Question

root2/root6-root2-root3/root6+root2 simplify

Answer 1

Step-by-step explanation:

$\frac{ \sqrt{2} }{ \sqrt{6} } - \sqrt{2} - \frac{ \sqrt{3} }{ \sqrt{6} } + \sqrt{2} \\ = \sqrt{2} ( \frac{1}{ \sqrt{3}} - 1 - \frac{ \sqrt{3} }{ \sqrt{3} } + 1) \\ = \sqrt{2} ( \frac{1}{ \sqrt{3} } - \frac{ \sqrt{3} }{ \sqrt{3} } ) \\ = \sqrt{6} (1 - 1) \\ = \sqrt{6} \\ \\ \\ hope \: it \: helps \: you........... \\ please \: mark \: my \: answer \: as \: \\ brainleist..........$

Answer 2

Answer:

Given expression is,

\frac{\sqrt{6} }{\sqrt{2} + \sqrt{3} } + \frac{3\sqrt{2} }{\sqrt{6}+\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6}+\sqrt{2} }2+36+6+332−6+243

By rationalizing each term,

=\frac{\sqrt{6}(-\sqrt{2} + \sqrt{3}) }{(\sqrt{3})^2 - (\sqrt{2})^2 } + \frac{3\sqrt{2}(\sqrt{6}-\sqrt{3}) }{(\sqrt{6})^2-(\sqrt{3})^2 } - \frac{4\sqrt{3}(\sqrt{6}-\sqrt{2}) }{(\sqrt{6})^2-(\sqrt{2})^2 }=(3)2−(2)26(−2+3)+(6)2−(3)232(6−3)−(6)2−(2)243(6−2)

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

=-\sqrt{12} + \sqrt{18} + \sqrt{12} - \sqrt{6} - \sqrt{18} + \sqrt{6}=−12+18+12−6−18+6

=0=0

Hence,

\frac{\sqrt{6} }{\sqrt{2} + \sqrt{3} } + \frac{3\sqrt{2} }{\sqrt{6}+\sqrt{3} } - \frac{4\sqrt{3} }{\sqrt{6}+\sqrt{2} }=02+36+6+332−