Skillnad mellan versioner av "1.4 Lösning 9b"

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<math> = \, {(a-3)^2 \over b^6}\, \cdot  \,{b^5 \over a - 3} \, = \, {(a-3)^2 \cdot b^5 \over b^6 \cdot (a - 3)} \, = {a-3 \over b} </math>
 
<math> = \, {(a-3)^2 \over b^6}\, \cdot  \,{b^5 \over a - 3} \, = \, {(a-3)^2 \cdot b^5 \over b^6 \cdot (a - 3)} \, = {a-3 \over b} </math>
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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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<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. 23.17

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


\( = \, {(a-3)^2 \over b^6}\, \cdot \,{b^5 \over a - 3} \, = \, {(a-3)^2 \cdot b^5 \over b^6 \cdot (a - 3)} \, = {a-3 \over b} \)

Hämtad från "https://matte5.mathonline.se/index.php?title=1.4_Lösning_9b&oldid=6500"