Lösung 3.1:6a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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We use the standard method and augment the fraction with the conjugate of the denominator <math>\sqrt{5}+2</math>. Then the formula for the difference of two squares gives | We use the standard method and augment the fraction with the conjugate of the denominator <math>\sqrt{5}+2</math>. Then the formula for the difference of two squares gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{\sqrt{2}+3}{\sqrt{5}-2} | \frac{\sqrt{2}+3}{\sqrt{5}-2} | ||
&= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] | &= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] | ||
Version vom 08:38, 22. Okt. 2008
We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the formula for the difference of two squares gives
| \displaystyle \begin{align}
\frac{\sqrt{2}+3}{\sqrt{5}-2} &= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] &= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt] &= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt] &= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt] &= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.} \end{align} |
