Lösung 2.2:3d
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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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 | |
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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 | + | |
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| + | {{Displayed math||<math>\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}</math>}} | ||
Collecting up terms, the left-hand side becomes | Collecting up terms, the left-hand side becomes | ||
| + | {{Displayed math||<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] | ||
| + | &\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}</math>}} | ||
| - | + | 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 | |
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| - | and because | + | |
| - | <math>33=3\ | + | |
| - | <math>9=3\ | + | |
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| - | <math>4=2\ | + | |
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| + | {{Displayed math||<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 | ||
| + | {{Displayed math||<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>. | |
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| - | and then we see that the equation has the solution | + | |
| - | <math>x= | + | |
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| + | Finally, we substitute <math>x=1/2</math> into the original equation to check that we have calculated correctly. | ||
| - | <math>\begin{align} | + | {{Displayed math||<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] |
| - | & | + | &\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}</math> | + | \end{align}</math>}} |
Version vom 10:18, 24. Sep. 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
Collecting up terms, the left-hand side becomes
and because \displaystyle 33=3\cdot 11, \displaystyle 9=3\cdot 3 and \displaystyle 4=2\cdot 2, the whole equation can rewritten as
Taking out common factors, we get
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.
