Question

Rationalise the denominator 1 upon root5 plus root3 minus root2

Answer 1

Answer:

$\boxed{\dfrac{-2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{3}}$

Step-by-step explanation:

$\dfrac{1}{(\sqrt{5} + \sqrt{3}) - \sqrt{2}}\\\\\\\text{Multiply Numerator and Denominator by } (\sqrt{5} + \sqrt{3}) + \sqrt{2}.\\\\\\= \dfrac{1((\sqrt{5} + \sqrt{3}) + \sqrt{2})}{((\sqrt{5} + \sqrt{3}) - \sqrt{2})((\sqrt{5} + \sqrt{3}) + \sqrt{2})}\\\\\\= \dfrac{\sqrt{5} + \sqrt{3} + \sqrt{2}}{(\sqrt{5} + \sqrt{3})^2 - (\sqrt{2})^2}\\\\\\= \dfrac{\sqrt{5} + \sqrt{3} + \sqrt{2}}{5 + 3 + 2(\sqrt{15}) - 2}\\\\\\= \dfrac{\sqrt{5} + \sqrt{3} + \sqrt{2}}{6 + 2\sqrt{15}}$

$\text{Again multiplying numerator and denominator by } 6 - 2\sqrt{15},\\\\\\= \dfrac{(\sqrt{5} + \sqrt{3} + \sqrt{2})(6 - 2\sqrt{15})}{(6+2\sqrt{15})(6-2\sqrt{15})}\\\\\\= \dfrac{6\sqrt{5} + 6\sqrt{3} + 6\sqrt{2} - 10\sqrt{3} - 6\sqrt{5} - 2\sqrt{30}}{6^2 - (2\sqrt{15})^2}\\\\\\= \dfrac{-4\sqrt{3} + 6\sqrt{2} - 2\sqrt{30}}{36 - 30}\\\\\\= \dfrac{2(-2\sqrt{3} + 3\sqrt{2} - \sqrt{30})}{6}\\\\\\= \boxed{\dfrac{-2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{3}}$

Answer 2

Step-by-step explanation:

thus this is the answer ..

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