Skillnad mellan versioner av "1.4 Lösning 7c"

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<big><big><math> {1\,-\,x\,y \over (x\,y)^2\,-\,x\,y} \; = \; {1\,-\,x\,y \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1)\,\cdot\,(-1\,+\,x\,y) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; </math></big></big>
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<math> \displaystyle{{1\,-\,x\,y \over (x\,y)^2\,-\,x\,y} \; = \; {1\,-\,x\,y \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1)\,\cdot\,(-1\,+\,x\,y) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \;} </math>
  
  
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<big><big><math> = \; {(-1)\,\cdot\,(x\,y\,-\,1) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1) \over x\,y} \; = \; - {1 \over x\,y} </math></big></big>
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<math> \displaystyle{= \; {(-1)\,\cdot\,(x\,y\,-\,1) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1) \over x\,y} \; = \; - {1 \over x\,y} }</math>

Nuvarande version från 17 september 2015 kl. 11.03

\( \displaystyle{{1\,-\,x\,y \over (x\,y)^2\,-\,x\,y} \; = \; {1\,-\,x\,y \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1)\,\cdot\,(-1\,+\,x\,y) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \;} \)


\( \displaystyle{= \; {(-1)\,\cdot\,(x\,y\,-\,1) \over x\,y\,\cdot\,(x\,y\,-\,1)} \; = \; {(-1) \over x\,y} \; = \; - {1 \over x\,y} }\)

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