\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(C=\sqrt{4+5}\)

\(C=3\)

a) \(\frac{2}{4-3\sqrt{2}}-\frac{2}{4+3\sqrt{2}}\)

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

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

\(=\frac{12\sqrt{2}}{-2}\)

\(=-6\sqrt{2}\)

b) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{4\sqrt{35}}{2}\)

\(=2\sqrt{35}\)

a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)

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

Ta có: 

\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)

\(\Leftrightarrow A^2=10-2\sqrt{25-17}=10-4\sqrt{2}\)

\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)

Ta lại có:

\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)

\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)

\(\Leftrightarrow B=\sqrt{2}\)

Thế vô biểu thức ban đầu ta được

\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)

\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)

\(\sqrt{2}\)

ta có :

\(A=\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}\)

\(\Leftrightarrow A^2=10-2\sqrt{25-17=10-4\sqrt{2}}\)

\(\Leftrightarrow A=\sqrt{10-4\sqrt{2}}\)

ta lại có :

\(B=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)

\(\Leftrightarrow B^2=6-2\sqrt{9-5}=2\)

\(\Leftrightarrow B=\sqrt{2}\)

the vo bieu thuc ban dau ta duoc

\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2=\sqrt{2}}\)

\(=\frac{\sqrt{10-4\sqrt{2}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{2}+2-\sqrt{2}}=\frac{4}{2}=2\)

a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)

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

b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

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

d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)