Question

√2 ÷ √6-√2 -√3÷√6+√2

Answer 1

hope this answer might help you

Answer 2

$hi..$
$\frac{ \sqrt{2} }{ \sqrt{6} } - \sqrt{2} - \frac{ \sqrt{3} }{ \sqrt{6} } + \sqrt{2}$

$= \frac{1}{ \sqrt{3} } - \sqrt{2} - \frac{1}{ \sqrt{2} } + \sqrt{2}$

$= \frac{1}{ \sqrt{3} } - \sqrt{2} - \frac{1 + 2}{ \sqrt{2} } .......(l.c.m)$

$= \frac{1}{ \sqrt{3} } - \sqrt{2} - \frac{3}{ \sqrt{2} }$

$= \frac{1 - \sqrt{6} }{ \sqrt{3} } - \frac{3}{ \sqrt{2} } ............(l.c.m)$

$= \frac{ \sqrt{2}(1 - \sqrt{6} ) - 3 \sqrt{3} }{ \sqrt{6} } .........(l.c.m)$

$= \frac{ \sqrt{2} - \sqrt{12} - 3 \sqrt{3} }{ \sqrt{6} }$

$= \frac{ \sqrt{2} - \sqrt{3 \times 4} - 3 \sqrt{3} }{ \sqrt{6} }$


$= \frac{ \sqrt{2} - 2 \sqrt{3} - 3 \sqrt{3} }{ \sqrt{6} }$

$= \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{6} }$

$hope \: \: this \: \: helps..$