Skillnad mellan versioner av "1.5a Lösning 10a"
Från Mathonline
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
| − | <math> f(x) = {3\,x^2 + 6\,x + 12 \over x^2 - 4} = {3\,(x^2 + 2\,x + 4) \over (x + 2)\,(x - 2)} = </math> | + | <math> f(x) = {3\,x^2 + 6\,x + 12 \over x^2 - 4} = {3\,(x^2 + 2\,x + 4) \over (x + 2)\,(x - 2)} = {3\,(x + 2)^2 \over (x + 2)\,(x - 2)} = </math> |
| − | <math> = {3\,(x + 2) | + | <math> = {3\,{\color{Red}(x + 2)}\,(x + 2) \over {\color{Red}(x + 2)}\,(x - 2)} = </math> |
<math> {x^2 - 9 \over x-3} = {(x+3)\,{\color{Red} (x-3)} \over {\color{Red} (x-3)}} = x+3 </math> | <math> {x^2 - 9 \over x-3} = {(x+3)\,{\color{Red} (x-3)} \over {\color{Red} (x-3)}} = x+3 </math> | ||
Versionen från 16 juli 2014 kl. 23.27
\( f(x) = {3\,x^2 + 6\,x + 12 \over x^2 - 4} = {3\,(x^2 + 2\,x + 4) \over (x + 2)\,(x - 2)} = {3\,(x + 2)^2 \over (x + 2)\,(x - 2)} = \)
\( = {3\,{\color{Red}(x + 2)}\,(x + 2) \over {\color{Red}(x + 2)}\,(x - 2)} = \)
\( {x^2 - 9 \over x-3} = {(x+3)\,{\color{Red} (x-3)} \over {\color{Red} (x-3)}} = x+3 \)
