Lösung 2.3:3f - Versionsgeschichte

Moni um 09:33, 9. Aug. 2009

Moni — Sun, 09 Aug 2009 09:33:49 GMT

← Nächstältere Version		Version vom 09:33, 9. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir faktorisieren zuerst die beiden Terme in der linken Seite der Gleichung:	+	Wir faktorisieren zuerst die beiden Terme in der linken Seite der Gleichung

	<math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math>		<math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math>

Moni: Zeilenumbruch eingefügt

Moni — Wed, 05 Aug 2009 14:02:59 GMT

Zeilenumbruch eingefügt

← Nächstältere Version		Version vom 14:02, 5. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir faktorisieren zuerst die beiden Terme in der linken Seite der Gleichung;	+	Wir faktorisieren zuerst die beiden Terme in der linken Seite der Gleichung:

	<math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math>		<math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math>
		+	<br>
	<math>x\cdot (2-x) = -x(x-2)</math>.		<math>x\cdot (2-x) = -x(x-2)</math>.

Markus.bez: Syntax

Markus.bez — Tue, 09 Jun 2009 12:50:25 GMT

Syntax

← Nächstältere Version		Version vom 12:50, 9. Jun. 2009
Zeile 19:		Zeile 19:
	geschrieben werden. Diese Gleichung ist erfüllt, wenn einer der Faktoren <math>x</math>, <math>x-2</math> oder <math>x-1</math> null ist. Also sind die Lösungen <math>x=0</math>, <math>x=2</math> und <math>x=1</math>.		geschrieben werden. Diese Gleichung ist erfüllt, wenn einer der Faktoren <math>x</math>, <math>x-2</math> oder <math>x-1</math> null ist. Also sind die Lösungen <math>x=0</math>, <math>x=2</math> und <math>x=1</math>.

-	Nachdem es nicht ganz offensichtlich ist, ob <math>x=1</math> eine Lösung ist, kontrollieren wir, ob so der Fall ist	+	Nachdem es nicht ganz offensichtlich ist, ob <math>x=1</math> eine Lösung ist, kontrollieren wir, ob dies hierbei der Fall ist:

-	~~:*''x'' = 1:~~ <math>\ \text{Linke Seite} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{Rechte Seite.}</math>	+	<math>\begin{align}
		+	\text{Linke Seite}
		+	&= 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) \\[5pt]
		+	&= 1\cdot (-1) + 1\cdot 1 = 0 = \text{Rechte Seite.}\end{align}</math>

Markus.bez: Sprache und Formulierung

Markus.bez — Tue, 09 Jun 2009 12:47:25 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 12:47, 9. Jun. 2009
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	Die Gleichung kann jetzt ~~wie~~	+	Die Gleichung kann jetzt als

	{{Abgesetzte Formel\|\|<math>x(x-2)(x-1) = 0</math>}}		{{Abgesetzte Formel\|\|<math>x(x-2)(x-1) = 0</math>}}

-	geschrieben werden. Diese Gleichung ist erfüllt wenn einer der Faktoren <math>x</math>, <math>x-2</math> oder <math>x-1</math> null ist. Also sind die Lösungen <math>x=0</math>, <math>x=2</math> und <math>x=1</math>.	+	geschrieben werden. Diese Gleichung ist erfüllt, wenn einer der Faktoren <math>x</math>, <math>x-2</math> oder <math>x-1</math> null ist. Also sind die Lösungen <math>x=0</math>, <math>x=2</math> und <math>x=1</math>.

