Lösung 3.4:1d - Versionsgeschichte

Sekretariat1 um 11:35, 4. Sep. 2009

Sekretariat1 — Fri, 04 Sep 2009 11:35:41 GMT

← Nächstältere Version		Version vom 11:35, 4. Sep. 2009
Zeile 1:		Zeile 1:
-	Wir beginnen damit <math>x^2</math> zu addieren und subtrahieren, sodass wir <math>x^3+x^2 = x^2(x+1)</math> im Zähler erhalten,	+	Wir beginnen damit, <math>x^2</math> zu addieren und subtrahieren, sodass wir <math>x^3+x^2 = x^2(x+1)</math> im Zähler erhalten

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

	Danach addieren und subtrahieren wir		Danach addieren und subtrahieren wir
-	<math>-x</math>, sodass wir <math>-x^2-x = -x(x+1)</math> erhalten, ~~nachdem~~ dies durch	+	<math>-x</math>, sodass wir <math>-x^2-x = -x(x+1)</math> erhalten, da dies durch
-	<math>x+1</math> teilbar ist,	+	<math>x+1</math> teilbar ist

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 25:		Zeile 25:
	{{Abgesetzte Formel\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}

-	Wir testen ob	+	Wir testen, ob

-	{{Abgesetzte Formel\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{,}</math>}}

-	indem wir kontrollieren ob	+	indem wir kontrollieren, ob

-	{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}	+	{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math> .}}

-	Wir erweitern die rechte Seite und sehen, dass alles stimmt,	+	Wir erweitern die rechte Seite und sehen, dass alles stimmt

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}		(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Moni um 19:07, 23. Aug. 2009

Moni — Sun, 23 Aug 2009 19:07:17 GMT

← Nächstältere Version		Version vom 19:07, 23. Aug. 2009
Zeile 10:		Zeile 10:

	Danach addieren und subtrahieren wir		Danach addieren und subtrahieren wir
-	<math>-x</math> sodass wir <math>-x^2-x = -x(x+1)</math> erhalten, nachdem dies durch	+	<math>-x</math>, sodass wir <math>-x^2-x = -x(x+1)</math> erhalten, nachdem dies durch
	<math>x+1</math> teilbar ist,		<math>x+1</math> teilbar ist,

Zeile 33:		Zeile 33:
	{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}		{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}

-	Wir erweitern die rechte Seite, und sehen dass alles stimmt,	+	Wir erweitern die rechte Seite und sehen, dass alles stimmt,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}		(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Martin um 13:36, 21. Mai 2009

Martin — Thu, 21 May 2009 13:36:48 GMT

← Nächstältere Version		Version vom 13:36, 21. Mai 2009
Zeile 1:		Zeile 1:
-	~~We start by adding and taking away~~ <math>x^2</math> ~~in the numerator~~, ~~so that, in combination with <math>x^3</math>, we obtain the expression~~ <math>x^3+x^2 = x^2(x+1)</math> ~~which can be simplified with the denominator <math>x+1</math>~~,	+	Wir beginnen damit <math>x^2</math> zu addieren und subtrahieren, sodass wir <math>x^3+x^2 = x^2(x+1)</math> im Zähler erhalten,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 9:		Zeile 9:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The term <math>-x^2</math> in the remaining quotient needs to complemented with~~	+	Danach addieren und subtrahieren wir
-	<math>-x</math> ~~so that we get~~ <math>-x^2-x = -x(x+1)</math>, ~~which is divisible by~~	+	<math>-x</math> sodass wir <math>-x^2-x = -x(x+1)</math> erhalten, nachdem dies durch
-	<math>x+1</math>,	+	<math>x+1</math> teilbar ist,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 21:		Zeile 21:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The last quotient divides perfectly and we obtain~~	+	Wir erhalten

	{{Abgesetzte Formel\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}

-	~~A quick check of whether~~	+	Wir testen ob

	{{Abgesetzte Formel\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}

-	~~is the correct answer is to investigate whether~~	+	indem wir kontrollieren ob

	{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}		{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}

-	~~holds. If we expand the right-hand side~~, ~~we see that the relation really does hold~~	+	Wir erweitern die rechte Seite, und sehen dass alles stimmt,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
	(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}		(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tekbot: Solution 3.4:1d moved to Lösung 3.4:1d: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:51:15 GMT

