Lösung 1.2:2d
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K |
K |
||
| Zeile 8: | Zeile 8: | ||
the expression can be written as | the expression can be written as | ||
| - | {{Displayed math||<math>\frac{ | + | {{Displayed math||<math>\frac{2}{3\cdot 3\cdot 5}+\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 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 | ||
| Zeile 14: | Zeile 14: | ||
{{Displayed math||<math>\begin{align} | {{Displayed math||<math>\begin{align} | ||
| - | \frac{2}{ | + | \frac{2}{3\cdot 3\cdot 5}\cdot \frac{5}{5}+\frac{1}{3\cdot 5\cdot 5}\cdot |
| - | \frac{3}{3} &=\frac{2}{ | + | \frac{3}{3} &=\frac{2}{3\cdot 3\cdot 5\cdot 5} |
+\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | +\frac{3}{3\cdot 5\cdot 5\cdot 3}\\[10pt] | ||
&= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | &= \frac{10}{225}+\frac{3}{225}\,\textrm{.}\\ | ||
Version vom 12:21, 22. 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.
