Skillnad mellan versioner av "1.5 Lösning 4c"
Från Mathonline
Taifun (Diskussion | bidrag) m (Created page with "<math> {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5...") |
Taifun (Diskussion | bidrag) m |
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| − | <math> {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5\,z/3)}} = </math> | + | :<math> {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5\,z/3)}} = </math> |
| − | <math> = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2+3\,z} \over 3^{5\,z}} = {3^{5\,z+2} \over 3^{5\,z}} = {3^{5\,z+2-5\,z}} = 3^2 = 9 </math> | + | :<math> = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2+3\,z} \over 3^{5\,z}} = {3^{5\,z+2} \over 3^{5\,z}} = {3^{5\,z+2-5\,z}} = 3^2 = 9 </math> |
Nuvarande version från 7 juli 2015 kl. 16.06
\[ {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5\,z/3)}} = \]
\[ = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2+3\,z} \over 3^{5\,z}} = {3^{5\,z+2} \over 3^{5\,z}} = {3^{5\,z+2-5\,z}} = 3^2 = 9 \]
