Lösung 3.4:1a - Versionsgeschichte

Sekretariat1 um 11:16, 4. Sep. 2009

Sekretariat1 — Fri, 04 Sep 2009 11:16:10 GMT

← Nächstältere Version		Version vom 11:16, 4. Sep. 2009
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-	Der Zähler kann durch ~~einer~~ der binomischen Formeln in Faktoren zerlegt werden, sodass <math>x^2-1=(x+1)(x-1)</math>~~, und wir~~ erhalten dadurch	+	Der Zähler kann durch eine der binomischen Formeln in Faktoren zerlegt werden, sodass <math>x^2-1=(x+1)(x-1)</math>. Wir erhalten dadurch

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

Martin — Thu, 21 May 2009 12:42:36 GMT

← Nächstältere Version		Version vom 12:42, 21. Mai 2009
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-	~~The numerator can be factorized using the formula for the difference of two squares to give~~ <math>x^2-1=(x+1)(x-1)</math> ~~and then we see that the numerator and denominator have a common factor which we can eliminate~~	+	Der Zähler kann durch einer der binomischen Formeln in Faktoren zerlegt werden, sodass <math>x^2-1=(x+1)(x-1)</math>, und wir erhalten dadurch

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

Tekbot — Wed, 11 Mar 2009 10:50:15 GMT

← Nächstältere Version

Version vom 10:50, 11. Mär. 2009

Tekbot — Tue, 10 Mar 2009 13:14:34 GMT

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

← Nächstältere Version		Version vom 13:14, 10. Mär. 2009
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	The numerator can be factorized using the formula for the difference of two squares to give <math>x^2-1=(x+1)(x-1)</math> and then we see that the numerator and denominator have a common factor which we can eliminate		The numerator can be factorized using the formula for the difference of two squares to give <math>x^2-1=(x+1)(x-1)</math> and then we see that the numerator and denominator have a common factor which we can eliminate

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

Tek — Fri, 31 Oct 2008 11:44:02 GMT

← Nächstältere Version		Version vom 11:44, 31. Okt. 2008
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-	The numerator can be factorized using the ~~conjugate rule~~ to give	+	The numerator can be factorized using the formula for the difference of two squares to give <math>x^2-1=(x+1)(x-1)</math> and then we see that the numerator and denominator have a common factor which we can eliminate
-	<math>x^{2}-1=~~\left~~( x+1 ~~\right~~)~~\left~~( x-1 ~~\right~~)</math>	+
-	and then we see that the numerator and denominator have a common factor which we can eliminate	+

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

Ian — Sun, 26 Oct 2008 12:30:04 GMT

← Nächstältere Version		Version vom 12:30, 26. Okt. 2008
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-	~~{{NAVCONTENT_START}}~~	+	The numerator can be factorized using the conjugate rule to give
-	<~~center~~> ~~[[Image:3_4_1a.gif]]~~ </~~center~~>	+	<math>x^{2}-1=\left( x+1 \right)\left( x-1 \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and then we see that the numerator and denominator have a common factor which we can eliminate
		+
		+
		+	<math>\frac{x^{2}-1}{x-1}=\frac{\left( x+1 \right)\left( x-1 \right)}{x-1}=x+1</math>

Tekbot — Tue, 16 Sep 2008 15:10:41 GMT

← Nächstältere Version

Version vom 15:10, 16. Sep. 2008

Tekbot — Thu, 21 Aug 2008 08:15:04 GMT

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

Oskar — Mon, 07 Apr 2008 15:04:04 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_4_1a.gif </center> {{NAVCONTENT_STOP}}

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