Solution 3.4:1d moved to Lösung 3.4:1d: Robot: moved page

← Nächstältere Version

Version vom 10:51, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:15:04 GMT

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

← Nächstältere Version		Version vom 13:15, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,		We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{x^3+x+2}{x+1}		\frac{x^3+x+2}{x+1}
	&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]		&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]
Zeile 13:		Zeile 13:
	<math>x+1</math>,		<math>x+1</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^2 + \frac{-x^2+x+2}{x+1}		x^2 + \frac{-x^2+x+2}{x+1}
	&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]		&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]
Zeile 23:		Zeile 23:
	The last quotient divides perfectly and we obtain		The last quotient divides perfectly and we obtain

-	{{~~Displayed math~~\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}

	A quick check of whether		A quick check of whether

-	{{~~Displayed math~~\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}

	is the correct answer is to investigate whether		is the correct answer is to investigate whether

-	{{~~Displayed math~~\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}	+	{{Abgesetzte Formel\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}

	holds. If we expand the right-hand side, we see that the relation really does hold		holds. If we expand the right-hand side, we see that the relation really does hold

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}		(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 12:29, 31. Okt. 2008

Tek — Fri, 31 Oct 2008 12:29:23 GMT

← Nächstältere Version		Version vom 12:29, 31. Okt. 2008
Zeile 1:		Zeile 1:
-	We start by adding and taking away	+	We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>,
-	<math>x^{2}</math>	+
-	in the numerator, so that, in combination with	+
-	<math>x^{3}</math>, we obtain the expression	+
-	<math>x^{3}+x^{2}=x^{2~~}\left~~( x+1 ~~\right~~)</math>	+
-	which can be simplified with the denominator	+
-	<math>x+1</math>,	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{x^3+x+2}{x+1}
		+	&= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt]
		+	&= \frac{x^3+x^2}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt]
		+	&= \frac{x^2(x+1)}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt]
		+	&= x^2 + \frac{-x^2+x+2}{x+1}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	The term <math>-x^2</math> in the remaining quotient needs to complemented with
-	~~& \frac{x^{3}+x+2}{x+1}=\frac{x^{3}+x^{2}-x^{2}+x+2}{x+1} \\~~	+	<math>-x</math> so that we get <math>-x^2-x = -x(x+1)</math>, which is divisible by
-	~~& =\frac{x^{3}+x^{2}}{x+1}+\frac{-x^{2}+x+2}{x+1}=\frac{x^{2}\left( x+1 \right)}{x+1}+\frac{-x^{2}+x+2}{x+1} \\~~	+
-	~~& =x^{2}+\frac{-x^{2}+x+2}{x+1} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	The term	+
-	<math>-x^{2}</math>	+
-	in the remaining quotient needs to complemented with	+
-	<math>-x</math>	+
-	so that we get	+
-	<math>-x^{2}-x=-x~~\left~~( x+1 ~~\right~~)</math>, which is divisible by	+
	<math>x+1</math>,		<math>x+1</math>,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x^2 + \frac{-x^2+x+2}{x+1}
-	& x^{2}+\frac{-x^{2}+x+2}{x+1}=x^{2}+\frac{-x^{2}-x+x+x+2}{x+1} \\	+	&= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt]
-	& =x^{2}+\frac{-x^{2}-x}{x+1}+\frac{2x+2}{x+1} \\	+	&= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt]
-	& =x^{2}+\frac{-x~~\left~~( x+1 ~~\right~~)}{x+1}+\frac{2x+2}{x+1} \\	+	&= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt]
-	& =x^{2}-x+\frac{2x+2}{x+1} \\	+	&= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}
-		+

	The last quotient divides perfectly and we obtain		The last quotient divides perfectly and we obtain

-		+	{{Displayed math\|\|<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}}
-	<math>x^{2}-x+\frac{2x+2}{x+1}=x^{2}-x+2.</math>	+
-		+

	A quick check of whether		A quick check of whether

-		+	{{Displayed math\|\|<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}}
-	<math>\frac{x^{3}+x+2}{x+1}=x^{2}-x+2.</math>	+
-		+

	is the correct answer is to investigate whether		is the correct answer is to investigate whether

