Normalform eines Wurzelterms

Bringen Sie den Wurzelterm auf die Normalform.

$\begin{equation*} \frac{\sqrt{3}-\sqrt{5}-\sqrt{2}}{\sqrt{3}-\sqrt{5}+\sqrt{2}} \end{equation*}$

Lösung

$\begin{eqnarray*} \frac{\sqrt{3}-\sqrt{5}-\sqrt{2}}{\sqrt{3}-\sqrt{5}+\sqrt{2}} & = & \frac{\sqrt{3}-\sqrt{5}-\sqrt{2}}{\left( \sqrt{3}-\sqrt{5}\right)+\sqrt{2}}\cdot\frac{\left( \sqrt{3}-\sqrt{5}\right)-\sqrt{2}}{\left( \sqrt{3}-\sqrt{5}\right)-\sqrt{2}} \\ & = & \frac{\left( \sqrt{3}-\sqrt{5}-\sqrt{2}\right)^2}{\left( \sqrt{3}-\sqrt{5}\right)^2-2} \\ & = & \frac{\left( \sqrt{3}-\sqrt{5}-\sqrt{2}\right)^2}{3-2\sqrt{15}+5-2} \\ & = & \frac{\left( \sqrt{3}-\sqrt{5}-\sqrt{2}\right)^2}{6-2\sqrt{15}} \\ & = & \frac{\left( \sqrt{3}-\sqrt{5}-\sqrt{2}\right)^2}{2\left( 3-\sqrt{15}\right)}\cdot\frac{3+\sqrt{15}}{3+\sqrt{15}} \\ & = & \frac{\left( \sqrt{3}-\sqrt{5}-\sqrt{2}\right)^2\left( 3+\sqrt{15}\right)}{2\left(9-15\right)} \end{eqnarray*}$

$\begin{equation*} = \frac{\left( 3+5+2-2\sqrt{15}-2\sqrt{6}+2\sqrt{10}\right)\left( 3+\sqrt{15}\right)}{-12} \end{equation*}$

$\begin{equation*} = \frac{\left( 10-2\sqrt{15}-2\sqrt{6}+2\sqrt{10}\right)\left( 3+\sqrt{15}\right)}{-12} \end{equation*}$

$\begin{equation*} = \frac{\left( 5-\sqrt{15}-\sqrt{6}+\sqrt{10}\right)\left( 3+\sqrt{15}\right)}{-6} \end{equation*}$

$\begin{equation*} = \frac{15+5\sqrt{15}-3\sqrt{15}-15-3\sqrt{6}-\sqrt{90}+3\sqrt{10}+\sqrt{150}}{-6} \end{equation*}$

$\begin{equation*} = \frac{2\sqrt{15}-3\sqrt{6}-3\sqrt{10}+3\sqrt{10}+5\sqrt{6}}{-6} = \underline{\underline{-\frac{1}{3}\sqrt{6}-\frac{1}{3}\sqrt{15}}} \end{equation*}$

Bringen Sie den Wurzelterm auf die Normalform.

$\begin{equation*} \frac{2\sqrt{6}-1}{\sqrt{2}+\sqrt{3}+\sqrt{6}} \end{equation*}$

Lösung

$\begin{eqnarray*} \frac{2\sqrt{6}-1}{\sqrt{2}+\sqrt{3}+\sqrt{6}} & = & \frac{2\sqrt{6}-1}{\left( \sqrt{2}+\sqrt{3}\right) + \sqrt{6}}\cdot\frac{\left( \sqrt{2}+\sqrt{3}\right) - \sqrt{6}}{\left( \sqrt{2}+\sqrt{3}\right) - \sqrt{6}} \\ & = & \frac{\left( 2\sqrt{6}-1\right)\left( \sqrt{2}+\sqrt{3}-\sqrt{6}\right)}{\left(\sqrt{2}+\sqrt{3}\right)^2 - 6} \\ & = & \frac{2\sqrt{12}+2\sqrt{18}-12-\sqrt{2}-\sqrt{3}+\sqrt{6}}{2+2\sqrt{6}+3-6} \\ & = & \frac{4\sqrt{3}+6\sqrt{2}-12-\sqrt{2}-\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1} \\ & = & \frac{5\sqrt{2}+3\sqrt{3}+\sqrt{6}-12}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1} \\ & = & \frac{\left( 5\sqrt{2}+3\sqrt{3}+\sqrt{6}-12\right)\left( 2\sqrt{6}+1\right)}{23} \end{eqnarray*}$

$\begin{equation*} = \frac{10\sqrt{12}+5\sqrt{2}+6\sqrt{18}+3\sqrt{3}+12+\sqrt{6}-24\sqrt{6}-12}{23} \end{equation*}$

$\begin{equation*} = \frac{20\sqrt{3}+5\sqrt{2}+18\sqrt{2}+3\sqrt{3}-23\sqrt{6}}{23} = \underline{\underline{\sqrt{2}+\sqrt{3}-\sqrt{6}}} \end{equation*}$

Schlagwort: Normalform eines Wurzelterms

138l – FW Algebra

Lösung

138k – FW Algebra

Lösung