Answers

2014-08-22T00:39:54+05:30
\frac{6}{2 \sqrt{3} -  \sqrt{6} } =  \frac{6(2 \sqrt{3} +  \sqrt{6} )}{(2 \sqrt{3} -  \sqrt{6} )(2 \sqrt{3} +  \sqrt{6} )} = \frac{6(2 \sqrt{3} +  \sqrt{6} )}{12 - 6} = \frac{6(2 \sqrt{3} +  \sqrt{6} )}{6} = 2 \sqrt{3} + \sqrt{6}

\frac{ \sqrt{6} }{ \sqrt{3} +  \sqrt{2} } =  \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ (\sqrt{3} +  \sqrt{2}) (\sqrt{3} - \sqrt{2}) } = \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ (3 - 2) } = \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ 1 } = \sqrt{6}(\sqrt{3} - \sqrt{2})

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

So,
(2 \sqrt{3} +  \sqrt{6}) + ( \sqrt{6} ( \sqrt{3} -  \sqrt{2} )) - ( \sqrt{3} ( \sqrt{6} +  \sqrt{2} )) \\  \\ (2 \sqrt{3} +  \sqrt{6}) + ( \sqrt{6}\sqrt{3} -  \sqrt{6}\sqrt{2} ) - ( \sqrt{3}\sqrt{6} +  \sqrt{3}\sqrt{2} ) \\  \\ 2 \sqrt{3} +  \sqrt{6} + \sqrt{6}\sqrt{3} -  \sqrt{6}\sqrt{2} - \sqrt{3}\sqrt{6} - \sqrt{3}\sqrt{2} \\  \\ 2 \sqrt{3} +  \sqrt{6} + \sqrt{6}\sqrt{3} -  \sqrt{3}\sqrt{2}\sqrt{2} - \sqrt{3}\sqrt{6} - \sqrt{3}\sqrt{2}

2 \sqrt{3} +  \sqrt{6} + \sqrt{3}\sqrt{6} -  2\sqrt{3} - \sqrt{3}\sqrt{6} - \sqrt{6}


You can see each terms is cancelled here.
So, it gives 0 as final sum.


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