Aus Online Mathematik Brückenkurs 1

Version vom 08:20, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 1.2:6 moved to Solution 1.2:6: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 14:48, 11. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:
-	<center> [[Image:1_2_6-1(2).gif]] </center>	+
-	<center> [[Image:1_2_6-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>\frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}</math>
		+	and
		+	<math>\frac{3}{2-\frac{2}{7}}</math>
		+
		+
		+	We can do this by multiplying the top and bottom of each fraction by
		+	<math>2</math>
		+	,
		+	<math>12</math>
		+	and
		+	<math>7</math>
		+	respectively, so as to get rid of the partial fractions:
		+
		+
		+	<math>\begin{align}
		+	& \frac{2}{3+\frac{1}{2}}=\ \frac{2\centerdot 2}{\left( 3+\frac{1}{2} \right)\centerdot 2}=\frac{4}{3\centerdot 2+\frac{1}{2}\centerdot 2}=\frac{4}{6+1}=\frac{4}{7} \\
		+	& \\
		+	& \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}=\frac{\frac{1}{2}\centerdot 12}{\left( \frac{1}{4}-\frac{1}{3} \right)\centerdot 12}=\frac{6}{\frac{12}{4}-\frac{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\
		+	& \\
		+	& \frac{3}{2-\frac{2}{7}}=\frac{3\centerdot 7}{\left( 2-\frac{2}{7} \right)\centerdot 7}=\frac{21}{2\centerdot 7-\frac{2}{7}\centerdot 7}=\frac{21}{14-2}=\frac{21}{12} \\
		+	\end{align}</math>
		+
		+	The whole expression therefore equals
		+
		+
		+	<math>\frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}</math>
		+
		+
		+
		+	If we multiply the tops and bottoms of the fractions
		+	<math></math>
		+
		+	<math>{4}/{7}\;</math>
		+	,
		+	<math>{1}/{2}\;</math>
		+	and
		+	<math>{21}/{12}\;</math>
		+	in the main fraction by
		+	their lowest common denominator,
		+	<math>7\centerdot 12</math>
		+	, we obtain integers in the numerator and denominator:
		+
		+
		+	<math>\begin{align}
		+	& \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\
		+	& \\
		+	& =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\
		+	\end{align}</math>
		+
		+
		+	By factorizing
		+	<math>12</math>
		+	,
		+	<math>15</math>
		+	and
		+	<math>38</math>
		+	,
		+
		+
		+	<math>\begin{align}
		+	& 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\
		+	& 15=3\centerdot 5 \\
		+	& 38=2\centerdot 19 \\
		+	\end{align}</math>
		+
		+	the answer can be simplified to
		+
		+
		+	<math>\frac{38\centerdot 12}{15\centerdot 7}=\frac{2\centerdot 19\centerdot 2\centerdot 2\centerdot 3}{3\centerdot 5\centerdot 7}=\frac{153}{35}</math>

---

Version vom 14:48, 11. Sep. 2008

When we work with large expressions, it is often best to proceed step by step. A first step on the way can be to simplify all the parts:

\displaystyle \frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}} and \displaystyle \frac{3}{2-\frac{2}{7}}

We can do this by multiplying the top and bottom of each fraction by \displaystyle 2 , \displaystyle 12 and \displaystyle 7 respectively, so as to get rid of the partial fractions:

\displaystyle \begin{align} & \frac{2}{3+\frac{1}{2}}=\ \frac{2\centerdot 2}{\left( 3+\frac{1}{2} \right)\centerdot 2}=\frac{4}{3\centerdot 2+\frac{1}{2}\centerdot 2}=\frac{4}{6+1}=\frac{4}{7} \\ & \\ & \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}=\frac{\frac{1}{2}\centerdot 12}{\left( \frac{1}{4}-\frac{1}{3} \right)\centerdot 12}=\frac{6}{\frac{12}{4}-\frac{12}{3}}=\frac{6}{3-4}=\frac{6}{-1}=-6 \\ & \\ & \frac{3}{2-\frac{2}{7}}=\frac{3\centerdot 7}{\left( 2-\frac{2}{7} \right)\centerdot 7}=\frac{21}{2\centerdot 7-\frac{2}{7}\centerdot 7}=\frac{21}{14-2}=\frac{21}{12} \\ \end{align}

The whole expression therefore equals

\displaystyle \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}

If we multiply the tops and bottoms of the fractions \displaystyle

\displaystyle {4}/{7}\; , \displaystyle {1}/{2}\; and \displaystyle {21}/{12}\; in the main fraction by their lowest common denominator, \displaystyle 7\centerdot 12 , we obtain integers in the numerator and denominator:

\displaystyle \begin{align} & \frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}=\frac{\left( \frac{4}{7}-6 \right)\centerdot 7\centerdot 12}{\left( \frac{1}{2}-\frac{21}{12} \right)\centerdot 7\centerdot 12}=\frac{4\centerdot 12-6\centerdot 7\centerdot 12}{7\centerdot 6-21\centerdot 7} \\ & \\ & =\frac{\left( 4-6\centerdot 7 \right)\centerdot 12}{\left( 6-21 \right)\centerdot 7}=\frac{-38\centerdot 12}{-15\centerdot 7}=\frac{38\centerdot 12}{15\centerdot 7} \\ \end{align}

By factorizing \displaystyle 12 , \displaystyle 15 and \displaystyle 38 ,

\displaystyle \begin{align} & 12=2\centerdot 6=2\centerdot 2\centerdot 3 \\ & 15=3\centerdot 5 \\ & 38=2\centerdot 19 \\ \end{align}

the answer can be simplified to

\displaystyle \frac{38\centerdot 12}{15\centerdot 7}=\frac{2\centerdot 19\centerdot 2\centerdot 2\centerdot 3}{3\centerdot 5\centerdot 7}=\frac{153}{35}