Lösung 3.1:7b
Aus Online Mathematik Brückenkurs 1
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Version vom 14:25, 22. Okt. 2008
We multiply the top and bottom of the fraction by the conjugate of the denominator, \displaystyle \sqrt{7}+\sqrt{5}, and see what it leads to,
| \displaystyle \begin{align}
\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} &= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt] &= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt] &= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt] &= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt] &= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt] &= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt] &= -\sqrt{35}\,\textrm{.} \end{align} |
