Lösung 1.2:4c - Versionsgeschichte

Moni um 13:18, 8. Aug. 2009

Moni — Sat, 08 Aug 2009 13:18:12 GMT

← Nächstältere Version		Version vom 13:18, 8. Aug. 2009
Zeile 24:		Zeile 24:
	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

-	Den kleinsten gemeinsamen Nenner erhalten wir indem wir den ersten Bruch mit 5, und den zweiten Bruch mit 4 erweitern	+	Den kleinsten gemeinsamen Nenner erhalten wir indem wir den ersten Bruch mit 5 und den zweiten Bruch mit 4 erweitern

	{{Abgesetzte Formel\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}

-	Nachdem <math>10=2\cdot 5</math> und <math>4=2\cdot 2</math>, kürzen wir den Bruch mit den gemeinsamen Faktoren 2 und 5, und erhalten	+	Nachdem <math>10=2\cdot 5</math> und <math>4=2\cdot 2</math> kürzen wir den Bruch mit den gemeinsamen Faktoren 2 und 5 und erhalten

	{{Abgesetzte Formel\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}

Martin um 12:51, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 12:51:40 GMT

← Nächstältere Version		Version vom 12:51, 26. Okt. 2008
Zeile 5:		Zeile 5:
	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	Wir erweitern jetzt den Hauptbruch mit ~~den~~ Kehrwert des unteren Bruches, also <math>{10}/{3}\,</math>,	+	Wir erweitern jetzt den Hauptbruch mit dem Kehrwert des unteren Bruches, also <math>{10}/{3}\,</math>,

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}
Zeile 20:		Zeile 20:
	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	Und erweitern beide Hauptbrüche mit ~~den~~ Kehrwert des unteren Bruches, also <math>10/3</math>	+	Und erweitern beide Hauptbrüche mit dem Kehrwert des unteren Bruches, also <math>10/3</math>

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

Martin um 12:44, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 12:44:33 GMT

← Nächstältere Version		Version vom 12:44, 26. Okt. 2008
Zeile 5:		Zeile 5:
	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	Wir erweitern jetzt den Hauptbruch mit ~~dem~~ Kehrwert des unteren Bruches, also <math>{10}/{3}\,</math>,	+	Wir erweitern jetzt den Hauptbruch mit den Kehrwert des unteren Bruches, also <math>{10}/{3}\,</math>,

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}
Zeile 20:		Zeile 20:
	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	Und erweitern beide Hauptbrüche mit ~~dem~~ Kehrwert des unteren Bruches, also <math>10/3</math>	+	Und erweitern beide Hauptbrüche mit den Kehrwert des unteren Bruches, also <math>10/3</math>

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

Martin um 12:35, 26. Okt. 2008

Martin — Sun, 26 Oct 2008 12:35:18 GMT

← Nächstältere Version		Version vom 12:35, 26. Okt. 2008
Zeile 1:		Zeile 1:
-	~~Method 1~~	+	Erste Methode:

-	~~If we calculate the numerator~~ in ~~the main fraction first~~, ~~we get~~	+	Wenn wir zuerst den Zähler in den Hauptbruch berechnen, bekommen wir

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	~~The double fraction on the right-hand side becomes~~, ~~after multiplying top and bottom by~~ <math>{10}/{3}\,</math>,	+	Wir erweitern jetzt den Hauptbruch mit dem Kehrwert des unteren Bruches, also <math>{10}/{3}\,</math>,

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}

-	~~Then, we remove the common factor~~ 10,	+	Danach kürzen wir den Bruch mit den gemeinsamen Faktor 10,

	{{Abgesetzte Formel\|\|<math>\dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,</math>.}}


-	~~Method 2~~	+	Zweite Methode:

-	~~Another way to calculate the expression is to divide it up into two separate terms~~	+	Wir können auch den Hauptbruch als zwei Brüche schreiben,

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	~~We simplify both double fractions on the right-hand side by multiplying top and bottom by~~ 10/3	+	Und erweitern beide Hauptbrüche mit dem Kehrwert des unteren Bruches, also <math>10/3</math>

	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

-	~~Instead of multiplying, respectively, by <math>4\cdot 3</math> and <math>~~5~~\cdot 3</math>~~, ~~we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by 5 and the second by~~ 4~~, we obtain the common denominator~~	+	Den kleinsten gemeinsamen Nenner erhalten wir indem wir den ersten Bruch mit 5, und den zweiten Bruch mit 4 erweitern

	{{Abgesetzte Formel\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}

-	~~Because~~ <math>10=2\cdot 5</math> ~~and~~ <math>4=2\cdot 2</math>, ~~we can cancel out the common factors~~ 2 ~~and~~ 5 ~~and obtain the answer~~	+	Nachdem <math>10=2\cdot 5</math> und <math>4=2\cdot 2</math>, kürzen wir den Bruch mit den gemeinsamen Faktoren 2 und 5, und erhalten

