\(A=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{20+4\sqrt{5}+1}}}}{\sqrt{10}-\sqrt{2}}\)\(=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(2\sqrt{5}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\left(2\sqrt{5}+1\right)}}}{\sqrt{10}-\sqrt{2}}=\)\(=\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5}-1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{6-2\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{5-2\sqrt{5}+1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\left(\sqrt{5}-1\right)}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{10}-\sqrt{2}}=1\)

a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)

d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)

\(\sqrt{3-2\sqrt{2}}\)

GIẢI 

\(M=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)

        \(=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

         \(=\frac{\sqrt[2]{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\left(\sqrt{5}+1\right)}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

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

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d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)

\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)

\(=\dfrac{3}{x-y}\)

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)

\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)

=7-2

=5

d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)

\(=4\sqrt{7}\)

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

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

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

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

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

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

=16-5=11

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

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

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

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