\(a,=\sqrt{\dfrac{81}{25}}=\dfrac{9}{5}\\ b,\approx6,39\\ c,=\sqrt{8,1\cdot20\cdot8}=\sqrt{81\cdot16}=\sqrt{81}\cdot\sqrt{16}=9\cdot4=36\\ d,=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\\ =\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}=2\sqrt{5}\)

a) \(\sqrt{3\dfrac{6}{25}}=\sqrt{\dfrac{81}{25}}=\dfrac{9}{5}\)

b) \(\sqrt[3]{216}=6\)

c) \(\sqrt{8,1}.\sqrt{20}.\sqrt{8}=\dfrac{9\sqrt{10}}{10}.2\sqrt{5}.2\sqrt{2}=36\)

d) \(\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}=2\sqrt{5}\)

1.

 Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)

\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)

Đặt \(A=\sqrt{11-2\sqrt{30}}-\sqrt{11+2\sqrt{30}}\)

\(\Leftrightarrow A^2=11-2\sqrt{30}+11+2\sqrt{30}-2\sqrt{\left(11-2\sqrt{30}\right)\left(11+2\sqrt{30}\right)}\)

\(\Leftrightarrow A^2=22-2\sqrt{11^2-\left(2\sqrt{30}\right)^2}\)

\(\Leftrightarrow A^2=22-2=20\)

\(\Leftrightarrow A=\pm\sqrt{20}\)

Vì \(\sqrt{11-2\sqrt{30}}< \sqrt{11+2\sqrt{30}}\)

Nên A chỉ nhận giá trị \(-\sqrt{20}\)

\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)

\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)

\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)

\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)

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

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

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

\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)

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

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

\(\sqrt{11-2\sqrt{30}}-\sqrt{11+2\sqrt{30}}\)

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

\(=\sqrt{6}-\sqrt{5}-\sqrt{6}-\sqrt{5}\)

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

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

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

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

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

\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{3}-1+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}+1-4}\)

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

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

\(A=\dfrac{\sqrt{2}-\sqrt{6}-\sqrt{6}+3\sqrt{2}+\sqrt{2}+\sqrt{6}+\sqrt{6}+3\sqrt{2}}{1-3}=\dfrac{8\sqrt{2}}{-2}=-4\sqrt{2}\)

* \(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\) \(=\dfrac{\sqrt{6+2.\sqrt{6}.\sqrt{5}+5}-\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}{\sqrt{5}}\)\(=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}{\sqrt{5}}\)

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

* \(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)

Đặt:\(y=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\Rightarrow y^2=4+\sqrt{15}+4-\sqrt{15}+2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=8+2=10\Rightarrow y=\sqrt{10}\)

Suy ra: \(C=\sqrt{12+4\sqrt{5}}-y=\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}-\sqrt{10}=\sqrt{10}+\sqrt{2}-\sqrt{10}=\sqrt{2}\)* \(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)+\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{1}=4\)

\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{7-2\sqrt{10}}\)

\(=\dfrac{1}{\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}-\dfrac{3\left(7+2\sqrt{10}\right)}{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)

\(=\dfrac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\dfrac{21+6\sqrt{10}}{49-40}\)

\(=\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{21+6\sqrt{10}}{9}\)

\(=\dfrac{\left(\sqrt{6}+\sqrt{5}\right).9-21-6\sqrt{10}}{9}\)

\(=\dfrac{9\sqrt{6}+9\sqrt{5}-21-6\sqrt{10}}{9}=\dfrac{3\sqrt{6}+3\sqrt{5}-7-2\sqrt{10}}{3}\)

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

\(B=\sqrt{16+8\sqrt{7}+7}-\sqrt{7}\)

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

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

\(C=\sqrt{6+2\sqrt{30}+5}-\sqrt{6-2\sqrt{30}+5}\\ =\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\\ =\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}\\ =2\sqrt{5}\)