Question

2 under root 6 upon under root 2 + under root 3 + 6 under root 2 upon under root 6 + under root 3 minus under root 3 upon under root 6 + under root 2 ​

Answer 1

Answer:

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Answer 2

Answer: The answer is 0.

Step-by-step explanation:

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} }

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 }

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

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

=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} }=0