Aus Online Mathematik Brückenkurs 1

Version vom 09:14, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.1:6d moved to Solution 3.1:6d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 09:50, 23. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as
-	<center> [[Image:3_1_6d-1(2).gif]] </center>	+	<math>\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}</math>
-	{{NAVCONTENT_STOP}}	+	and multiply the top and bottom of the fraction by the conjugate-like expression
-	{{NAVCONTENT_START}}	+	<math>\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}</math>
-	<center> [[Image:3_1_6d-2(2).gif]] </center>	+	Then, at least
-	{{NAVCONTENT_STOP}}	+	<math>\sqrt{6}</math>
		+	will be squared away using the conjugate rule
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}}\centerdot \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}=\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-\left( \sqrt{6} \right)^{2}} \\
		+	& =\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6} \\
		+	\end{align}</math>
		+
		+
		+	We expand the remaining quadratic,
		+	<math>\left( \sqrt{2}+\sqrt{3} \right)^{2}</math>, using the squaring rule
		+
		+
		+	<math>\begin{align}
		+	& \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{\left( \sqrt{2} \right)^{2}+2\sqrt{2}\sqrt{3}+\left( \sqrt{3} \right)^{2}-6} \\
		+	& =\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\centerdot 3}+3-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} \\
		+	\end{align}</math>
		+
		+
		+	This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate
		+	<math>2\sqrt{6}+1</math>,
		+
		+	<math>\begin{align}
		+	& \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\centerdot \frac{2\sqrt{6}+1}{2\sqrt{6}+1}=\frac{\left( \sqrt{2}+\sqrt{3}-\sqrt{6} \right)\left( 2\sqrt{6}+1 \right)}{\left( 2\sqrt{6} \right)^{2}-1^{2}} \\
		+	& =\frac{\sqrt{2}\centerdot 2\sqrt{6}+\sqrt{2}\centerdot 1+\sqrt{3}\centerdot 2\sqrt{6}+\sqrt{3}\centerdot 1-\sqrt{6}\centerdot 2\sqrt{6}-\sqrt{6}\centerdot 1}{2^{2}\left( \sqrt{6} \right)^{2}-1^{2}} \\
		+	& =\frac{\sqrt{2}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{2}+\sqrt{3}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{3}-2\left( \sqrt{6} \right)^{2}-\sqrt{6}}{4\centerdot 6-1^{2}} \\
		+	& =\frac{2\left( \sqrt{2} \right)^{2}\sqrt{3}+\sqrt{2}+2\left( \sqrt{3} \right)^{2}\sqrt{2}+\sqrt{3}-2\centerdot 6-\sqrt{6}}{24-1} \\
		+	& =\frac{2\centerdot 2\centerdot \sqrt{3}+\sqrt{2}+2\centerdot 3\centerdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23} \\
		+	& =\frac{\left( 1+2\centerdot 3 \right)\sqrt{2}+\left( 2\centerdot 2+1 \right)\sqrt{3}-12-\sqrt{6}}{23} \\
		+	& =\frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23} \\
		+	\end{align}</math>

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Version vom 09:50, 23. Sep. 2008

The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as \displaystyle \left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6} and multiply the top and bottom of the fraction by the conjugate-like expression \displaystyle \left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6} Then, at least \displaystyle \sqrt{6} will be squared away using the conjugate rule

\displaystyle \begin{align} & \frac{1}{\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}}\centerdot \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}=\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-\left( \sqrt{6} \right)^{2}} \\ & =\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6} \\ \end{align}

We expand the remaining quadratic, \displaystyle \left( \sqrt{2}+\sqrt{3} \right)^{2}, using the squaring rule

\displaystyle \begin{align} & \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{\left( \sqrt{2} \right)^{2}+2\sqrt{2}\sqrt{3}+\left( \sqrt{3} \right)^{2}-6} \\ & =\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\centerdot 3}+3-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} \\ \end{align}

This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate \displaystyle 2\sqrt{6}+1,

\displaystyle \begin{align} & \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\centerdot \frac{2\sqrt{6}+1}{2\sqrt{6}+1}=\frac{\left( \sqrt{2}+\sqrt{3}-\sqrt{6} \right)\left( 2\sqrt{6}+1 \right)}{\left( 2\sqrt{6} \right)^{2}-1^{2}} \\ & =\frac{\sqrt{2}\centerdot 2\sqrt{6}+\sqrt{2}\centerdot 1+\sqrt{3}\centerdot 2\sqrt{6}+\sqrt{3}\centerdot 1-\sqrt{6}\centerdot 2\sqrt{6}-\sqrt{6}\centerdot 1}{2^{2}\left( \sqrt{6} \right)^{2}-1^{2}} \\ & =\frac{\sqrt{2}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{2}+\sqrt{3}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{3}-2\left( \sqrt{6} \right)^{2}-\sqrt{6}}{4\centerdot 6-1^{2}} \\ & =\frac{2\left( \sqrt{2} \right)^{2}\sqrt{3}+\sqrt{2}+2\left( \sqrt{3} \right)^{2}\sqrt{2}+\sqrt{3}-2\centerdot 6-\sqrt{6}}{24-1} \\ & =\frac{2\centerdot 2\centerdot \sqrt{3}+\sqrt{2}+2\centerdot 3\centerdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23} \\ & =\frac{\left( 1+2\centerdot 3 \right)\sqrt{2}+\left( 2\centerdot 2+1 \right)\sqrt{3}-12-\sqrt{6}}{23} \\ & =\frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23} \\ \end{align}