Lösung 3.1:6d

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The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as
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The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as <math>(\sqrt{2}+\sqrt{3})+\sqrt{6}</math> and multiply the top and bottom of the fraction by the conjugate-like expression <math>(\sqrt{2}+\sqrt{3})-\sqrt{6}\,</math>. Then, at least <math>\sqrt{6}</math> will be squared away using the formula for the difference of two squares
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<math>\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}</math>
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and multiply the top and bottom of the fraction by the conjugate-like expression
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<math>\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}</math>
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Then, at least
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<math>\sqrt{6}</math>
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will be squared away using the conjugate rule
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{{Displayed math||<math>\begin{align}
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\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{6}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})-\sqrt{6}}
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&= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{6})^{2}}\\[10pt]
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&= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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We expand the remaining quadratic, <math>(\sqrt{2}+\sqrt{3})^{2}</math>, using the formula for the difference of two squares,
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& \frac{1}{\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}}\centerdot \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}=\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-\left( \sqrt{6} \right)^{2}} \\
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& =\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6}
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&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{3}+(\sqrt{3})^{2}-6}\\[10pt]
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&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\cdot 3}+3-6}\\[10pt]
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&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\,\textrm{.}
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\end{align}</math>}}
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We expand the remaining quadratic,
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This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate <math>2\sqrt{6}+1</math>,
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<math>\left( \sqrt{2}+\sqrt{3} \right)^{2}</math>, using the squaring rule
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}
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& \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{\left( \sqrt{2} \right)^{2}+2\sqrt{2}\sqrt{3}+\left( \sqrt{3} \right)^{2}-6} \\
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&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1}\\[10pt]
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& =\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\centerdot 3}+3-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} \\
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&= \frac{(\sqrt{2}+\sqrt{3}-\sqrt{6})(2\sqrt{6}+1)}{(2\sqrt{6})^{2}-1^{2}}\\[10pt]
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\end{align}</math>
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&= \frac{\sqrt{2}\cdot 2\sqrt{6}+\sqrt{2}\cdot 1+\sqrt{3}\cdot 2\sqrt{6}+\sqrt{3}\cdot 1-\sqrt{6}\cdot 2\sqrt{6}-\sqrt{6}\cdot 1}{2^{2}(\sqrt{6})^{2}-1^{2}}\\[10pt]
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&= \frac{\sqrt{2}\cdot 2\sqrt{2\cdot 3}+\sqrt{2}+\sqrt{3}\cdot 2\sqrt{2\cdot 3}+\sqrt{3}-2(\sqrt{6})^{2}-\sqrt{6}}{4\cdot 6-1^{2}}\\[10pt]
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&= \frac{2(\sqrt{2})^{2}\sqrt{3}+\sqrt{2}+2(\sqrt{3})^{2}\sqrt{2}+\sqrt{3}-2\cdot 6-\sqrt{6}}{24-1}\\[10pt]
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This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate
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&= \frac{2\cdot 2\cdot \sqrt{3}+\sqrt{2}+2\cdot 3\cdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23}\\[10pt]
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<math>2\sqrt{6}+1</math>,
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&= \frac{(1+2\cdot 3)\sqrt{2}+(2\cdot 2+1)\sqrt{3}-12-\sqrt{6}}{23}\\[10pt]
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&= \frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23}\,\textrm{.}
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<math>\begin{align}
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\end{align}</math>}}
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& \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\centerdot \frac{2\sqrt{6}+1}{2\sqrt{6}+1}=\frac{\left( \sqrt{2}+\sqrt{3}-\sqrt{6} \right)\left( 2\sqrt{6}+1 \right)}{\left( 2\sqrt{6} \right)^{2}-1^{2}} \\
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& =\frac{\sqrt{2}\centerdot 2\sqrt{6}+\sqrt{2}\centerdot 1+\sqrt{3}\centerdot 2\sqrt{6}+\sqrt{3}\centerdot 1-\sqrt{6}\centerdot 2\sqrt{6}-\sqrt{6}\centerdot 1}{2^{2}\left( \sqrt{6} \right)^{2}-1^{2}} \\
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& =\frac{\sqrt{2}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{2}+\sqrt{3}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{3}-2\left( \sqrt{6} \right)^{2}-\sqrt{6}}{4\centerdot 6-1^{2}} \\
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& =\frac{2\left( \sqrt{2} \right)^{2}\sqrt{3}+\sqrt{2}+2\left( \sqrt{3} \right)^{2}\sqrt{2}+\sqrt{3}-2\centerdot 6-\sqrt{6}}{24-1} \\
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& =\frac{2\centerdot 2\centerdot \sqrt{3}+\sqrt{2}+2\centerdot 3\centerdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23} \\
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& =\frac{\left( 1+2\centerdot 3 \right)\sqrt{2}+\left( 2\centerdot 2+1 \right)\sqrt{3}-12-\sqrt{6}}{23} \\
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& =\frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23} \\
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\end{align}</math>
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Version vom 13:57, 30. Sep. 2008

The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as \displaystyle (\sqrt{2}+\sqrt{3})+\sqrt{6} and multiply the top and bottom of the fraction by the conjugate-like expression \displaystyle (\sqrt{2}+\sqrt{3})-\sqrt{6}\,. Then, at least \displaystyle \sqrt{6} will be squared away using the formula for the difference of two squares

Vorlage:Displayed math

We expand the remaining quadratic, \displaystyle (\sqrt{2}+\sqrt{3})^{2}, using the formula for the difference of two squares,

Vorlage:Displayed math

This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate \displaystyle 2\sqrt{6}+1,

Vorlage:Displayed math

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