Lösung 1.2:2d
Aus Online Mathematik Brückenkurs 1
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If we divide up the denominators into their smallest possible integer factors, | If we divide up the denominators into their smallest possible integer factors, | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | 45&=5\cdot 9=5\cdot 3\cdot 3\,, \\ |
| - | + | 75&=3\cdot 25=3\cdot 5\cdot 5\,, \\ | |
| - | + | \end{align}</math>}} | |
| - | \end{align}</math> | + | |
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the expression can be written as | the expression can be written as | ||
| + | {{Displayed math||<math>\frac{1}{5\cdot 3\cdot 3}+\frac{1}{3\cdot 5\cdot 5}</math>}} | ||
| - | + | and then we see that the denominators have <math>3\cdot 5</math> as a common factor. Therefore, if we multiply the top and bottom of the first fraction by 5 | |
| - | + | and the second by 3, the result is the lowest possible denominator | |
| - | and then we see that the denominators have | + | |
| - | <math>3\ | + | |
| - | as a common factor. Therefore, if we multiply the top and bottom of | + | |
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| - | and the second by | + | |
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| - | , the result is the lowest possible denominator | + | |
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| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{2}{5\cdot 3\cdot 3}\cdot \frac{5}{5}+\frac{1}{3\cdot 5\cdot 5}\cdot | ||
| + | \frac{3}{3} &=\frac{2}{5\cdot 3\cdot 3\cdot 5} | ||
| + | +\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | ||
| + | &= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | ||
| + | \end{align}</math>}} | ||
| - | The lowest common denominator is | + | The lowest common denominator is 225. |
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| - | . | + | |
Version vom 07:52, 19. Sep. 2008
If we divide up the denominators into their smallest possible integer factors,
the expression can be written as
and then we see that the denominators have \displaystyle 3\cdot 5 as a common factor. Therefore, if we multiply the top and bottom of the first fraction by 5 and the second by 3, the result is the lowest possible denominator
The lowest common denominator is 225.
