Lösung 2.1:6b
Aus Online Mathematik Brückenkurs 1
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| Zeile 1: | Zeile 1: | ||
The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator | The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{x}{x-2}+\frac{x}{x+3}-2 | \frac{x}{x-2}+\frac{x}{x+3}-2 | ||
&= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] | &= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] | ||
| Zeile 11: | Zeile 11: | ||
Now, collect the terms in the numerator | Now, collect the terms in the numerator | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] | \frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] | ||
&= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} | &= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} | ||
Version vom 08:24, 22. Okt. 2008
The lowest common denominator for the three terms is \displaystyle (x-2)(x+3) and we expand each term so that all terms have the same denominator
| \displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] &= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] &= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt] &= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.} \end{align} |
Now, collect the terms in the numerator
| \displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] &= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} \end{align} |
Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.
