Question

express the following with rational denominator
$\frac{ 2 \sqrt{3} }{ \sqrt{6 } + 2 }$

Answer 1

$\boxed{\sf{ \frac{2 \sqrt{3} }{ \sqrt{6} + 2}}}$

$\implies$ $\sf{\frac{2 \sqrt{3} }{ \sqrt{6} + 2 } \times \frac{ \sqrt{6} - 2 }{ \sqrt{6} - 2}}$

$\implies$ $\sf{ \frac{2 \sqrt{3} \times ( \sqrt{6} - 2)}{6 - 4}}$

$\implies$ $\sf{ \frac{2 \sqrt{3} \times ( \sqrt{6} - 2) }{2}}$

$\implies$ $\sf{\sqrt{3} \times ( \sqrt{6} - 2)}$

$\implies$ $\sf{ \sqrt{18} - 2 \sqrt{3}}$

$\implies$ $\sf{3 \sqrt{2} - 2 \sqrt{3}}$

$\implies$ $\sf{0.778539}$

Note: Check this attachment.

Answer 2

Answer:

$\huge{\boxed{\boxed{\sf{ \frac{ 2 \sqrt{3} }{ \sqrt{6 } + 2 } =0.77 }}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

$\frac{ 2 \sqrt{3} }{ \sqrt{6 } + 2 } \\ = )\frac{ 2 \sqrt{3} }{ \sqrt{6 } + 2 } \times \frac{ (\sqrt{6 } - 2)}{ \sqrt{6} - 2} \\ we \: use \: formula \: in \: denominator \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ = ) \frac{2 \sqrt{18} - 4 \sqrt{3} }{ (\sqrt{6})^{2} - ( {2})^{2} } \\ = ) \frac{2 \times 3 \sqrt{2} - 4 \sqrt{3} }{6 - 4} \\ = ) \frac{6 \sqrt{2} - 4 \sqrt{3} }{2} \\ = ) \frac{2(3 \sqrt{2} - 2 \sqrt{3}) }{2} \\ = ) 3 \times 1.41 - 2 \times 1.73 \\ = )4.23 - 3.46 \\ = )0.77</p><p></p><p>$

$\huge{\boxed{\boxed{\sf{ \frac{ 2 \sqrt{3} }{ \sqrt{6 } + 2 } =0.77 }}}}$