Lösung 2.2:3d
Aus Online Mathematik Brückenkurs 1
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There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side | There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr) | \biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr) | ||
&= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt] | &= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt] | ||
| Zeile 17: | Zeile 17: | ||
Collecting up terms, the left-hand side becomes | Collecting up terms, the left-hand side becomes | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
&\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt] | &\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt] | ||
&\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt] | &\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt] | ||
| Zeile 26: | Zeile 26: | ||
and because <math>33=3\cdot 11</math>, <math>9=3\cdot 3</math> and <math>4=2\cdot 2</math>, the whole equation can rewritten as | and because <math>33=3\cdot 11</math>, <math>9=3\cdot 3</math> and <math>4=2\cdot 2</math>, the whole equation can rewritten as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{11}{3\cdot 2\cdot 2}\cdot \frac{1}{x}-\frac{3\cdot 11}{2\cdot 3\cdot 3}=0\,\textrm{.}</math>}} |
Taking out common factors, we get | Taking out common factors, we get | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{11}{3\cdot 2}\biggl(\frac{1}{2x}-1\biggr)=0</math>}} |
and then we see that the equation has the solution <math>x=1/2</math>. | and then we see that the equation has the solution <math>x=1/2</math>. | ||
| Zeile 36: | Zeile 36: | ||
Finally, we substitute <math>x=1/2</math> into the original equation to check that we have calculated correctly. | Finally, we substitute <math>x=1/2</math> into the original equation to check that we have calculated correctly. | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
& \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt] | & \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt] | ||
&\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt] | &\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt] | ||
Version vom 08:27, 22. Okt. 2008
There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side
| \displaystyle \begin{align}
\biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr) &= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt] &= \frac{1}{2x^{2}} + \frac{1}{x} - \frac{3}{4x} - \frac{3}{2}\\[5pt] &= \frac{1}{2x^{2}} + \frac{1}{4x} - \frac{3}{2}\,,\\[15pt] \biggl(\frac{1}{2x}-\frac{2}{3}\biggr)^{2} &= \frac{1}{(2x)^{2}} - 2\cdot\frac{1}{2x}\cdot\frac{2}{3} + \biggl(\frac{2}{3}\biggr)^{2}\\[5pt] &= \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\,,\\[15pt] \biggl(\frac{1}{2x}+\frac{1}{3}\biggr)\biggl(\frac{1}{2x}-\frac{1}{3}\biggr) &= \bigl\{\,\text{difference of two squares}\,\}\\[5pt] &= \frac{1}{(2x)^{2}}-\frac{1}{3^{2}}\\[5pt] &= \frac{1}{4x^{2}}-\frac{1}{9}\,\textrm{.} \end{align} |
Collecting up terms, the left-hand side becomes
| \displaystyle \begin{align}
&\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt] &\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt] &\qquad\quad{}= \frac{2-1-1}{4}\,\frac{1}{x^{2}} + \frac{3+2\cdot 4}{3\cdot 4}\,\frac{1}{x} + \frac{-3\cdot 9 - 4\cdot 2 + 1\cdot 2}{2\cdot 9}\\[7pt] &\qquad\quad{}=\frac{11}{3\cdot 4}\cdot \frac{1}{x}-\frac{33}{2\cdot 9} \end{align} |
and because \displaystyle 33=3\cdot 11, \displaystyle 9=3\cdot 3 and \displaystyle 4=2\cdot 2, the whole equation can rewritten as
| \displaystyle \frac{11}{3\cdot 2\cdot 2}\cdot \frac{1}{x}-\frac{3\cdot 11}{2\cdot 3\cdot 3}=0\,\textrm{.} |
Taking out common factors, we get
| \displaystyle \frac{11}{3\cdot 2}\biggl(\frac{1}{2x}-1\biggr)=0 |
and then we see that the equation has the solution \displaystyle x=1/2.
Finally, we substitute \displaystyle x=1/2 into the original equation to check that we have calculated correctly.
| \displaystyle \begin{align}
& \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt] &\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt] &\qquad\quad{}= 1 - \biggl(\frac{1}{3}\biggr)^{2} - \frac{4}{3}\cdot\frac{2}{3}\\[5pt] &\qquad\quad{}= 1 - \frac{1}{9} - \frac{8}{9}\\[5pt] &\qquad\quad{}= 0\,\textrm{.} \end{align} |
