a) Ta có: \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)
\(=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-\dfrac{6\left(\sqrt{10}+2\right)}{\left(\sqrt{10}+2\right)\cdot\left(\sqrt{10}-2\right)}-\dfrac{20}{\sqrt{10}}\)
\(=\sqrt{10}-6\sqrt{10}-12-2\sqrt{10}\)
\(=-7\sqrt{10}-12\)
b) Ta có: \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
\(=\left(\sqrt{5}-1-2\right)\left(\sqrt{5}-1+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)
=5-9=-4
c) Ta có: \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=8-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
d) Ta có: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3-\sqrt{7}}{2}-2\sqrt{7}-4-\dfrac{\sqrt{7}-5}{2}\)
\(=\sqrt{11}-2\sqrt{7}+\dfrac{3-\sqrt{7}-\sqrt{7}+5}{2}\)
\(=\dfrac{2\sqrt{11}-4\sqrt{7}+8-2\sqrt{7}}{2}\)
\(=\dfrac{2\sqrt{11}-6\sqrt{7}+8}{2}\)
\(=\sqrt{11}-3\sqrt{7}+4\)
e) Ta có: \(\dfrac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\dfrac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)
\(=\dfrac{\sqrt{6}\left(2\sqrt{2}-1\right)}{\sqrt{3}\left(2\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(2\sqrt{5}+1\right)}{\sqrt{3}\left(2\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}+\sqrt{15}}{3}\)
