# Get rid of the square roots of the denominator: <mstyle displaystyle="true" scriptlevel="0">

Ryan Robertson 2022-07-02 Answered
Get rid of the square roots of the denominator: $\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
I squared the whole denominator, but that didn't help.
Also I searched for a propriety or identity like ${A}^{2}-{B}^{2}$, but I didn't see one that could fit. Any help is appreciated.
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## Answers (2)

iskakanjulc
Answered 2022-07-03 Author has 18 answers
$\begin{array}{rl}\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}& =\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\cdot \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{\sqrt{7}-2\sqrt{5}-\sqrt{3}}\\ & =\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{\left(\sqrt{7}-2\sqrt{5}{\right)}^{2}-3}\\ & =\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{4\left(6-\sqrt{35}\right)}\cdot \frac{6+\sqrt{35}}{6+\sqrt{35}}\\ & =\frac{\left(\sqrt{7}-2\sqrt{5}-\sqrt{3}\right)\left(6+\sqrt{35}\right)}{4}\end{array}$
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tripes3h
Answered 2022-07-04 Author has 5 answers
To get rid of the square roots of the denominator, you may use ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
$\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{\left(\sqrt{7}-2\sqrt{5}{\right)}^{2}-3}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{7-4\sqrt{35}+20-3}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{24-4\sqrt{35}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}-2\sqrt{5}-\sqrt{3}\right)\left(24+4\sqrt{35}\right)}{{24}^{2}-16\ast 35}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}-2\sqrt{5}-\sqrt{3}\right)\left(24+4\sqrt{35}\right)}{16}\phantom{\rule{0ex}{0ex}}=\frac{-24\sqrt{3}-20\sqrt{5}-16\sqrt{7}-4\sqrt{105}}{16}\phantom{\rule{0ex}{0ex}}=\frac{-6\sqrt{3}-5\sqrt{5}-4\sqrt{7}-\sqrt{105}}{4}$
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