A-32=0\)
\(\Leftrightarrow A^3-2A^2+2A^2-4A+16A-32=0\)
\(\Leftrightarrow A^2\left(A-2\right)+2A\left(A-2\right)+16\left(A-2\right)=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+16\right)=0\)
mà \(A^2+2A+16>0\)
nên A-2=0
hay A=2
Vậy: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)