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Rationalize the denominator of $\frac{1}{\sqrt{2} + \sqrt{5} + \sqrt{7}}$.

 Jan 23, 2021

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 #1
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\(\quad \dfrac1{\sqrt2 +\sqrt5+\sqrt7}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{(\sqrt2+\sqrt7)^2-\sqrt5^2}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{4+2\sqrt{14}}\\ =\dfrac{(\sqrt2-\sqrt5+\sqrt7)(2-\sqrt{14})}{(4+2\sqrt{14})(2-\sqrt{14})}\\ =\dfrac{\sqrt{70}-2\sqrt5-5\sqrt2}{2(2^2 - \sqrt{14}^2)}\\ = \dfrac{2\sqrt5 + 5\sqrt 2 - \sqrt{70}}{20}\)

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 Jan 23, 2021
 #1
avatar+8456 
+1
Best Answer

\(\quad \dfrac1{\sqrt2 +\sqrt5+\sqrt7}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{(\sqrt2+\sqrt7)^2-\sqrt5^2}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{4+2\sqrt{14}}\\ =\dfrac{(\sqrt2-\sqrt5+\sqrt7)(2-\sqrt{14})}{(4+2\sqrt{14})(2-\sqrt{14})}\\ =\dfrac{\sqrt{70}-2\sqrt5-5\sqrt2}{2(2^2 - \sqrt{14}^2)}\\ = \dfrac{2\sqrt5 + 5\sqrt 2 - \sqrt{70}}{20}\)

MaxWong Jan 23, 2021

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