Skillnad mellan versioner av "1.4 Lösning 8c"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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<math> \begin{align} {15\cdot 2^2 - 2\cdot 2 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 2\cdot 2^2 \over 3} - {z \over 2} \\ | <math> \begin{align} {15\cdot 2^2 - 2\cdot 2 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 2\cdot 2^2 \over 3} - {z \over 2} \\ | ||
| − | {15\cdot 4 - 4 - 6 \over 6}& = {2 - 3\,z \over 2} - {z - 2\cdot 4 \over 3} - {z \over 2} \\ | + | {15\cdot 4 - 4 - 6 \over 6}& = {2 - 3\,z \over 2} - {z - 2\cdot 4 \over 3} - {z \over 2} \\ |
| − | {60 - 4 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ | + | {60 - 4 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ |
| − | {50 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ | + | {50 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ |
| − | {25 \over 3} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \quad & &\,| \; \cdot 6\\ | + | {25 \over 3} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \quad & &\,| \; \cdot 6 \\ |
| − | 50 & = 3\,(2 - 3\,z) - 2\,(z - 8) - 3\,z \ | + | 50 & = 3\,(2 - 3\,z) - 2\,(z - 8) - 3\,z \\ |
| + | 50 & = 6 - 9\,z) - 2\,z + 16 - 3\,z \\ | ||
\end{align} </math> | \end{align} </math> | ||
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Versionen från 20 september 2012 kl. 15.28
Vi sätter in \( x=2\, \) i ekvationen\[ \begin{align} {15\cdot 2^2 - 2\cdot 2 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 2\cdot 2^2 \over 3} - {z \over 2} \\ {15\cdot 4 - 4 - 6 \over 6}& = {2 - 3\,z \over 2} - {z - 2\cdot 4 \over 3} - {z \over 2} \\ {60 - 4 - 6 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ {50 \over 6} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \\ {25 \over 3} & = {2 - 3\,z \over 2} - {z - 8 \over 3} - {z \over 2} \quad & &\,| \; \cdot 6 \\ 50 & = 3\,(2 - 3\,z) - 2\,(z - 8) - 3\,z \\ 50 & = 6 - 9\,z) - 2\,z + 16 - 3\,z \\ \end{align} \]
