Lösung 3.1:7b
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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We multiply the top and bottom of the fraction by the conjugate of the denominator, | We multiply the top and bottom of the fraction by the conjugate of the denominator, | ||
| - | <math>\sqrt{7}+\sqrt{5}</math> | + | <math>\sqrt{7}+\sqrt{5}</math>, and see what it leads to, |
| - | , and see what it leads to | + | |
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} |
| - | & \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}\ | + | &= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt] |
| - | & =\frac{5 | + | &= \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\ | + | &= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt] |
| - | & =\frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}=\frac{ | + | &= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt] |
| - | & =-\sqrt{35} \\ | + | &= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt] |
| - | \end{align}</math> | + | &= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt] |
| + | &= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt] | ||
| + | &= -\sqrt{35}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Version vom 14:15, 30. Sep. 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,