-	Nachdem es nicht ganz ~~offenbar~~ ist, ob <math>x=1</math> eine Lösung ist, kontrollieren wir ob so der Fall ist	+	Nachdem es nicht ganz offensichtlich ist, ob <math>x=1</math> eine Lösung ist, kontrollieren wir, ob so der Fall ist

	:*''x'' = 1: <math>\ \text{Linke Seite} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{Rechte Seite.}</math>		:*''x'' = 1: <math>\ \text{Linke Seite} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{Rechte Seite.}</math>

Martin um 15:30, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 15:30:13 GMT

← Nächstältere Version		Version vom 15:30, 16. Mär. 2009
Zeile 1:		Zeile 1:
-	~~We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket,~~ <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> ~~and writing the other term as~~ <math>x\cdot (2-x) = -x(x-2)</math>. ~~From this we see that both terms contain~~ <math>x(x-2)</math> ~~as common factors and~~, ~~if we take out those, the left-hand side becomes~~	+	Wir faktorisieren zuerst die beiden Terme in der linken Seite der Gleichung;
		+
		+	<math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math>
		+	<math>x\cdot (2-x) = -x(x-2)</math>.
		+
		+	Nachdem wir den gemeinsamen Faktor <math>x(x-2)</math> haben, faktorisieren wir den ganzen Ausdruck auf der linken Seite der Gleichung

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 8:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The whole equation can be written as~~	+	Die Gleichung kann jetzt wie

	{{Abgesetzte Formel\|\|<math>x(x-2)(x-1) = 0</math>}}		{{Abgesetzte Formel\|\|<math>x(x-2)(x-1) = 0</math>}}

-	~~and this equation is satisfied only when one of the three factors~~ <math>x</math>, <math>x-2</math> or <math>x-1</math> ~~is zero, i.e~~. ~~the solutions are~~ <math>x=0</math>, <math>x=2</math> ~~and~~ <math>x=1</math>.	+	geschrieben werden. Diese Gleichung ist erfüllt wenn einer der Faktoren <math>x</math>, <math>x-2</math> oder <math>x-1</math> null ist. Also sind die Lösungen <math>x=0</math>, <math>x=2</math> und <math>x=1</math>.

-	~~Because it is not completely obvious that <math>x=1</math> is a solution of the equation~~, ~~we check that~~ <math>x=1</math> ~~satisfies the equation~~, ~~i.e. that we haven't calculated incorrectly:~~	+	Nachdem es nicht ganz offenbar ist, ob <math>x=1</math> eine Lösung ist, kontrollieren wir ob so der Fall ist

-	:*''x'' = 1: <math>\ \text{~~LHS~~} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{~~RHS~~.}</math>	+	:*''x'' = 1: <math>\ \text{Linke Seite} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{Rechte Seite.}</math>

Tekbot: hat „Solution 2.3:3f“ nach „Lösung 2.3:3f“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:10:05 GMT

hat „Solution 2.3:3f“ nach „Lösung 2.3:3f“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:10, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:32:58 GMT

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

← Nächstältere Version		Version vom 08:32, 22. Okt. 2008
Zeile 1:		Zeile 1:
	We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes		We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x(x^{2}-2x) + x(2-x)		x(x^{2}-2x) + x(2-x)
	&= x^{2}(x-2) - x(x-2)\\[5pt]		&= x^{2}(x-2) - x(x-2)\\[5pt]
Zeile 10:		Zeile 10:
	The whole equation can be written as		The whole equation can be written as

-	{{~~Displayed math~~\|\|<math>x(x-2)(x-1) = 0</math>}}	+	{{Abgesetzte Formel\|\|<math>x(x-2)(x-1) = 0</math>}}

	and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>.		and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>.

