Solution 1.3:4e
From Förberedande kurs i matematik 1
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| - | {{  | + | Because <math>5^{9} = 5^{8+1} = 5^{8}\cdot 5^{1} = 5^{8}\cdot 5</math>, the two terms inside the brackets have <math>5^{8}</math> as a common factor and can therefore be taken outside the bracket  | 
| - | <  | + | |
| - | {{  | + | {{Displayed math||<math>\begin{align}  | 
| + | \bigl(5^{8}+5^{9}\bigr)^{-1} &= \bigl(5^{8}+5^{8}\cdot 5\bigr)^{-1} = \bigl(5^{8}\cdot (1+5)\bigr)^{-1}\\[5pt]   | ||
| + | &= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}.   | ||
| + | \end{align}</math>}}  | ||
| + | |||
| + | Furthermore, <math>625 = 5\cdot 125 = 5\cdot 5\cdot 25 = 5\cdot 5\cdot 5\cdot 5 = 5^{4}</math> and we obtain  | ||
| + | |||
| + | {{Displayed math||  | ||
| + | <math>\begin{align}  | ||
| + | 625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt]  | ||
| + | &= 5^{-4}\cdot 6^{-1} = \frac{1}{5^{4}}\cdot \frac{1}{6}\\[5pt]  | ||
| + | &= \frac{1}{5^{4}\cdot 6} = \frac{1}{5\cdot 5\cdot 5\cdot 5\cdot 6}\\[5pt]   | ||
| + | &= \frac{1}{3750}\,\textrm{.}  | ||
| + | \end{align}</math>}}  | ||
Current revision
Because \displaystyle 5^{9} = 5^{8+1} = 5^{8}\cdot 5^{1} = 5^{8}\cdot 5, the two terms inside the brackets have \displaystyle 5^{8} as a common factor and can therefore be taken outside the bracket
| \displaystyle \begin{align}
 \bigl(5^{8}+5^{9}\bigr)^{-1} &= \bigl(5^{8}+5^{8}\cdot 5\bigr)^{-1} = \bigl(5^{8}\cdot (1+5)\bigr)^{-1}\\[5pt] &= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}. \end{align}  | 
Furthermore, \displaystyle 625 = 5\cdot 125 = 5\cdot 5\cdot 25 = 5\cdot 5\cdot 5\cdot 5 = 5^{4} and we obtain
| 
 \displaystyle \begin{align} 625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt] &= 5^{-4}\cdot 6^{-1} = \frac{1}{5^{4}}\cdot \frac{1}{6}\\[5pt] &= \frac{1}{5^{4}\cdot 6} = \frac{1}{5\cdot 5\cdot 5\cdot 5\cdot 6}\\[5pt] &= \frac{1}{3750}\,\textrm{.} \end{align}  | 
