Lösung 2.1:5a - Versionsgeschichte

Sekretariat1 um 11:28, 27. Aug. 2009

2009-08-27T11:28:50Z

← Nächstältere Version		Version vom 11:28, 27. Aug. 2009
Zeile 10:		Zeile 10:
	\end{align}</math>}}		\end{align}</math>}}

-	~~also~~ ist	+	Also ist

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
Zeile 16:		Zeile 16:
	Schreiben wir die Wurzeln als Potenzen, erhalten wir		Schreiben wir die Wurzeln als Potenzen, erhalten wir

-	{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,</math>}}

	mit der Stammfunktion:		mit der Stammfunktion:

Moni um 22:48, 21. Aug. 2009

2009-08-21T22:48:40Z

← Nächstältere Version		Version vom 22:48, 21. Aug. 2009
Zeile 1:		Zeile 1:
	Wir erweitern den Bruch mit den Konjugat		Wir erweitern den Bruch mit den Konjugat
-	<math>\sqrt{x+9}+\sqrt{x}</math>, und mit der binomischen Formel erhalten wir	+	<math>\sqrt{x+9}+\sqrt{x}</math> und mit der binomischen Formel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 10:		Zeile 10:
	\end{align}</math>}}		\end{align}</math>}}

-	~~und~~ also ist,	+	also ist

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}

-	Schreiben wir die Wurzeln ~~wir~~ Potenzen erhalten wir	+	Schreiben wir die Wurzeln als Potenzen, erhalten wir

	{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
Zeile 28:		Zeile 28:
	\end{align}</math>}}		\end{align}</math>}}

-	wo C eine beliebige Konstante ist.	+	wobei C eine beliebige Konstante ist.

	Dies ist dasselbe wie		Dies ist dasselbe wie
Zeile 35:		Zeile 35:


-	Hinweis: Um zu testen ob wir richtig gerechnet haben, leiten wir unsere Funktion ab, und vergleichen mit ~~den~~ Integrand.	+	Hinweis: Um zu testen, ob wir richtig gerechnet haben, leiten wir unsere Funktion ab und vergleichen mit dem Integrand.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Martin um 19:51, 28. Apr. 2009

2009-04-28T19:51:08Z

← Nächstältere Version		Version vom 19:51, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If we multiply top and bottom of the fraction by the conjugate expression~~	+	Wir erweitern den Bruch mit den Konjugat
-	<math>\sqrt{x+9}+\sqrt{x}</math> ~~then the formula for the difference of two squares gives that denominator's root is squared away~~,	+	<math>\sqrt{x+9}+\sqrt{x}</math>, und mit der binomischen Formel erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 10:		Zeile 10:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Thus~~,	+	und also ist,

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}

-	~~If we write the square roots in power form,~~	+	Schreiben wir die Wurzeln wir Potenzen erhalten wir

	{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}

-	~~we see that we have a standard integral and can write down the primitive functions directly,~~	+	mit der Stammfunktion:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 28:		Zeile 28:
	\end{align}</math>}}		\end{align}</math>}}

-	~~where~~ C ~~is an arbitrary constant~~.	+	wo C eine beliebige Konstante ist.

-	~~This can also be written with square roots as~~	+	Dies ist dasselbe wie

	{{Abgesetzte Formel\|\|<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}


-	~~Note~~: ~~To be completely certain that we have done everything correctly~~, ~~we differentiate the answer and see if we get back the integrand~~,	+	Hinweis: Um zu testen ob wir richtig gerechnet haben, leiten wir unsere Funktion ab, und vergleichen mit den Integrand.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: Solution 2.1:5a moved to Lösung 2.1:5a: Robot: moved page

2009-03-11T10:19:35Z

Solution 2.1:5a moved to Lösung 2.1:5a: Robot: moved page

← Nächstältere Version

Version vom 10:19, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2009-03-10T13:00:14Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 13:00, 10. Mär. 2009
Zeile 2:		Zeile 2:
	<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,		<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1}{\sqrt{x+9}-\sqrt{x}}		\frac{1}{\sqrt{x+9}-\sqrt{x}}
	&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]		&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
Zeile 12:		Zeile 12:
	Thus,		Thus,

-	{{~~Displayed math~~\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}

	If we write the square roots in power form,		If we write the square roots in power form,

-	{{~~Displayed math~~\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}

	we see that we have a standard integral and can write down the primitive functions directly,		we see that we have a standard integral and can write down the primitive functions directly,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx		\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
	&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]		&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
Zeile 32:		Zeile 32:
	This can also be written with square roots as		This can also be written with square roots as

-	{{~~Displayed math~~\|\|<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}


	Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,		Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)		\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
	&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]		&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
	&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}		&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 09:36, 28. Okt. 2008

2008-10-28T09:36:43Z

← Nächstältere Version		Version vom 09:36, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	~~(HINT: multiply the top and bottom by the conjugate of the denominator.)~~	+	If we multiply top and bottom of the fraction by the conjugate expression
-		+	<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
-	If we multiply top and bottom of the fraction by the conjugate expression,	+
-	<math>\sqrt{x+9}+\sqrt{x}</math>, then the ~~conjugate rule~~ gives that denominator's root is squared away:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \frac{1}{\sqrt{x+9}-\sqrt{x}}=\frac{1}{\sqrt{x+9}-\sqrt{x}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\~~	+
-	~~& =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\~~	+
-	~~& =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\~~	+
-	~~& =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1}{\sqrt{x+9}-\sqrt{x}}
		+	&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
		+	&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt]
		+	&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt]
		+	&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.}
		+	\end{align}</math>}}

