Lösung 3.1:4c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.1:4c moved to Solution 3.1:4c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
-
<center> [[Image:3_1_4c.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\begin{align}
 +
& 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\
 +
& 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\
 +
& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\
 +
& 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\
 +
\end{align}</math>
 +
 
 +
and then taking the squares out from under the root sign.
 +
 
 +
 
 +
<math>\begin{align}
 +
& \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\
 +
& \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\
 +
& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\
 +
& \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\
 +
\end{align}</math>
 +
 
 +
 
 +
All together, we get
 +
 
 +
 
 +
<math>\begin{align}
 +
& \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\
 +
& =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\
 +
& =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\
 +
& =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\
 +
\end{align}</math>

Version vom 14:11, 22. Sep. 2008

Each term in the expression can be simplified by breaking down the number under the root sign into its factors,


\displaystyle \begin{align} & 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\ & 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\ & 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\ & 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\ \end{align}

and then taking the squares out from under the root sign.


\displaystyle \begin{align} & \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\ & \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\ & \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\ & \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\ \end{align}


All together, we get


\displaystyle \begin{align} & \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\ & =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\ & =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\ & =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\ \end{align}

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