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Rationalize (sqrt(2) + sqrt(3))/(sqrt(10) + sqrt(14) + sqrt(15) + sqrt(21))

 Feb 14, 2022
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\(\frac{\sqrt2+\sqrt3}{\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}}\)

Looking at the denominator, notice that \(\sqrt{10}\) and \(\sqrt{14}\) are both divisible by \(\sqrt{2}\) and\( \sqrt{15}\), \(\sqrt{21}\) are divisible by \(\sqrt{2}\)

We can factor those out to get \(\frac{\sqrt2+\sqrt3}{\sqrt2(\sqrt5+\sqrt7)+\sqrt{3}(\sqrt5+\sqrt7)} = \frac{\sqrt2+\sqrt3}{(\sqrt2+\sqrt{3})(\sqrt5+\sqrt7)}\). Divide top and bottom by \(\sqrt2+\sqrt3\)\(\frac{1}{\sqrt5+\sqrt7}\).

Moving the radicals to the numerator: \(\frac{1}{\sqrt5+\sqrt7} = \frac{1\cdot(\sqrt5-\sqrt7)}{(\sqrt5+\sqrt7)(\sqrt5-\sqrt7)} = \frac{\sqrt{7}-\sqrt{5}}{2}\).

So the final result is: \(\boxed{\frac{\sqrt7 - \sqrt5}{2}}\). Hope this helped smiley

 Feb 14, 2022
edited by asiandude  Feb 14, 2022
edited by asiandude  Feb 14, 2022

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