Question

rationalise the denominator of 1/root5+root6-root2

Answer 1

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Answer 2

Answer:

On rationalizing the denominator of the denominator we get $=\frac{\sqrt{6}+3\sqrt{5}+4\sqrt{15}-9\sqrt{2}}{39}$

Step-by-step explanation:

$\frac{1}{\sqrt{5}+\sqrt{6}-\sqrt{2}}$

$=\frac{1}{(\sqrt{5}+\sqrt{6})-\sqrt{2}}\times\frac{(\sqrt{5}+\sqrt{6})+\sqrt{2}}{(\sqrt{5}+\sqrt{6})+\sqrt{2}}$

$=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{(\sqrt{5}+\sqrt{6})^2-(\sqrt{2})^2}$

$=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{5+6+2\sqrt{30}-2}$

$=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{9+2\sqrt{30}}$

$=\frac{\sqrt{5}+\sqrt{6}+\sqrt{2}}{9+2\sqrt{30}}\times\frac{9-2\sqrt{30}}{9-2\sqrt{30}}$

$=\frac{(\sqrt{5}+\sqrt{6}+\sqrt{2})(9-2\sqrt{30})}{9^2-(2\sqrt{30})^2}$

$=\frac{9\sqrt{5}+9\sqrt{6}+9\sqrt{2}-10\sqrt{6}-12\sqrt{5}-4\sqrt{5}}{81-4\times30}$

$=\frac{9\sqrt{2}-\sqrt{6}-3\sqrt{5}-4\sqrt{15}}{-39}$

$=\frac{\sqrt{6}+3\sqrt{5}+4\sqrt{15}-9\sqrt{2}}{39}$