Lösung 3.1:3b

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Aktuelle Version (10:10, 9. Aug. 2009) (bearbeiten) (rückgängig)
 
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When simplifying a root expression, a common technique is to divide up the numbers under the root sign into their smallest possible integer factors and then take out the squares and see if common factor cancel each other out or can be combined together in a new way.
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Wir zerlegen alle Zahlen im Ausdruck in ihre Primfaktoren, um den Ausdruck zu vereinfachen.
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By successively dividing by
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Durch mehrfache Division mit 2 und 3 erhalten wir
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<math>2</math>
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and
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<math>3</math>, we see that
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{{Abgesetzte Formel||<math>\begin{align}
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96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\
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&= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt]
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18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}.
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\end{align}</math>}}
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<math>\begin{align}
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Also
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& 96=2\centerdot 48=2\centerdot 2\centerdot 24=2\centerdot 2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 2\centerdot 6 \\
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& =2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{5}\centerdot 3, \\
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& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2}. \\
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\end{align}</math>
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{{Abgesetzte Formel||<math>\begin{align}
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\sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt]
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\sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,.
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\end{align}</math>}}
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Thus,
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Für den Bruch gilt also
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{{Abgesetzte Formel||<math>\frac{\sqrt{96}}{\sqrt{18}} = \frac{2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}}{3\cdot \sqrt{2}} = \frac{4\sqrt{3}}{3}\,\textrm{.}</math>}}
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<math>\begin{align}
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Hinweis: Wir können auch mit Potenzen rechnen
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& \sqrt{96}=\sqrt{2^{5}\centerdot 3}=\sqrt{2^{2}\centerdot 2^{2}\centerdot 2\centerdot 3}=2\centerdot 2\centerdot \sqrt{2}\centerdot \sqrt{3} \\
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& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\centerdot \sqrt{2} \\
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\end{align}</math>
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{{Abgesetzte Formel||<math>\begin{align}
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and the whole quotient can be written as
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\frac{\sqrt{96}}{\sqrt{18}}
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&= \frac{96^{1/2}}{18^{1/2}}
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= \frac{(2^{5}\cdot 3)^{1/2}}{(2\cdot 3^{2})^{1/2}}
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<math>\frac{\sqrt{96}}{\sqrt{18}}=\frac{2\centerdot 2\centerdot \sqrt{2}\centerdot \sqrt{3}}{3\centerdot \sqrt{2}}=\frac{4\sqrt{3}}{3}</math>
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= \frac{2^{5\cdot\frac{1}{2}}\cdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\cdot 3^{2\cdot \frac{1}{2}}}\\[5pt]
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&= 2^{\frac{5}{2}-\frac{1}{2}}\cdot 3^{\frac{1}{2}-1}
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= 2^{2}\cdot 3^{-\frac{1}{2}}
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NOTE: If it is difficult to work with the root sign, it is possible instead to write everything in power form
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= \frac{4}{\sqrt{3}}
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= \frac{4\sqrt{3}}{3}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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& \frac{\sqrt{96}}{\sqrt{18}}=\frac{\left( 96 \right)^{\frac{1}{2}}}{\left( 18 \right)^{{1}/{2}\;}}=\frac{\left( 2^{5}\centerdot 3 \right)^{\frac{1}{2}}}{\left( 2\centerdot 3^{2} \right)^{{1}/{2}\;}}=\frac{2^{5\centerdot \frac{1}{2}}\centerdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\centerdot 3^{2\centerdot \frac{1}{2}}} \\
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& =2^{\frac{5}{2}-\frac{1}{2}}\centerdot 3^{\frac{1}{2}-1}=2^{2}\centerdot 3^{-\frac{1}{2}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3} \\
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\end{align}</math>
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(in the last line, we multiply top and bottom by
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<math>\sqrt{3}</math>
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).
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Aktuelle Version

Wir zerlegen alle Zahlen im Ausdruck in ihre Primfaktoren, um den Ausdruck zu vereinfachen.

Durch mehrfache Division mit 2 und 3 erhalten wir

\displaystyle \begin{align}

96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\ &= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt] 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}. \end{align}

Also

\displaystyle \begin{align}

\sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt] \sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,. \end{align}

Für den Bruch gilt also

\displaystyle \frac{\sqrt{96}}{\sqrt{18}} = \frac{2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}}{3\cdot \sqrt{2}} = \frac{4\sqrt{3}}{3}\,\textrm{.}

Hinweis: Wir können auch mit Potenzen rechnen

\displaystyle \begin{align}

\frac{\sqrt{96}}{\sqrt{18}} &= \frac{96^{1/2}}{18^{1/2}} = \frac{(2^{5}\cdot 3)^{1/2}}{(2\cdot 3^{2})^{1/2}} = \frac{2^{5\cdot\frac{1}{2}}\cdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\cdot 3^{2\cdot \frac{1}{2}}}\\[5pt] &= 2^{\frac{5}{2}-\frac{1}{2}}\cdot 3^{\frac{1}{2}-1} = 2^{2}\cdot 3^{-\frac{1}{2}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:3b“