a: \(\frac{\sqrt7-5}{2}-\frac{6-2\sqrt7}{4}+\frac{6}{\sqrt7-2}-\frac{5}{4+\sqrt7}\)
\(=\frac{2\left(\sqrt7-5\right)-6+2\sqrt7}{4}+\frac{6\left(\sqrt7+2\right)}{\left(\sqrt7-2\right)\left(\sqrt7+2\right)}-\frac{5\left(4-\sqrt7\right)}{16-7}\)
\(=\frac{4\sqrt7-16}{4}+\frac{6\left(\sqrt7+2\right)}{7-4}-\frac{5\left(4-\sqrt7\right)}{9}\)
\(=\sqrt7-4+2\sqrt7+4-\frac{5\left(4-\sqrt7\right)}{9}=\frac{27\sqrt7-20+5\sqrt7}{9}=\frac{32\sqrt7-20}{9}\)
b: \(\frac{2}{\sqrt6-2}+\frac{2}{\sqrt6+2}+\frac{5}{\sqrt6}\)
\(=\frac{2\left(\sqrt6+2\right)+2\left(\sqrt6-2\right)}{\left(\sqrt6+2\right)\left(\sqrt6-2\right)}+\frac{5\sqrt6}{6}=\frac{2\sqrt6+4+2\sqrt6-4}{6-4}+\frac{5\sqrt6}{6}\)
\(=2\sqrt6+\frac{5\sqrt6}{6}=\frac{17\sqrt6}{6}\)
c: \(\frac{1}{\sqrt3+\sqrt2-\sqrt5}-\frac{1}{\sqrt3+\sqrt2+\sqrt5}\)
\(=\frac{\sqrt3+\sqrt2+\sqrt5-\sqrt3-\sqrt2+\sqrt5}{\left(\sqrt3+\sqrt2\right)^2-5}\)
\(=\frac{2\sqrt5}{5+2\sqrt6-5}=\frac{2\sqrt5}{2\sqrt6}=\sqrt{\frac56}=\frac{\sqrt{30}}{6}\)
d: \(\left(\frac{\sqrt6-\sqrt2}{1-\sqrt3}-\frac{5}{\sqrt5}\right):\frac{1}{\sqrt5-\sqrt2}\)
\(=\left(-\frac{\sqrt2\left(1-\sqrt3\right)}{1-\sqrt3}-\sqrt5\right)\left(\sqrt5-\sqrt2\right)=-\left(\sqrt5+\sqrt2\right)\left(\sqrt5-\sqrt2\right)\)
=-(5-2)
=-3
e: \(\frac{1}{\sqrt3}+\frac{1}{3\sqrt2}+\frac{1}{\sqrt3}\cdot\sqrt{\frac{5}{12}-\frac{1}{\sqrt6}}\)
\(=\frac{\sqrt6+1}{3\sqrt2}+\frac{1}{\sqrt3}\cdot\sqrt{\frac{5-2\sqrt6}{12}}=\frac{\sqrt{12}+\sqrt2}{6}+\frac{1}{\sqrt3}\cdot\frac{\sqrt3-\sqrt2}{2\sqrt3}\)
\(=\frac{\sqrt{12}+\sqrt2+\sqrt3-\sqrt2}{6}=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2}\)
f: \(2\cdot\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}\)
\(=2\cdot\sqrt{3-\sqrt{3+\sqrt{13+2\sqrt{12}}}}\)
\(=2\cdot\sqrt{3-\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)
\(=2\cdot\sqrt{3-\sqrt{3+\sqrt{\left(2\sqrt3+1\right)^2}}}\)
\(=2\cdot\sqrt{3-\sqrt{3+\left(2\sqrt3+1\right)^{}}}\)
\(=2\cdot\sqrt{3-\sqrt{4+2\sqrt3^{}}}\)
\(=2\cdot\sqrt{3-\sqrt{\left(\sqrt3+1\right)^2}}\)
\(=2\cdot\sqrt{3-\left(\sqrt3+1\right)^{}}=2\cdot\sqrt{2-\sqrt3}=\sqrt2\cdot\sqrt{4-2\sqrt3}=\sqrt2\left(\sqrt3-1\right)\)
Ta có: \(\frac{2\cdot\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt6-\sqrt2}\)
\(=\frac{\sqrt2\left(\sqrt3-1\right)}{\sqrt2\left(\sqrt3-1\right)}\)
=1
