Skillnad mellan versioner av "1.4 Lösning 6c"

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<math> \left({2\,a - 4 \over a^2}\right)\, \Bigg / \,\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>
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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)\, \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>
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<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. 22.39

\( \left({2\,a - 4 \over a^2}\right)\, \Big / \,\left({a^2 - 4 \over a^4}\right) \, = \, \left({2\,a - 4 \over a^2}\right)\, \cdot \,\left({a^4 \over a^2 - 4}\right) \, = \, {(2\,a - 4) \cdot a^4 \over a^2 \cdot (a^2 - 4)} \, = \)


\( = \; {(2\,a - 4) \cdot a^2 \over (a^2 - 4)} \; = \; {2\,(a - 2) \cdot a^2 \over (a + 2) \cdot (a-2)} \; = \; {2\,a^2 \over (a + 2)} \)