Lösung 3.1:6d
Aus Online Mathematik Brückenkurs 1
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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 | 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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{6}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})-\sqrt{6}} | \frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{6}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})-\sqrt{6}} | ||
&= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{6})^{2}}\\[10pt] | &= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{6})^{2}}\\[10pt] | ||
| Zeile 9: | Zeile 9: | ||
We expand the remaining quadratic, <math>(\sqrt{2}+\sqrt{3})^{2}</math>, using the formula for the difference of two squares, | We expand the remaining quadratic, <math>(\sqrt{2}+\sqrt{3})^{2}</math>, using the formula for the difference of two squares, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6} | \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6} | ||
&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{3}+(\sqrt{3})^{2}-6}\\[10pt] | &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{3}+(\sqrt{3})^{2}-6}\\[10pt] | ||
| Zeile 18: | Zeile 18: | ||
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>, | 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>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} | \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} | ||
&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1}\\[10pt] | &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1}\\[10pt] | ||
Version vom 08:39, 22. Okt. 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
| \displaystyle \begin{align}
\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{6}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})-\sqrt{6}} &= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{6})^{2}}\\[10pt] &= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6}\,\textrm{.} \end{align} |
We expand the remaining quadratic, \displaystyle (\sqrt{2}+\sqrt{3})^{2}, using the formula for the difference of two squares,
| \displaystyle \begin{align}
\frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6} &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{3}+(\sqrt{3})^{2}-6}\\[10pt] &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\cdot 3}+3-6}\\[10pt] &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\,\textrm{.} \end{align} |
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,
| \displaystyle \begin{align}
\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1}\\[10pt] &= \frac{(\sqrt{2}+\sqrt{3}-\sqrt{6})(2\sqrt{6}+1)}{(2\sqrt{6})^{2}-1^{2}}\\[10pt] &= \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] &= \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] &= \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] &= \frac{2\cdot 2\cdot \sqrt{3}+\sqrt{2}+2\cdot 3\cdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23}\\[10pt] &= \frac{(1+2\cdot 3)\sqrt{2}+(2\cdot 2+1)\sqrt{3}-12-\sqrt{6}}{23}\\[10pt] &= \frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23}\,\textrm{.} \end{align} |
