Lösung 3.1:6c
Aus Online Mathematik Brückenkurs 1
Zuerst schreiben wir die Brüche \displaystyle 1/\!\sqrt{3}, \displaystyle 1/\!\sqrt{5} und \displaystyle 1/\!\sqrt{2}, sodass sie nur Wurzeln im Zähler enthalten
| \displaystyle \frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.} |
Wir erweitern den gesamten Bruch mit 2, um die Nenner im Nenner los zu werden
| \displaystyle \frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.} |
Jetzt erweitern wir den ganzen Bruch mit den konjugierten Nenner \displaystyle \sqrt{2}+1 und vereinfachen
| \displaystyle \begin{align}
\frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1} &= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt] &= \frac{\Bigl(\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}\Bigr)(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\\[10pt] &= \frac{\dfrac{2\sqrt{3}\sqrt{2}}{3}+\dfrac{2\sqrt{3}\cdot 1}{3}-\dfrac{2\sqrt{5}\sqrt{2}}{5}-\dfrac{2\sqrt{5}\cdot 1}{5}}{2-1}\\[10pt] &= \frac{\dfrac{2}{3}\sqrt{3\cdot 2}+\dfrac{2}{3}\sqrt{3}-\dfrac{2}{5}\sqrt{2\cdot 5}-\dfrac{2}{5}\sqrt{5}}{1}\\[10pt] &= \frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}\,\textrm{.} \end{align} |
