Question

$\sqrt{3(} \frac{1}{ \sqrt{3} - 1 } + \frac{1}{ \sqrt{3} + 1 } ) =$
​

Answer 1

 

$\displaystyle\bf\\\sqrt{3}\left(\frac{1}{\sqrt{3}-1}+\frac{1}{\sqrt{3}+1} \right)=\\\\\\=\sqrt{3}\left(\frac{1\times\Big(\sqrt{3}+1\Big)}{\Big(\sqrt{3}-1\Big)\Big(\sqrt{3}+1\Big)}+\frac{1\times\Big(\sqrt{3}-1\Big)}{\Big(\sqrt{3}+1\Big)\Big(\sqrt{3}-1\Big)} \right)=$

.

$\displaystyle\bf\\=\sqrt{3}\left(\frac{\sqrt{3}+1}{\Big(\sqrt{3}\Big)^2-1^2}+\frac{\sqrt{3}-1}{{\Big(\sqrt{3}\Big)^2-1^2}}\right)=\\\\\\=\sqrt{3}\left(\frac{\sqrt{3}+1}{3-1}+\frac{\sqrt{3}-1}{{3-1}}\right)=\\\\\\=\sqrt{3}\left(\frac{\sqrt{3}+1}{2}+\frac{\sqrt{3}-1}{{2}}\right)=\\\\\\=\sqrt{3}\times\frac{\sqrt{3}+1+\sqrt{3}-1}{2}=\\\\\\=\sqrt{3}\times\frac{\sqrt{3}+\sqrt{3}+1-1}{2}=\\\\\\=\sqrt{3}\times\frac{2\sqrt{3}}{2}=\\\\\\=\sqrt{3}\times\sqrt{3}=\boxed{\bf3}$