	{{Abgesetzte Formel\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}		{{Abgesetzte Formel\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}

Tekbot: hat „Solution 1.2:4c“ nach „Lösung 1.2:4c“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 13:29:25 GMT

hat „Solution 1.2:4c“ nach „Lösung 1.2:4c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 13:29, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:15:28 GMT

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

Tek um 12:26, 22. Sep. 2008

Tek — Mon, 22 Sep 2008 12:26:32 GMT

← Nächstältere Version		Version vom 12:26, 22. Sep. 2008
Zeile 30:		Zeile 30:
	Because <math>10=2\cdot 5</math> and <math>4=2\cdot 2</math>, we can cancel out the common factors 2 and 5 and obtain the answer		Because <math>10=2\cdot 5</math> and <math>4=2\cdot 2</math>, we can cancel out the common factors 2 and 5 and obtain the answer

-	{{Displayed math\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{6}\,</math>.}}	+	{{Displayed math\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}

Tek um 11:07, 19. Sep. 2008

Tek — Fri, 19 Sep 2008 11:07:25 GMT

← Nächstältere Version		Version vom 11:07, 19. Sep. 2008
Zeile 3:		Zeile 3:
	If we calculate the numerator in the main fraction first, we get		If we calculate the numerator in the main fraction first, we get

		+	{{Displayed math\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	<math>~~\frac~~{~~\frac{1~~}~~{4}-\frac{1}{5}}{\frac~~{3}~~{16}}=~~\~~frac{\frac{1\centerdot 5}{4\centerdot 5}-\frac{1\centerdot 4}{5\centerdot 4}}{\frac{3}{16}}=\frac{\frac{5}{20}-\frac{4}{20}}{\frac{3}{10}}=\frac{\frac{1}{20}}{\frac{3}{10}}~~</math>	+	The double fraction on the right-hand side becomes, after multiplying top and bottom by <math>{10}/{3}\,</math>,

-	~~The double fraction on the right-hand side becomes, after multiplying top and bottom by~~	+	{{Displayed math\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}
-	<math>{10}/{3}\;</math>	+
-	,	+

		+	Then, we remove the common factor 10,

-	<math>\~~frac{\frac{1}{20}}{\frac{3}{10}}=\frac{\frac~~{1}{20}\~~centerdot~~ \~~frac~~{10}{3}}{\~~frac~~{3}{~~10}\centerdot~~ \~~frac~~{10}~~{3}}=~~\~~frac~~{~~1}{20}~~\~~centerdot \frac{10~~}~~{3}</math>~~	+	{{Displayed math\|\|<math>\dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,</math>.}}
-		+
-		+
-	~~Then, we remove the common factor~~ 10:	+
-		+
-		+
-	~~<math>\frac{1}{20~~}\~~centerdot~~ \~~frac~~{~~10}{3}=~~\~~frac~~{~~1}{2~~\~~centerdot 10~~}~~\centerdot \frac{~~10}{3}=\~~frac~~{1}{2\~~centerdot~~ 3}=\~~frac~~{1}{6}</math>	+
-		+


	Method 2		Method 2

-	Another way to calculate the expression is to divide it up into two separate terms:	+	Another way to calculate the expression is to divide it up into two separate terms
-		+
-		+
-	~~<math>\frac{\frac{1}{4}-\frac{1}{5}}{\frac{3}{16}}=\frac{\frac{1}{4}}{\frac{3}{10}}-\frac{\frac{1}{5}}{\frac{3}{10}}</math>~~	+
-		+
-	~~We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3:~~	+
-		+

-	<math>\frac{\~~frac~~{1}{4}}{\~~frac~~{3}{10}}-\~~frac~~{\~~frac{1}{5~~}}{\~~frac~~{3}{10}}=\frac{\~~frac~~{1}{4}\~~centerdot~~ \~~frac{10}{3~~}}{\~~frac~~{3}{10}\~~centerdot~~ \~~frac{10}{3~~}}-\frac{\~~frac~~{1}{5}\~~centerdot~~ \~~frac{10}{3~~}}{\~~frac~~{3}{10}\~~centerdot \frac~~{~~10}{3}}=~~\~~frac{1}{4~~}\~~centerdot \frac{10~~}~~{3}-~~\~~frac{1}{5}\centerdot \frac{10}{3}~~</math>	+	{{Displayed math\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