Tek um 08:57, 29. Sep. 2008

Tek — Mon, 29 Sep 2008 08:57:38 GMT

← Nächstältere Version		Version vom 08:57, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	We can split up the first term on the left-hand side,	+	We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes
-	<math>x~~\left~~( x^{2}-2x ~~\right~~)</math>	+
-	, into factors by taking	+
-	<math>x</math>	+
-	outside the bracket,	+
-	<math>x~~\left~~( x^{2}-2x ~~\right~~)=x\~~centerdot~~ x\~~centerdot \left~~( x-2 ~~\right~~)</math>	+
-	and writing the other term as	+
-	<math>x\~~centerdot \left~~( 2-x ~~\right~~)=-x~~\left~~( x-2 ~~\right~~)</math>. From this we see that both terms contain	+
-	<math>x~~\left~~( x-2 ~~\right~~)</math>	+
-	as common factors and, if we take out those, the left-hand side becomes	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\~~	+
-	~~& =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\~~	+
-	~~\end{align}</math>~~	+

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

	The whole equation can be written as		The whole equation can be written as

		+	{{Displayed math\|\|<math>x(x-2)(x-1) = 0</math>}}

-	~~<math>x\left( x-2 \right)\left( x-1 \right)=0</math>~~	+	and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>.
-		+
-		+
-	and this equation is satisfied only when one of the three factors	+
-	<math>x</math>,	+
-	<math>x-~~\text{~~2}</math>	+
-	or	+
-	<math>x-~~\text{~~1}</math>	+
-	is zero, i.e. the solutions are	+
-	<math>x=0</math>,	+
-	<math>x=~~\text{~~2 }</math>	+
-	and	+
-	<math>x=~~\text{~~1}</math>.	+

-	Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:	+	Because it is not completely obvious that <math>x=1</math> is a solution of the equation, we check that <math>x=1</math> satisfies the equation, i.e. that we haven't calculated incorrectly:

-	x=1: ~~LHS~~	+	:*''x'' = 1: <math>\ \text{LHS} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{RHS.}</math>
-	<math>=1\~~centerdot \left~~( 1^{2}-2\~~centerdot~~ 1 ~~\right~~)+1\~~centerdot \left~~( 2-1 ~~\right~~)=1\~~centerdot \left~~( -1 ~~\right~~)+1\~~centerdot~~ 1=0=</math>	+
-	~~RHS~~	+

Ian um 15:28, 20. Sep. 2008

Ian — Sat, 20 Sep 2008 15:28:58 GMT

← Nächstältere Version		Version vom 15:28, 20. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We can split up the first term on the left-hand side,
-	<~~center~~> ~~[[Image:2_3_3f~~.~~gif]]~~ </~~center~~>	+	<math>x\left( x^{2}-2x \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, into factors by taking
		+	<math>x</math>
		+	outside the bracket,
		+	<math>x\left( x^{2}-2x \right)=x\centerdot x\centerdot \left( x-2 \right)</math>
		+	and writing the other term as
		+	<math>x\centerdot \left( 2-x \right)=-x\left( x-2 \right)</math>. From this we see that both terms contain
		+	<math>x\left( x-2 \right)</math>
		+	as common factors and, if we take out those, the left-hand side becomes
		+
		+
		+	<math>\begin{align}
		+	& x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\
		+	& =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\
		+	\end{align}</math>
		+
		+
		+	The whole equation can be written as
		+
		+
		+	<math>x\left( x-2 \right)\left( x-1 \right)=0</math>
		+
		+
		+	and this equation is satisfied only when one of the three factors
		+	<math>x</math>,
		+	<math>x-\text{2}</math>
		+	or
		+	<math>x-\text{1}</math>
		+	is zero, i.e. the solutions are
		+	<math>x=0</math>,
		+	<math>x=\text{2 }</math>
		+	and
		+	<math>x=\text{1}</math>.
		+
		+	Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:
		+
		+	x=1: LHS
		+	<math>=1\centerdot \left( 1^{2}-2\centerdot 1 \right)+1\centerdot \left( 2-1 \right)=1\centerdot \left( -1 \right)+1\centerdot 1=0=</math>
		+	RHS

Tekbot: Lösning 2.3:3f moved to Solution 2.3:3f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:59:43 GMT

Lösning 2.3:3f moved to Solution 2.3:3f: Robot: moved page

← Nächstältere Version

Version vom 08:59, 9. Sep. 2008