		+	{{Displayed math\|\|<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}}

-	~~<math>x^{3}+x+2=\left( x^{2}-x+2 \right)\left( x+1 \right)</math>~~	+	holds. If we expand the right-hand side, we see that the relation really does hold
-		+
-	holds. If we expand the right-hand side, we see that the relation really does hold:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	~~& \left~~( x^{2}-x+2 ~~\right~~)~~\left~~( x+1 ~~\right~~)=x^{3}+x^{2}-x^{2}-x+2x+2 \\	+	(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.}
-	& =x^{3}+x+2 \\	+	\end{align}</math>}}
-	\end{align}</math>	+

Ian um 13:22, 26. Okt. 2008

Ian — Sun, 26 Oct 2008 13:22:34 GMT

← Nächstältere Version		Version vom 13:22, 26. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We start by adding and taking away
-	<~~center~~> ~~[[Image:3_4_1d~~.~~gif]]~~ </~~center~~>	+	<math>x^{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	in the numerator, so that, in combination with
		+	<math>x^{3}</math>, we obtain the expression
		+	<math>x^{3}+x^{2}=x^{2}\left( x+1 \right)</math>
		+	which can be simplified with the denominator
		+	<math>x+1</math>,
		+
		+
		+	<math>\begin{align}
		+	& \frac{x^{3}+x+2}{x+1}=\frac{x^{3}+x^{2}-x^{2}+x+2}{x+1} \\
		+	& =\frac{x^{3}+x^{2}}{x+1}+\frac{-x^{2}+x+2}{x+1}=\frac{x^{2}\left( x+1 \right)}{x+1}+\frac{-x^{2}+x+2}{x+1} \\
		+	& =x^{2}+\frac{-x^{2}+x+2}{x+1} \\
		+	\end{align}</math>
		+
		+
		+	The term
		+	<math>-x^{2}</math>
		+	in the remaining quotient needs to complemented with
		+	<math>-x</math>
		+	so that we get
		+	<math>-x^{2}-x=-x\left( x+1 \right)</math>, which is divisible by
		+	<math>x+1</math>,
		+
		+
		+	<math>\begin{align}
		+	& x^{2}+\frac{-x^{2}+x+2}{x+1}=x^{2}+\frac{-x^{2}-x+x+x+2}{x+1} \\
		+	& =x^{2}+\frac{-x^{2}-x}{x+1}+\frac{2x+2}{x+1} \\
		+	& =x^{2}+\frac{-x\left( x+1 \right)}{x+1}+\frac{2x+2}{x+1} \\
		+	& =x^{2}-x+\frac{2x+2}{x+1} \\
		+	\end{align}</math>
		+
		+
		+	The last quotient divides perfectly and we obtain
		+
		+
		+	<math>x^{2}-x+\frac{2x+2}{x+1}=x^{2}-x+2.</math>
		+
		+
		+	A quick check of whether
		+
		+
		+	<math>\frac{x^{3}+x+2}{x+1}=x^{2}-x+2.</math>
		+
		+
		+	is the correct answer is to investigate whether
		+
		+
		+	<math>x^{3}+x+2=\left( x^{2}-x+2 \right)\left( x+1 \right)</math>
		+
		+	holds. If we expand the right-hand side, we see that the relation really does hold:
		+
		+
		+	<math>\begin{align}
		+	& \left( x^{2}-x+2 \right)\left( x+1 \right)=x^{3}+x^{2}-x^{2}-x+2x+2 \\
		+	& =x^{3}+x+2 \\
		+	\end{align}</math>

Tekbot: Lösning 3.4:1d moved to Solution 3.4:1d: Robot: moved page

Tekbot — Tue, 16 Sep 2008 15:11:41 GMT

Lösning 3.4:1d moved to Solution 3.4:1d: Robot: moved page

← Nächstältere Version

Version vom 15:11, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:15:34 GMT

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

← Nächstältere Version		Version vom 08:15, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_4_1d.gif]] </center>	+	<center> [[Image:3_4_1d.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_4_1d.gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 15:05:52 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_1d.gif </center> {{NAVCONTENT_STOP}}

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