Lösung 3.1:6a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.1:6a moved to Solution 3.1:6a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
We use the standard method and augment the fraction with the conjugate of the denominator
-
<center> [[Image:3_1_6a.gif]] </center>
+
<math>\sqrt{5}+2</math>. Then the conjugate rule gives
-
{{NAVCONTENT_STOP}}
+
 
 +
 
 +
<math>\begin{align}
 +
& \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\
 +
& =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\
 +
& =6+2\sqrt{2}+3\sqrt{5}+10 \\
 +
\end{align}</math>

Version vom 08:50, 23. Sep. 2008

We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the conjugate rule gives


\displaystyle \begin{align} & \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\ & =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\ & =6+2\sqrt{2}+3\sqrt{5}+10 \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:6a“