`a.`\(\dfrac{2\sqrt{3}-\sqrt{21}}{2-\sqrt{7}}+\sqrt{\dfrac{6}{2-\sqrt{3}}}-\dfrac{6}{\sqrt{3}}\)
`=`\(\dfrac{\sqrt{3}\left(2-\sqrt{7}\right)}{2-\sqrt{7}}+\sqrt{\dfrac{6}{2-\sqrt{3}}}-\dfrac{6}{\sqrt{3}}\)
`=`\(\sqrt{3}+\sqrt{\dfrac{6}{2-\sqrt{3}}}-\dfrac{6}{\sqrt{3}}\)
`=`\(\sqrt{\dfrac{6}{2-\sqrt{3}}}-\sqrt{3}\)
\(=\dfrac{\sqrt{6}}{\sqrt{2-\sqrt{3}}}-\sqrt{3}\)
\(=\dfrac{2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}-\sqrt{3}\)
\(=\dfrac{2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{3}\)
\(=\dfrac{2\sqrt{3}}{\sqrt{3}-1}-\sqrt{3}\)
\(=3\)
`b.`\(\sqrt{16x-32}=6\sqrt{\dfrac{x-2}{9}}+8\)
\(\Leftrightarrow4\sqrt{x-2}=2\sqrt{x-2}+8\) ; \(\left(x\ge2\right)\)
\(\Leftrightarrow2\sqrt{x-2}=8\)
\(\Leftrightarrow\sqrt{x-2}=4\)
`<=>x-2=16`
`<=>x=18`
Vậy \(S=\left\{18\right\}\)
