From Förberedande kurs i matematik 1

Revision as of 14:03, 1 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (10:54, 30 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	By writing <math>0\textrm{.}027</math> as <math>27\cdot 10^{-3}</math>, where
-	<center> [[Bild:3_1_4b.gif]] </center>	+	<math>27 = 3\cdot 3\cdot 3 = 3^3</math> and <math>10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3</math> we see that
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\begin{align}
		+	\sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt]
		+	&= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,}
		+	\end{align}</math>}}
		+
		+	where we have used <math>\sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}</math>

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Current revision

By writing \displaystyle 0\textrm{.}027 as \displaystyle 27\cdot 10^{-3}, where \displaystyle 27 = 3\cdot 3\cdot 3 = 3^3 and \displaystyle 10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3 we see that

\displaystyle \begin{align} \sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt] &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} \end{align}

where we have used \displaystyle \sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}