Skillnad mellan versioner av "1.5 Lösning 1d"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<math> {(x^{-2})^6 \cdot \sqrt{y} \over y^{0,5} \cdot (x^{-4})^3} = {x^{-12} \cdot \sqrt{y} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over ...") |
Taifun (Diskussion | bidrag) m |
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| − | <math> {(x^{-2})^6 \cdot \sqrt{y} \over y^{0,5} \cdot (x^{-4})^3} = {x^{-12} \cdot \sqrt{y} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot x^{-12}} = 1 </math> | + | :<math> {(x^{-2})^6 \cdot \sqrt{y} \over y^{0,5} \cdot (x^{-4})^3} = {x^{-12} \cdot \sqrt{y} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot x^{-12}} = 1 </math> |
Nuvarande version från 24 mars 2015 kl. 23.16
\[ {(x^{-2})^6 \cdot \sqrt{y} \over y^{0,5} \cdot (x^{-4})^3} = {x^{-12} \cdot \sqrt{y} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot (x^{-4})^3} = {x^{-12} \cdot y^{1 \over 2} \over y^{1 \over 2} \cdot x^{-12}} = 1 \]
