Lösung 3.1:6a

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We use the standard method and augment the fraction with the conjugate of the denominator
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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
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<math>\sqrt{5}+2</math>. Then the conjugate rule gives
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{\sqrt{2}+3}{\sqrt{5}-2}
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& \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\
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&= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt]
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& =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\
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&= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt]
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& =6+2\sqrt{2}+3\sqrt{5}+10 \\
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&= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt]
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\end{align}</math>
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&= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt]
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&= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.}
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\end{align}</math>}}

Version vom 11:41, 30. Sep. 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

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Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:6a“