Lösung 1.2:6

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
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:
+
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
 +
{{Displayed math||<math>\frac{2}{\,3+\dfrac{1}{2}\vphantom{\Biggl(}\,}\ ,\quad \frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\,\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}\,}\quad\text{and}\quad\frac{3}{\,2-\dfrac{2}{7}\vphantom{\Biggl(}\,}\,</math>.}}
-
<math>\frac{2}{3+\frac{1}{2}}\ ,\quad \frac{\frac{1}{2}}{\frac{1}{4}-\frac{1}{3}}</math>
+
We can do this by multiplying the top and bottom of each fraction by 2, 12 and 7 respectively, so as to get rid of the partial fractions
-
and
+
-
<math>\frac{3}{2-\frac{2}{7}}</math>
+
-
 
+
{{Displayed math||<math>\begin{align}
-
We can do this by multiplying the top and bottom of each fraction by
+
\frac{2}{3+\dfrac{1}{2}\vphantom{\Biggl(}} &= \frac{2\cdot 2}{\left( 3+\dfrac{1}{2} \right)\cdot 2\vphantom{\Biggl(}} = \frac{4}{3\cdot 2+\dfrac{1}{2}\cdot 2\vphantom{\Biggl(}} = \frac{4}{6+1} = \frac{4}{7}\,,\\[5pt]
-
<math>2</math>
+
\frac{\dfrac{1}{2}\vphantom{\Biggl(}}{\dfrac{1}{4}-\dfrac{1}{3}\vphantom{\Biggl(}} &= \frac{\dfrac{1}{2}\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{4}-\dfrac{1}{3} \right)\cdot 12\vphantom{\Biggl(}} = \frac{6}{\dfrac{12}{4}-\dfrac{12}{3}\vphantom{\Biggl(}} = \frac{6}{3-4} = \frac{6}{-1} = -6\,,\\[10pt]
-
,
+
\frac{3}{2-\dfrac{2}{7}\vphantom{\Biggl(}} &= \frac{3\cdot 7}{\left( 2-\dfrac{2}{7} \right)\cdot 7\vphantom{\Biggl(}} = \frac{21}{2\cdot 7-\dfrac{2}{7}\cdot 7\vphantom{\Biggl(}} = \frac{21}{14-2} = \frac{21}{12}\,\textrm{.}
-
<math>12</math>
+
\end{align}</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
The whole expression therefore equals
 +
{{Displayed math||<math>\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}}\,</math>.}}
-
<math>\frac{\frac{4}{7}-6}{\frac{1}{2}-\frac{21}{12}}</math>
+
If we multiply the tops and bottoms of the fractions 4/7, 1/2 and 21/12 in the main fraction by their lowest common denominator, <math>7\cdot 12</math>, we obtain integers in the numerator and denominator
 +
{{Displayed math||<math>\begin{align}
 +
\frac{\dfrac{4}{7}-6\vphantom{\Biggl(}}{\dfrac{1}{2}-\dfrac{21}{12}\vphantom{\Biggl(}} &= \frac{\left( \dfrac{4}{7}-6 \right)\cdot 7\cdot 12\vphantom{\Biggl(}}{\left( \dfrac{1}{2}-\dfrac{21}{12} \right)\cdot 7\cdot 12\vphantom{\Biggl(}} = \frac{4\cdot 12-6\cdot 7\cdot 12}{7\cdot 6-21\cdot 7}\\[10pt]
 +
& =\frac{( 4-6\cdot 7)\cdot 12}{( 6-21)\cdot 7} = \frac{-38\cdot 12}{-15\cdot 7} = \frac{38\cdot 12}{15\cdot 7}\,\textrm{.}
 +
\end{align}</math>}}
 +
By factorizing 12, 15 and 38,
-
If we multiply the tops and bottoms of the fractions
+
{{Displayed math||<math>\begin{align}
-
<math></math>
+
12 &= 2\cdot 6 = 2\cdot 2\cdot 3\,,\\
-
 
+
15 &= 3\cdot 5\,,\\
-
<math>{4}/{7}\;</math>
+
38 &= 2\cdot 19\,,\\
-
,
+
\end{align}</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
the answer can be simplified to
-
 
+
{{Displayed math||<math>\frac{38\cdot 12}{15\cdot 7}=\frac{2\cdot 19\cdot 2\cdot 2\cdot{}\rlap{/}3}{\rlap{/}3\cdot 5\cdot 7}=\frac{153}{35}\,</math>.}}
-
<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 14:22, 19. 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

Vorlage:Displayed math

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

Vorlage:Displayed math

The whole expression therefore equals

Vorlage:Displayed math

If we multiply the tops and bottoms of the fractions 4/7, 1/2 and 21/12 in the main fraction by their lowest common denominator, \displaystyle 7\cdot 12, we obtain integers in the numerator and denominator

Vorlage:Displayed math

By factorizing 12, 15 and 38,

Vorlage:Displayed math

the answer can be simplified to

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:6“