Solution 3.1:5d

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We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression
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We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression <math>\sqrt{17}+\sqrt{13}</math>, and use the difference of two squares
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<math>\left( a-b \right)\left( a+b \right)=a^{2}-b^{2}</math>, and use the conjugate rule
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{{Displayed math||<math>(a-b)(a+b) = a^2-b^2</math>}}
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with
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with <math>a=\sqrt{17}</math> and <math>b=\sqrt{13}</math>. Both roots are squared away and we get
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<math>a=\sqrt{17}</math>
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and
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<math>b=\sqrt{13}</math>. Both roots are squared away and we get
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{{Displayed math||<math>\begin{align}
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\frac{1}{\sqrt{17}-\sqrt{13}}
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&= \frac{1}{\sqrt{17}-\sqrt{13}}\cdot\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}}\\[5pt]
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&= \frac{\sqrt{17}+\sqrt{13}}{(\sqrt{17})^{2}-(\sqrt{13})^{2}}\\[5pt]
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&= \frac{\sqrt{17}+\sqrt{13}}{17-13}\\[5pt]
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&= \frac{\sqrt{17}+\sqrt{13}}{4}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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This expression cannot be simplified any further because neither 17 nor 13 contain any squares as factors.
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& \frac{1}{\sqrt{17}-\sqrt{13}}=\frac{1}{\sqrt{17}-\sqrt{13}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\
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& =\frac{\sqrt{17}+\sqrt{13}}{\left( \sqrt{17} \right)^{2}-\left( \sqrt{13} \right)^{2}}=\frac{\sqrt{17}+\sqrt{13}}{17-13}=\frac{\sqrt{17}+\sqrt{13}}{4}. \\
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\end{align}</math>
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This expression cannot be simplified any further because neither
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<math>\text{17}</math>
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nor
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<math>\text{13}</math>
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contain any squares as factors.
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Current revision

We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression \displaystyle \sqrt{17}+\sqrt{13}, and use the difference of two squares

\displaystyle (a-b)(a+b) = a^2-b^2

with \displaystyle a=\sqrt{17} and \displaystyle b=\sqrt{13}. Both roots are squared away and we get

\displaystyle \begin{align}

\frac{1}{\sqrt{17}-\sqrt{13}} &= \frac{1}{\sqrt{17}-\sqrt{13}}\cdot\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{(\sqrt{17})^{2}-(\sqrt{13})^{2}}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{17-13}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{4}\,\textrm{.} \end{align}

This expression cannot be simplified any further because neither 17 nor 13 contain any squares as factors.

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