Lösung 3.1:4d

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We start by factorizing the numbers under the root sign,
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<center> [[Image:3_1_4d.gif]] </center>
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<math>\begin{align}
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& 48=2\centerdot 24=2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 6=2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{4}\centerdot 3, \\
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& 12=2\centerdot 6=2\centerdot 2\centerdot 3=2^{2}\centerdot 3, \\
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& 3=3, \\
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& 75=3\centerdot 25=3\centerdot 5\centerdot 5=3\centerdot 5^{2} \\
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\end{align}</math>
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Now, we can take the squares out from under the root signs,
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<math>\begin{align}
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& \sqrt{48}=\sqrt{2^{4}\centerdot 3}=2^{2}\sqrt{3}=4\sqrt{3} \\
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& \sqrt{12}=\sqrt{2^{2}\centerdot 3}=2\sqrt{3} \\
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& \sqrt{3}=\sqrt{3} \\
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& \sqrt{75}=\sqrt{3\centerdot 5^{2}}=5\sqrt{3} \\
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\end{align}</math>
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and then simplify the whole expression:
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<math>\begin{align}
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& \sqrt{48}+\sqrt{12}+\sqrt{3}-\sqrt{75}=4\sqrt{3}+2\sqrt{3}+\sqrt{3}-5\sqrt{3} \\
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& =\left( 4+2+1-5 \right)\sqrt{3}=2\sqrt{3} \\
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\end{align}</math>

Version vom 14:20, 22. Sep. 2008

We start by factorizing the numbers under the root sign,


\displaystyle \begin{align} & 48=2\centerdot 24=2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 6=2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{4}\centerdot 3, \\ & 12=2\centerdot 6=2\centerdot 2\centerdot 3=2^{2}\centerdot 3, \\ & 3=3, \\ & 75=3\centerdot 25=3\centerdot 5\centerdot 5=3\centerdot 5^{2} \\ \end{align}

Now, we can take the squares out from under the root signs,


\displaystyle \begin{align} & \sqrt{48}=\sqrt{2^{4}\centerdot 3}=2^{2}\sqrt{3}=4\sqrt{3} \\ & \sqrt{12}=\sqrt{2^{2}\centerdot 3}=2\sqrt{3} \\ & \sqrt{3}=\sqrt{3} \\ & \sqrt{75}=\sqrt{3\centerdot 5^{2}}=5\sqrt{3} \\ \end{align}


and then simplify the whole expression:


\displaystyle \begin{align} & \sqrt{48}+\sqrt{12}+\sqrt{3}-\sqrt{75}=4\sqrt{3}+2\sqrt{3}+\sqrt{3}-5\sqrt{3} \\ & =\left( 4+2+1-5 \right)\sqrt{3}=2\sqrt{3} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:4d“