a)

\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2(\sqrt{6}+2+\sqrt{6}-2)}{(\sqrt{6}-2)(\sqrt{6}+2)}+\frac{5\sqrt{6}}{6}\)

\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)

b)

\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)

\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)

c)

\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)

\(=\left[\frac{\sqrt{2}(\sqrt{3}-1)}{1-\sqrt{3}}-\sqrt{5}\right].(\sqrt{5}-\sqrt{2})\)

\(=(-\sqrt{2}-\sqrt{5})(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)

\(=-(5-2)=-3\)

d)

\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)

\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)

\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

\(=\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{6\left(2+\sqrt{3}\right)}-4+\sqrt{2\left(2+\sqrt{3}\right)}}{2\sqrt{6}}\right)\)

\(=\frac{\sqrt{2+\sqrt{3}}}{2}.\left(\frac{2\sqrt{6}}{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}\right)\)

\(=\frac{\sqrt{6\left(2+\sqrt{3}\right)}}{\left|\sqrt{3}+3\right|-4+\left|\sqrt{3}+1\right|}\)

\(=\frac{\left|\sqrt{3}+3\right|}{\sqrt{3}+3-4+\sqrt{3}+1}\)

\(=\frac{\sqrt{3}+3}{2\sqrt{3}}\)

\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)

\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}\right)^2+6\sqrt{2}+9}-\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{6}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{10}+\left(\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}+\sqrt{2}}\)

\(=\frac{3}{2\sqrt{2}+1}\)

Bài rút gọn 

\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)

\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)

Bài gpt:

\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)

Đk:\(-1\le x\le3\)

\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)

Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm

Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)