Skillnad mellan versioner av "1.5 Lösning 1b"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<math> {2\,x^{-5} \over 3\,x^{-8}} \cdot (2\,x)^{-1} = {2\,x^{-5-(-8)} \over 3} \cdot (2\,x)^{-1} = {2\,x^{-5+8} \over 3} \cdot (2\,x)^{-1} = {2\,x^3 \over 3} \cdot (2\,x)^{-1} =...") |
Taifun (Diskussion | bidrag) m |
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| − | <math> {2\,x^{-5} \over 3\,x^{-8}} \cdot (2\,x)^{-1} = {2\,x^{-5-(-8)} \over 3} \cdot (2\,x)^{-1} = {2\,x^{-5+8} \over 3} \cdot (2\,x)^{-1} = | + | <math> {2\,x^{-5} \over 3\,x^{-8}} \cdot (2\,x)^{-1} = {2\,x^{-5-(-8)} \over 3} \cdot (2\,x)^{-1} = {2\,x^{-5+8} \over 3} \cdot (2\,x)^{-1} = </math> |
| − | <math> = {2\,x^3 \over 3} \cdot {1 \over 2\,x} = {2\,x^3 \cdot 1 \over 3 \cdot 2\,x} = {x^2 \over 3}</math> | + | <math> = {2\,x^3 \over 3} \cdot (2\,x)^{-1} = {2\,x^3 \over 3} \cdot {1 \over 2\,x} = {2\,x^3 \cdot 1 \over 3 \cdot 2\,x} = {x^2 \over 3}</math> |
Versionen från 9 mars 2011 kl. 22.08
\( {2\,x^{-5} \over 3\,x^{-8}} \cdot (2\,x)^{-1} = {2\,x^{-5-(-8)} \over 3} \cdot (2\,x)^{-1} = {2\,x^{-5+8} \over 3} \cdot (2\,x)^{-1} = \)
\( = {2\,x^3 \over 3} \cdot (2\,x)^{-1} = {2\,x^3 \over 3} \cdot {1 \over 2\,x} = {2\,x^3 \cdot 1 \over 3 \cdot 2\,x} = {x^2 \over 3}\)
