\(A=\sqrt[3]{3\sqrt{21}+8}-\sqrt[3]{3\sqrt{21}-8}\)

\(\Leftrightarrow A^3=3\sqrt{21}+8-3\sqrt{21}+8+3\cdot A\cdot\sqrt[3]{\left(3\sqrt{21}\right)^2-8^2}\)

\(\Leftrightarrow A^3=16+15A\)

\(\Leftrightarrow A^3-15A-16=0\)

hay \(A\simeq4.32\)

i) \(\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{16-6\sqrt{7}}{2}}+\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(3-\sqrt{7}\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|3-\sqrt{7}\right|}{\sqrt{2}}+\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{3-\sqrt{7}}{\sqrt{2}}+\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

j) \(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}=\sqrt{\dfrac{10+2\sqrt{21}}{2}}-\sqrt{\dfrac{10-2\sqrt{21}}{2}}\)

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

\(=\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

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

Ta có: \(\dfrac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)

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

\(=1+\sqrt{3}+\sqrt{5}\)

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

Biến đổi vế trái :

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

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

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

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

\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)

b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)

Biến đổi vế trái :

VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)

\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)

\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)

\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)

=> \(A=\sqrt{5}\)

1) ĐKXĐ: \(x\ge5\)

2) ĐKXĐ: \(\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)

5) ĐKXĐ: \(\left[{}\begin{matrix}x\le2\\x\ge3\end{matrix}\right.\)

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

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

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

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

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

\(=2\sqrt{2}\)

a: \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

b: \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{5}+2-4+\sqrt{5}\)

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

\(\sqrt{7+4\sqrt{3}}-\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{21-8\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\left|2-\sqrt{5}\right|-\sqrt{4^2-2\cdot4\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\left(2-\sqrt{5}\right)-\sqrt{\left(4-\sqrt{5}\right)^2}\)

\(=\left|2+\sqrt{3}\right|+2-\sqrt{5}-\left|4-\sqrt{5}\right|\)

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

\(=\sqrt{3}\) 

Lời giải:
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{8}}-2\sqrt{6+3\sqrt{3}}\)

\(=\sqrt{3+18-2\sqrt{3.18}}+\sqrt{8+1+2\sqrt{8.1}}-\sqrt{2}.\sqrt{12+6\sqrt{3}}\)

\(=\sqrt{(\sqrt{18}-\sqrt{3})^2}+\sqrt{(\sqrt{8}+1)^2}-\sqrt{2}.\sqrt{9+3+2\sqrt{9.3}}\)

\(=\sqrt{(\sqrt{18}-\sqrt{3})^2}+\sqrt{(\sqrt{8}+1)^2}-\sqrt{2}.\sqrt{(\sqrt{9}+\sqrt{3})^2}\)

\(=\sqrt{18}-\sqrt{3}+\sqrt{8}+1-\sqrt{2}(\sqrt{9}+\sqrt{3})\)

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