Skillnad mellan versioner av "1.4 Lösning 6c"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| − | <big><big><math> \left({2\,a - 4 \over a^2}\right)\, \Big / \,\left({a^2 - 4 \over a^4}\right) \, = \, \left({2\,a - 4 \over a^2}\right) | + | <big><big><math> \left({2\,a\,-\,4 \over a^2}\right)\, \Big / \,\left({a^2\,-\,4 \over a^4}\right) \, = \, \left({2\,a\,-\,4 \over a^2}\right) \cdot \left({a^4 \over a^2\,-\,4}\right) \, = \, {(2\,a\,-\,4) \cdot a^4 \over a^2 \cdot (a^2\,-\,4)} \, = </math> |
| − | <math> = \; {(2\,a - 4) \cdot a^2 \over (a^2 - 4)} \; = \; {2\,(a - 2) \cdot a^2 \over (a + 2) \cdot (a-2)} \; = \; {2\,a^2 \over (a + 2)} </math></big></big> | + | <math> = \; {(2\,a\,-\,4) \cdot a^2 \over (a^2\,-\,4)} \; = \; {2\,(a\,-\,2) \cdot a^2 \over (a\,+\,2) \cdot (a\,-\,2)} \; = \; {2\,a^2 \over (a\,+\,2)} </math></big></big> |
Versionen från 2 augusti 2014 kl. 22.43
\( \left({2\,a\,-\,4 \over a^2}\right)\, \Big / \,\left({a^2\,-\,4 \over a^4}\right) \, = \, \left({2\,a\,-\,4 \over a^2}\right) \cdot \left({a^4 \over a^2\,-\,4}\right) \, = \, {(2\,a\,-\,4) \cdot a^4 \over a^2 \cdot (a^2\,-\,4)} \, = \)
\( = \; {(2\,a\,-\,4) \cdot a^2 \over (a^2\,-\,4)} \; = \; {2\,(a\,-\,2) \cdot a^2 \over (a\,+\,2) \cdot (a\,-\,2)} \; = \; {2\,a^2 \over (a\,+\,2)} \)