	Thus,		Thus,

-		+	{{Displayed math\|\|<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
-	<math>\int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\~~left~~( \sqrt{x+9}+\sqrt{x} \~~right~~)}\,dx</math>	+
-		+

	If we write the square roots in power form,		If we write the square roots in power form,

		+	{{Displayed math\|\|<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}

-	~~<math>\frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx</math>~~,	+	we see that we have a standard integral and can write down the primitive functions directly,

-	~~we see that we have a standard integral and can write down the primitive functions directly:~~	+	{{Displayed math\|\|<math>\begin{align}
-		+	\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
-		+	&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
-	<math>\begin{align}	+	&= \frac{1}{9}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt]
-	& \frac{1}{9}\int{\~~left~~( ~~\left~~( x+9 ~~\right~~)^~~{\frac~~{1}{2}}+x^~~{\frac~~{1}{2}} \~~right~~)}\,dx=\frac{1}{9}\~~left~~( \frac{~~\left~~( x+9 ~~\right~~)^~~{\frac~~{1}{2}+1}}{\~~frac~~{1}{2}+1}+\frac{x^~~{\frac~~{1}{2}+1}}{\~~frac~~{1}{2}+1} \~~right~~)+C \\	+	&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt]
-	& =\frac{1}{9}\~~left~~( \frac{~~\left~~( x+9 ~~\right~~)^~~{\frac~~{3}{2}}~~}{\frac~~{3}{2}}+\frac{x^~~{\frac~~{3}{2}}}~~{\frac~~{3}{2}} \~~right~~)+C \\	+	&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
-	& =\frac{1}{9}\~~left~~( \frac{2}{3}~~\left~~( x+9 ~~\right~~)^~~{\frac~~{3}{2}}+\frac{2}{3}x^~~{\frac~~{3}{2}} \~~right~~)+C \\	+	\end{align}</math>}}
-	& =\frac{2}{27}~~\left~~( x+9 ~~\right~~)^~~{\frac~~{3}{2}}+\frac{2}{27}x^~~{\frac~~{3}{2}}+C \\	+
-	~~& \\~~	+
-	\end{align}</math>,	+

	where C is an arbitrary constant.		where C is an arbitrary constant.
Zeile 39:		Zeile 32:
	This can also be written with square roots as		This can also be written with square roots as

		+	{{Displayed math\|\|<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}

-	<math>\frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C</math>
-
-
-	To be completely certain that we have everything correctly, we differentiate the answer and see if we get back the integrand:

		+	Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{d}{dx}\~~left~~( \frac{2}{27}~~\left~~( x+9 ~~\right~~)^~~{\frac~~{3}{2}}+\frac{2}{27}x^~~{\frac~~{3}{2}}+C \~~right~~) \\	+	\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
-	& =\frac{2}{27}\~~centerdot~~ \frac{3}{2}~~\left~~( x+9 ~~\right~~)^~~{\frac~~{3}{2}-1}+\frac{2}{27}\~~centerdot~~ \frac{3}{2}x^~~{\frac~~{3}{2}-1}+0 \\	+	&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
-	& =\frac{1}{9}~~\left~~( x+9 ~~\right~~)^~~{\frac~~{1}{2}}+\frac{1}{9}x^~~{\frac~~{1}{2}} \\	+	&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 13:29, 18. Okt. 2008

2008-10-18T13:29:46Z

← Nächstältere Version		Version vom 13:29, 18. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	(HINT: multiply the top and bottom by the conjugate of the denominator.)
-	<~~center~~> ~~[[Image~~:~~2_1_5a~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we multiply top and bottom of the fraction by the conjugate expression,
-	{{~~NAVCONTENT_START~~}}	+	<math>\sqrt{x+9}+\sqrt{x}</math>, then the conjugate rule gives that denominator's root is squared away:
-	<~~center~~> ~~[[Image~~:~~2_1_5a-~~2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& \frac{1}{\sqrt{x+9}-\sqrt{x}}=\frac{1}{\sqrt{x+9}-\sqrt{x}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\
		+	\end{align}</math>
		+
		+
		+	Thus,
		+
		+
		+	<math>\int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\left( \sqrt{x+9}+\sqrt{x} \right)}\,dx</math>
		+
		+
		+	If we write the square roots in power form,
		+
		+
		+	<math>\frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx</math>,
		+
		+	we see that we have a standard integral and can write down the primitive functions directly:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx=\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right)+C \\
		+	& =\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right)+C \\
		+	& =\frac{1}{9}\left( \frac{2}{3}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{3}{2}} \right)+C \\
		+	& =\frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \\
		+	& \\
		+	\end{align}</math>,
		+
		+	where C is an arbitrary constant.
		+
		+	This can also be written with square roots as
		+
		+
		+	<math>\frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C</math>
		+
		+
		+	To be completely certain that we have everything correctly, we differentiate the answer and see if we get back the integrand:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\left( \frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \right) \\
		+	& =\frac{2}{27}\centerdot \frac{3}{2}\left( x+9 \right)^{\frac{3}{2}-1}+\frac{2}{27}\centerdot \frac{3}{2}x^{\frac{3}{2}-1}+0 \\
		+	& =\frac{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\
		+	\end{align}</math>

Tekbot: Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page

2008-09-16T14:40:01Z

Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page

← Nächstältere Version

Version vom 14:40, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T07:59:44Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

← Nächstältere Version		Version vom 07:59, 21. Aug. 2008
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2008-04-07T08:24:03Z

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