-	~~Instead of~~ multiplying~~, respectively, by~~	+	We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3
-	~~<math>4\centerdot 3</math>~~	+
-	~~and~~	+
-	~~<math>5\centerdot 3</math>~~	+
-	~~, we keep the numerators factorized and observe that if we multiply the~~ top and bottom ~~of the first fraction~~ by	+
-	~~<math>5</math>~~	+
-	~~and the second by~~	+
-	~~<math>4</math>~~	+
-	~~, we obtain the common denominator:~~	+

		+	{{Displayed math\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

-	<math>~~\frac{10}{~~4\~~centerdot~~ 3~~}-\frac{10}{~~5\~~centerdot~~ 3~~}=\frac{10\centerdot 5}{4\centerdot 3\centerdot 5}-\frac{10\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{50-40}{3\centerdot 4\centerdot 5}=\frac{10}{3\centerdot 4\centerdot 5}~~</math>	+	Instead of multiplying, respectively, by <math>4\cdot 3</math> and <math>5\cdot 3</math>, we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by 5 and the second by 4, we obtain the common denominator

-	~~Because~~	+	{{Displayed math\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}
-	<math>10=2\~~centerdot~~ 5~~</math>~~	+
-	~~and~~	+
-	~~<math>~~4=2\~~centerdot 2</math>~~	+
-	~~, we can cancel out the common factors~~	+
-	~~<math>2</math>~~	+
-	~~and~~	+
-	~~<math>~~5</math>	+
-	~~and obtain the answer:~~	+

		+	Because <math>10=2\cdot 5</math> and <math>4=2\cdot 2</math>, we can cancel out the common factors 2 and 5 and obtain the answer

-	<math>\frac{10}{3\~~centerdot~~ 4\~~centerdot~~ 5}=\frac{2\~~centerdot~~ 5}{3\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 5}=\frac{1}{6}</math>	+	{{Displayed math\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{6}\,</math>.}}

Ian um 13:31, 11. Sep. 2008

Ian — Thu, 11 Sep 2008 13:31:00 GMT

← Nächstältere Version		Version vom 13:31, 11. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Method 1
-	<~~center~~> ~~[[Image~~:~~1_2_4c-~~1(2~~).gif]]~~ </~~center~~>	+
-	<~~center~~> ~~[[Image~~:~~1_2_4c~~-2(2~~).gif]]~~ </~~center~~>	+	If we calculate the numerator in the main fraction first, we get
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\frac{\frac{1}{4}-\frac{1}{5}}{\frac{3}{16}}=\frac{\frac{1\centerdot 5}{4\centerdot 5}-\frac{1\centerdot 4}{5\centerdot 4}}{\frac{3}{16}}=\frac{\frac{5}{20}-\frac{4}{20}}{\frac{3}{10}}=\frac{\frac{1}{20}}{\frac{3}{10}}</math>
		+
		+	The double fraction on the right-hand side becomes, after multiplying top and bottom by
		+	<math>{10}/{3}\;</math>
		+	,
		+
		+
		+	<math>\frac{\frac{1}{20}}{\frac{3}{10}}=\frac{\frac{1}{20}\centerdot \frac{10}{3}}{\frac{3}{10}\centerdot \frac{10}{3}}=\frac{1}{20}\centerdot \frac{10}{3}</math>
		+
		+
		+	Then, we remove the common factor 10:
		+
		+
		+	<math>\frac{1}{20}\centerdot \frac{10}{3}=\frac{1}{2\centerdot 10}\centerdot \frac{10}{3}=\frac{1}{2\centerdot 3}=\frac{1}{6}</math>
		+
		+
		+
		+	Method 2
		+
		+	Another way to calculate the expression is to divide it up into two separate terms:
		+
		+
		+	<math>\frac{\frac{1}{4}-\frac{1}{5}}{\frac{3}{16}}=\frac{\frac{1}{4}}{\frac{3}{10}}-\frac{\frac{1}{5}}{\frac{3}{10}}</math>
		+
		+	We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3:
		+
		+
		+	<math>\frac{\frac{1}{4}}{\frac{3}{10}}-\frac{\frac{1}{5}}{\frac{3}{10}}=\frac{\frac{1}{4}\centerdot \frac{10}{3}}{\frac{3}{10}\centerdot \frac{10}{3}}-\frac{\frac{1}{5}\centerdot \frac{10}{3}}{\frac{3}{10}\centerdot \frac{10}{3}}=\frac{1}{4}\centerdot \frac{10}{3}-\frac{1}{5}\centerdot \frac{10}{3}</math>
		+
		+	Instead of multiplying, respectively, by
		+	<math>4\centerdot 3</math>
		+	and
		+	<math>5\centerdot 3</math>
		+	, we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by
		+	<math>5</math>
		+	and the second by
		+	<math>4</math>
		+	, we obtain the common denominator:
		+
		+
		+	<math>\frac{10}{4\centerdot 3}-\frac{10}{5\centerdot 3}=\frac{10\centerdot 5}{4\centerdot 3\centerdot 5}-\frac{10\centerdot 4}{5\centerdot 3\centerdot 4}=\frac{50-40}{3\centerdot 4\centerdot 5}=\frac{10}{3\centerdot 4\centerdot 5}</math>
		+
		+	Because
		+	<math>10=2\centerdot 5</math>
		+	and
		+	<math>4=2\centerdot 2</math>
		+	, we can cancel out the common factors
		+	<math>2</math>
		+	and
		+	<math>5</math>
		+	and obtain the answer:
		+
		+
		+	<math>\frac{10}{3\centerdot 4\centerdot 5}=\frac{2\centerdot 5}{3\centerdot 2\centerdot 2\centerdot 5}=\frac{1}{6}</math>

