Ta có: \(\frac{\sqrt{\sqrt5-1}-\sqrt{\sqrt5+2}}{\sqrt{\sqrt5+1}-\sqrt{\sqrt5-2}}\)
\(=\frac{\left(\sqrt{\sqrt5-1}-\sqrt{\sqrt5+2}\right)\left(\sqrt{\sqrt5+1}+\sqrt{\sqrt5-2}\right)}{\sqrt5+1-\sqrt5+2}\)
\(=\frac{\sqrt{5-1}+\sqrt{\left(\sqrt5-1\right)\left(\sqrt5-2\right)}-\sqrt{\left(\sqrt5+2\right)\left(\sqrt5+1\right)}-\sqrt{5-4}}{3}\)
\(=\frac{\sqrt4-\sqrt1+\sqrt{5-3\sqrt5+2}-\sqrt{5+3\sqrt5+2}}{3}=\frac{2-1+\sqrt{7-3\sqrt5}-\sqrt{7+3\sqrt5}}{3}\)
\(=\frac{\sqrt{7-3\sqrt5}-\sqrt{7+3\sqrt5}+1}{3}=\frac{\sqrt{14-6\sqrt5}-\sqrt{14+6\sqrt5}+\sqrt2}{3\sqrt2}\)
\(=\frac{\sqrt{\left(3-\sqrt5\right)^2}-\sqrt{\left(3+\sqrt5\right)^2}+\sqrt2}{3\sqrt2}=\frac{3-\sqrt5-3-\sqrt5+\sqrt2}{3\sqrt2}=\frac{-2\sqrt5+\sqrt2}{3\sqrt2}=\frac{-\sqrt{10}+1}{3}\)
Ta có: \(\sqrt{\frac{2}{\frac72-\sqrt{10}}}\)
\(=\sqrt{2:\frac{7-2\sqrt{10}}{2}}=\sqrt{\frac{4}{7-2\sqrt{10}}}\)
\(=\sqrt{\frac{2^2}{\left(\sqrt5-\sqrt2\right)^2}}=\frac{2}{\sqrt5-\sqrt2}=\frac{2\left(\sqrt5+\sqrt2\right)}{\left(\sqrt5-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}\)
\(=\frac{2\left(\sqrt5+\sqrt2\right)}{3}\)
Ta có: \(\left(\frac{\sqrt{\sqrt5-1}-\sqrt{\sqrt5+2}}{\sqrt{\sqrt5+1}-\sqrt{\sqrt5-2}}\right)\cdot\sqrt{\frac{2}{\frac72-\sqrt{10}}}\)
\(=\frac{-\sqrt{10}+1}{3}\cdot\frac{2\left(\sqrt5+\sqrt2\right)}{3}=\frac{\left(-\sqrt{10}+1\right)\left(2\sqrt5+2\sqrt2\right)}{9}\)
\(=\frac{-2\sqrt{50}-2\sqrt{20}+2\sqrt5+2\sqrt2}{9}=\frac{-10\sqrt2-4\sqrt5+2\sqrt5+2\sqrt2}{9}=\frac{-8\sqrt2-2\sqrt5}{9}\)
