Skillnad mellan versioner av "1.4 Lösning 9b"

Från Mathonline
Hoppa till: navigering, sök
m (Created page with "<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) \, = \...")
 
m
Rad 1: Rad 1:
 +
<math> \left({a^2 - 6\,a + 9 \over b^6}\right)\, \Bigg / \,\left({a - 3 \over b^5}\right) \, = \, \left({a^2 - 6\,a + 9 \over b^6}\right)\, \cdot  \,\left({b^5 \over a - 3}\right) \, = \, </math>
 +
 +
 
<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>
 
<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>
  
  
 
<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>
 
<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>

Versionen från 20 september 2012 kl. 23.03

\( \left({a^2 - 6\,a + 9 \over b^6}\right)\, \Bigg / \,\left({a - 3 \over b^5}\right) \, = \, \left({a^2 - 6\,a + 9 \over b^6}\right)\, \cdot \,\left({b^5 \over a - 3}\right) \, = \, \)


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


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