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

Đặt: \(A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\)

=> \(A^2=\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

=> \(A^2=2\sqrt{5}+2\sqrt{5-4}\)

=> \(A^2=2\sqrt{5}+2\)

=> \(A^2=2\left(\sqrt{5}+1\right)\)

=> \(A=\sqrt{2\left(\sqrt{5}+1\right)}\)

=> \(\frac{A}{\sqrt{\sqrt{5}+1}}=\frac{\sqrt{2\left(\sqrt{5}+1\right)}}{\sqrt{\sqrt{5}+1}}=\sqrt{2}\)

Đặt: \(B=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

=> \(VT=\frac{A}{\sqrt{\sqrt{5}+1}}-B=\sqrt{2}-\left(\sqrt{2}-1\right)=\sqrt{2}-\sqrt{2}+1=1\)

VẬY KẾT QUẢ CỦA PHÉP TÍNH = 1.

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

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

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)

\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)

c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)

d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)

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

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

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

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

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

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

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

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

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)

\(=\dfrac{\sqrt{1\cdot4}}{2}\)

\(=\dfrac{2}{2}\)

\(=1\)

\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

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

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{8}\)

\(=\frac{12\left(\sqrt{7}-5\right)}{24}-\frac{6\left(6+2\sqrt{7}\right)}{24}+\frac{8\left(6\sqrt{7}+12\right)}{24}-\frac{3\left(20-5\sqrt{7}\right)}{24}\)

\(=\frac{12\sqrt{7}-60-36-12\sqrt{7}+48\sqrt{7}+96-60+15\sqrt{7}}{24}\)

\(=\frac{-60+63\sqrt{7}}{24}\)

a/\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\)

b/Đề sai

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

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

d/ \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)