Skillnad mellan versioner av "1.4 Lösning 12"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 8: | Rad 8: | ||
v^2\,(x^2-u^2) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ | v^2\,(x^2-u^2) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ | ||
v^2\,x^2-v^2\,u^2 - u\,x + u^2 & = v^2\,x^2 + u\,v^2\,x - u^2\,v^2 & | \;\;\quad - v^2\,x^2 + v^2\,u^2 \\ | v^2\,x^2-v^2\,u^2 - u\,x + u^2 & = v^2\,x^2 + u\,v^2\,x - u^2\,v^2 & | \;\;\quad - v^2\,x^2 + v^2\,u^2 \\ | ||
| − | - u\,x + u^2 & = u\,v^2\,x | + | - u\,x + u^2 & = u\,v^2\,x \\ |
| − | + | - x + u & = v^2\,x \\ | |
| + | \end{align}</math> | ||
Versionen från 22 september 2012 kl. 18.53
\(\begin{align} v - {u \over u\,v + v\,x} & = {v\,x^2 \over x^2 - u^2} + {u\,v^2 \over v\,x + u\,v} \\ v - {u \over v\,(u + x)} & = {v\,x^2 \over (x+u)\,(x-u)} + {u\,v^2 \over v\,(x + u)} \\ v - {u \over v\,(x + u)} & = {v\,x^2 \over (x+u)\,(x-u)} + {u\,v^2 \over v\,(x + u)} & | \;\cdot v\,(x + u)\,(x-u)\\ v^2\,(x + u)\,(x-u) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ v^2\,(x^2-u^2) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ v^2\,x^2-v^2\,u^2 - u\,x + u^2 & = v^2\,x^2 + u\,v^2\,x - u^2\,v^2 & | \;\;\quad - v^2\,x^2 + v^2\,u^2 \\ - u\,x + u^2 & = u\,v^2\,x \\ - x + u & = v^2\,x \\ \end{align}\)
