Question

Date 1-12
simplify? V6
V2 +V3
+ 32
V6+ V3
– 412
V6 + V2​

Answer 1

Step-by-step explanation:

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

Rationalise all denominators of three fractions the we get

$\frac{ \sqrt{6}( \sqrt{3} - \sqrt{2})} {( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2})} + \frac{3 \sqrt{2}( \sqrt{6} - \sqrt{3})}{( \sqrt{6} - \sqrt{3})( \sqrt{6} + \sqrt{3})}$

$- \frac{4 \sqrt{3}( \sqrt{6} - \sqrt{2})}{( \sqrt{6} - \sqrt{2})( \sqrt{6} + \sqrt{2})} = >$

$\frac{ \sqrt{6}( \sqrt{3} - \sqrt{2})}{3 - 2} + \frac{3 \sqrt{2}( \sqrt{6} - \sqrt{3})}{6 - 3} - \frac{4 \sqrt{3}( \sqrt{6} - \sqrt{2}) }{6 - 2}$

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

$- \sqrt{3}( \sqrt{6} - \sqrt{2}) = >$

$\sqrt{18} - \sqrt{12} + \sqrt{12} - \sqrt{6} - \sqrt{18} +$

$\sqrt{6} = > 0$

Hope it helps you.