Lösung 1.2:5b

Aus Online Mathematik Brückenkurs 1

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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One solution is to calculate the numerator and denominator in the main fraction individually
One solution is to calculate the numerator and denominator in the main fraction individually
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt]
\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt]
\frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.}
\frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.}
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The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator
The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator
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{{Displayed math||<math>
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{{Abgesetzte Formel||<math>
\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,</math>.}}
\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,</math>.}}
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Another way to solve the exercise is to multiply the top and bottom of the main fraction by <math>3\cdot 2=6</math>, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step
Another way to solve the exercise is to multiply the top and bottom of the main fraction by <math>3\cdot 2=6</math>, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step
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{{Displayed math||<math>
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{{Abgesetzte Formel||<math>
\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}</math>}}
\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}</math>}}

Version vom 08:15, 22. Okt. 2008

Method 1

One solution is to calculate the numerator and denominator in the main fraction individually

\displaystyle \begin{align}

\frac{1}{2}+\frac{1}{3} &= \frac{1\cdot 3}{2\cdot 3}+\frac{1\cdot 2}{3\cdot 2} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6}\,,\\[10pt] \frac{1}{3}-\frac{1}{2} &= \frac{1\cdot 2}{3\cdot 2}-\frac{1\cdot 3}{2\cdot 3} = \frac{2}{6}-\frac{3}{6} = -\frac{1}{6}\,\textrm{.} \end{align}

The whole expression then reduces to a double fraction which we calculate by multiplying top and bottom by the reciprocal of the denominator

\displaystyle

\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{6}\vphantom{\Biggl(}\,}{\,-\dfrac{1}{6}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{5}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}{\,-\dfrac{1}{\rlap{/}6}\cdot{}\rlap{/}6\vphantom{\Biggl(}\,}=\frac{5}{-1}=-5\,.


Method 2

Another way to solve the exercise is to multiply the top and bottom of the main fraction by \displaystyle 3\cdot 2=6, so that all denominators in the partial fractions 1/2 and 1/3 can eliminated in one step

\displaystyle

\frac{\,\dfrac{1}{2}+\dfrac{1}{3}\vphantom{\Biggl(}\,}{\,\dfrac{1}{3}-\dfrac{1}{2}\vphantom{\Biggl(}\,} = \frac{\,\left( \dfrac{1}{2}+\dfrac{1}{3} \right)\cdot 6\vphantom{\Biggl(}\,}{\,\left( \dfrac{1}{3}-\dfrac{1}{2} \right)\cdot 6\vphantom{\Biggl(}\,}=\frac{\,\dfrac{6}{2}+\dfrac{6}{3}\vphantom{\Biggl(}\,}{\,\dfrac{6}{3}-\dfrac{6}{2}\vphantom{\Biggl(}\,}=\frac{3+2}{2-3}=\frac{5}{-1}=-5\,\textrm{.}

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