From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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
-	<center> [[Bild:1_3_4e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>}}

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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}