Lösung 1.2:3a

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The denominator in the expression has 10 as a common factor,
The denominator in the expression has 10 as a common factor,
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{{Displayed math||<math>\frac{3}{2\cdot 10}+\frac{7}{5\cdot 10}-\frac{1}{10}\,</math>,}}
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{{Abgesetzte Formel||<math>\frac{3}{2\cdot 10}+\frac{7}{5\cdot 10}-\frac{1}{10}\,</math>,}}
and it is therefore sufficient to multiply the top and bottom of each fraction by the other factors in the denominators in order to obtain a common denominator,
and it is therefore sufficient to multiply the top and bottom of each fraction by the other factors in the denominators in order to obtain a common denominator,
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{{Displayed math||<math>\frac{3\cdot 5}{20\cdot 5}+\frac{7\cdot 2}{50\cdot 2}-\frac{1\cdot 5\cdot 2}{10\cdot 5\cdot 2}=\frac{15}{100}+\frac{14}{100}-\frac{10}{100}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{3\cdot 5}{20\cdot 5}+\frac{7\cdot 2}{50\cdot 2}-\frac{1\cdot 5\cdot 2}{10\cdot 5\cdot 2}=\frac{15}{100}+\frac{14}{100}-\frac{10}{100}\,</math>.}}
The lowest common denominator (LCD) is therefore 100, and the expression is equal to
The lowest common denominator (LCD) is therefore 100, and the expression is equal to
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{{Displayed math||<math>\frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}\,</math>.}}
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{{Abgesetzte Formel||<math>\frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}\,</math>.}}

Version vom 08:14, 22. Okt. 2008

The denominator in the expression has 10 as a common factor,

\displaystyle \frac{3}{2\cdot 10}+\frac{7}{5\cdot 10}-\frac{1}{10}\,,

and it is therefore sufficient to multiply the top and bottom of each fraction by the other factors in the denominators in order to obtain a common denominator,

\displaystyle \frac{3\cdot 5}{20\cdot 5}+\frac{7\cdot 2}{50\cdot 2}-\frac{1\cdot 5\cdot 2}{10\cdot 5\cdot 2}=\frac{15}{100}+\frac{14}{100}-\frac{10}{100}\,.

The lowest common denominator (LCD) is therefore 100, and the expression is equal to

\displaystyle \frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}\,.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:3a“