Solution 1.3:1d
From Förberedande kurs i matematik 1
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| - | {{ | + | By using the power rules, we can rewrite the expression, |
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| - | {{ | + | {{Displayed math||<math>\left( \frac{2}{3} \right)^{-3} = \frac{2^{-3}}{3^{-3}} = \frac{\,\dfrac{1}{2^{3}}\,}{\,\dfrac{1}{3^{3}}\,} = \frac{\,\dfrac{1}{2^{3}}\cdot 3^{3}\,}{\,\dfrac{1}{\rlap{\,/}3^{3}}\cdot {}\rlap{\,/}3^{3}\,} = \frac{\,\dfrac{3^{3}}{2^{3}}\,}{1} = \frac{3^{3}}{2^{3}}\,,</math>}} |
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| + | and then carry out the calculation | ||
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| + | {{Displayed math||<math>\frac{3^{3}}{2^{3}} = \frac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2} = \frac{27}{8}\,</math>.}} | ||
Current revision
By using the power rules, we can rewrite the expression,
| \displaystyle \left( \frac{2}{3} \right)^{-3} = \frac{2^{-3}}{3^{-3}} = \frac{\,\dfrac{1}{2^{3}}\,}{\,\dfrac{1}{3^{3}}\,} = \frac{\,\dfrac{1}{2^{3}}\cdot 3^{3}\,}{\,\dfrac{1}{\rlap{\,/}3^{3}}\cdot {}\rlap{\,/}3^{3}\,} = \frac{\,\dfrac{3^{3}}{2^{3}}\,}{1} = \frac{3^{3}}{2^{3}}\,, |
and then carry out the calculation
| \displaystyle \frac{3^{3}}{2^{3}} = \frac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2} = \frac{27}{8}\,. |
