Skillnad mellan versioner av "1.4 Lösning 5a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | <big><big><math> \frac{x}{3} + {x \over 2} - {x \over 6} \; = \; {2\,x \over 2 \cdot 3} + {3\,x \over 3 \cdot 2} - {x \over 6} \; = \; {2\,x \over 6} + {3\,x \over 6} - {x \over 6} \; = \; {2\,x + 3\,x - x \over 6} \; = \; </math></big></big> | + | <big><big><math> \frac{x}{3} + {x \over 2} - {x \over 6} \; = \; {2\,x \over 2\,\cdot\,3} + {3\,x \over 3\,\cdot\,2} - {x \over 6} \; = \; {2\,x \over 6} + {3\,x \over 6} - {x \over 6} \; = \; {2\,x\,+\,3\,x\,-\,x \over 6} \; = \; </math></big></big> |
<big><big><math> = \; {4\,x \over 6} \; = \; {2\,x \over 3} \; = \; {2 \over 3} </math></big></big> <math>\!x </math> | <big><big><math> = \; {4\,x \over 6} \; = \; {2\,x \over 3} \; = \; {2 \over 3} </math></big></big> <math>\!x </math> | ||
Versionen från 2 augusti 2014 kl. 21.53
\( \frac{x}{3} + {x \over 2} - {x \over 6} \; = \; {2\,x \over 2\,\cdot\,3} + {3\,x \over 3\,\cdot\,2} - {x \over 6} \; = \; {2\,x \over 6} + {3\,x \over 6} - {x \over 6} \; = \; {2\,x\,+\,3\,x\,-\,x \over 6} \; = \; \)
\( = \; {4\,x \over 6} \; = \; {2\,x \over 3} \; = \; {2 \over 3} \) \(\!x \)
