Lösung 1.2:1d
Aus Online Mathematik Brückenkurs 1
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When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator, | When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,</math>.}} |
When all three fractions have a common denominator, they can easily be added up | When all three fractions have a common denominator, they can easily be added up | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,</math>.}} |
Version vom 08:13, 22. Okt. 2008
When we have three fractions involved in an addition, we need to multiply the top and bottom of each fraction by the product of the other fraction's denominators so that all fractions have a common denominator,
| \displaystyle \frac{1\cdot 4\cdot 5}{3\cdot 4\cdot 5}+\frac{1\cdot 3\cdot 5}{4\cdot 3\cdot 5}+\frac{1\cdot 3\cdot 4}{5\cdot 3\cdot 4}=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}\,. |
When all three fractions have a common denominator, they can easily be added up
| \displaystyle \frac{20}{60}+\frac{15}{60}+\frac{12}{60}=\frac{20+15+12}{60}=\frac{47}{60}\,. |
