Lösung 3.1:6c
Aus Online Mathematik Brückenkurs 1
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| - | The expression is so complicated that we first need to simplify it. We start with the three fractions, | + | The expression is so complicated that we first need to simplify it. We start with the three fractions, <math>1/\!\sqrt{3}</math>, <math>1/\!\sqrt{5}</math> and <math>1/\!\sqrt{2}</math>, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators |
| - | <math>\ | + | |
| - | and | + | |
| - | <math> | + | |
| + | {{Displayed math||<math>\frac{\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}}{\dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{2}}{\sqrt{2}}-\dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}}{\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}}\,\textrm{.}</math>}} | ||
| - | + | Then, multiply the top and bottom of the fraction by 2 so that we get rid of the fractions in the denominator | |
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| - | Then, multiply the top and bottom of the fraction by | + | |
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| - | so that we get rid of the fractions in the denominator | + | |
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| + | {{Displayed math||<math>\frac{\Bigl(\dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{5}}{5}\Bigr)\cdot 2}{\Bigl(\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\Bigr)\cdot 2} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\dfrac{2\sqrt{2}}{2}-\dfrac{2}{2}} = \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\,\textrm{.}</math>}} | ||
Now, we can multiply the top and bottom by the conjugate of the denominator | Now, we can multiply the top and bottom by the conjugate of the denominator | ||
<math>\sqrt{2}+1</math>, to get an expression without roots in the denominator. | <math>\sqrt{2}+1</math>, to get an expression without roots in the denominator. | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1} |
| - | + | &= \frac{\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}\\[10pt] | |
| - | & =\frac{\ | + | &= \frac{\Bigl(\dfrac{2\sqrt{3}}{3}-\dfrac{2\sqrt{5}}{5}\Bigr)(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\\[10pt] |
| - | & =\frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5} \\ | + | &= \frac{\dfrac{2\sqrt{3}\sqrt{2}}{3}+\dfrac{2\sqrt{3}\cdot 1}{3}-\dfrac{2\sqrt{5}\sqrt{2}}{5}-\dfrac{2\sqrt{5}\cdot 1}{5}}{2-1}\\[10pt] |
| - | \end{align}</math> | + | &= \frac{\dfrac{2}{3}\sqrt{3\cdot 2}+\dfrac{2}{3}\sqrt{3}-\dfrac{2}{5}\sqrt{2\cdot 5}-\dfrac{2}{5}\sqrt{5}}{1}\\[10pt] |
| + | &= \frac{2}{3}\sqrt{6}+\frac{2}{3}\sqrt{3}-\frac{2}{5}\sqrt{10}-\frac{2}{5}\sqrt{5}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Version vom 12:49, 30. Sep. 2008
The expression is so complicated that we first need to simplify it. We start with the three fractions, \displaystyle 1/\!\sqrt{3}, \displaystyle 1/\!\sqrt{5} and \displaystyle 1/\!\sqrt{2}, which contain root signs and multiply their numerators and denominators in such a way that the root signs end up only in the numerators
Then, multiply the top and bottom of the fraction by 2 so that we get rid of the fractions in the denominator
Now, we can multiply the top and bottom by the conjugate of the denominator \displaystyle \sqrt{2}+1, to get an expression without roots in the denominator.
