Lösung 3.1:5d
Aus Online Mathematik Brückenkurs 1
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 \left( a-b \right)\left( a+b \right)=a^{2}-b^{2}, and use the conjugate rule
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}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\
& =\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}. \\
\end{align}
This expression cannot be simplified any further because neither
\displaystyle \text{17}
nor
\displaystyle \text{13}
contain any squares as factors.