Lösung 1.2:3a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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| - | The denominator in the expression has | + | The denominator in the expression has 10 as a common factor, |
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| - | 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>,}} | ||
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, | ||
| + | {{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>.}} | ||
| - | + | The lowest common denominator (LCD) is therefore 100, and the expression is equal to | |
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| - | The lowest common denominator | + | |
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| - | , and the expression is equal to | + | |
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| - | <math>\frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}</math> | + | {{Displayed math||<math>\frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}\,</math>.}} |
Version vom 07:55, 19. Sep. 2008
The denominator in the expression has 10 as a common factor,
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,
The lowest common denominator (LCD) is therefore 100, and the expression is equal to
