Question

Rationalize the denomination of

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Answer 1

Answer:

Here we can take $(\sqrt{2} +\sqrt{3} )$ as a or $\sqrt{2}$ as a

and $\sqrt{10} or (\sqrt{3} +\sqrt{10} )$ as b

so I take $a=(\sqrt{2}+\sqrt{3} )$ and b = $\sqrt{10}$

$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{10} }*\frac{\sqrt{2}+\sqrt{3}-\sqrt{10}}{\sqrt{2}+\sqrt{3}-\sqrt{10}}\\\\ =>\frac{\sqrt{2}+\sqrt{3}-\sqrt{10}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{10} )^{2} } \\\\=>\frac{\sqrt{2}+\sqrt{3}-\sqrt{10} }{2+3+2\sqrt{6} -100}\\\\ =>\frac{\sqrt{2}+\sqrt{3}-\sqrt{10} }{2\sqrt{6}-95 }$

Hope helps u