Tekbot: Lösning 1.2:4c moved to Solution 1.2:4c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:19:03 GMT

Lösning 1.2:4c moved to Solution 1.2:4c: Robot: moved page

← Nächstältere Version

Version vom 08:19, 9. Sep. 2008

← Nächstältere Version		Version vom 08:15, 22. Okt. 2008
Zeile 3:		Zeile 3:
	If we calculate the numerator in the main fraction first, we get		If we calculate the numerator in the main fraction first, we get

-	{{~~Displayed math~~\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{1\cdot 5}{4\cdot 5}-\dfrac{1\cdot 4}{5\cdot 4}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} =\frac{\,\dfrac{5}{20}-\dfrac{4}{20}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

	The double fraction on the right-hand side becomes, after multiplying top and bottom by <math>{10}/{3}\,</math>,		The double fraction on the right-hand side becomes, after multiplying top and bottom by <math>{10}/{3}\,</math>,

-	{{~~Displayed math~~\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{20}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{20}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \dfrac{1}{20}\cdot \dfrac{10}{3}\,</math>.}}

	Then, we remove the common factor 10,		Then, we remove the common factor 10,

-	{{~~Displayed math~~\|\|<math>\dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\dfrac{1}{20}\cdot \dfrac{10}{3}=\dfrac{1}{2\cdot{}\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{3}=\dfrac{1}{2\cdot 3}=\dfrac{1}{6}\,</math>.}}


Zeile 18:		Zeile 18:
	Another way to calculate the expression is to divide it up into two separate terms		Another way to calculate the expression is to divide it up into two separate terms

-	{{~~Displayed math~~\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\dfrac{3}{10}\vphantom{\Biggl(}} = \frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}\,</math>.}}

	We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3		We simplify both double fractions on the right-hand side by multiplying top and bottom by 10/3

-	{{~~Displayed math~~\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{\,\dfrac{1}{4}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{3}{10}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{4}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,}-\frac{\,\dfrac{1}{5}\cdot \dfrac{10}{3}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}3}{\rlap{\,/}10}\cdot \dfrac{\rlap{\,/}10}{\rlap{/}3}\vphantom{\Biggl(}\,} = \frac{1}{4}\cdot \frac{10}{3}-\frac{1}{5}\cdot \frac{10}{3}\,</math>.}}

	Instead of multiplying, respectively, by <math>4\cdot 3</math> and <math>5\cdot 3</math>, we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by 5 and the second by 4, we obtain the common denominator		Instead of multiplying, respectively, by <math>4\cdot 3</math> and <math>5\cdot 3</math>, we keep the numerators factorized and observe that if we multiply the top and bottom of the first fraction by 5 and the second by 4, we obtain the common denominator

-	{{~~Displayed math~~\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{10}{4\cdot 3}-\frac{10}{5\cdot 3}=\frac{10\cdot 5}{4\cdot 3\cdot 5}-\frac{10\cdot 4}{5\cdot 3\cdot 4}=\frac{50-40}{3\cdot 4\cdot 5}=\frac{10}{3\cdot 4\cdot 5}\,</math>.}}

	Because <math>10=2\cdot 5</math> and <math>4=2\cdot 2</math>, we can cancel out the common factors 2 and 5 and obtain the answer		Because <math>10=2\cdot 5</math> and <math>4=2\cdot 2</math>, we can cancel out the common factors 2 and 5 and obtain the answer

-	{{~~Displayed math~~\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}	+	{{Abgesetzte Formel\|\|<math>\frac{10}{3\cdot 4\cdot 5}=\frac{\rlap{/}2{}\cdot{}\rlap{/}5}{3\cdot 2\cdot{}\rlap{/}2\cdot{}\rlap{/}5}=\frac{1}{3\cdot 2}=\frac{1}{6}\,</math>.}}