Solution 2.1:8a
From Förberedande kurs i matematik 1
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- | An expression which consists of several fraction signs can be rewritten in terms of one fraction sign | + | An expression which consists of several fraction signs can be rewritten in terms of one fraction sign by systematically eliminating all partial fractions. |
- | by systematically eliminating all partial fractions. | + | |
- | In our expression, we multiply the top and bottom of the main fraction by | + | In our expression, we multiply the top and bottom of the main fraction by <math>x+1</math> (so as to get rid of <math>x+1</math> from the numerator), |
- | <math>x+1</math> | + | |
- | (so as to get rid of | + | |
- | <math>x+1</math> | + | |
- | from the numerator), | + | |
- | + | {{Displayed math||<math>\frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.}</math>}} | |
- | <math>\frac{\ | + |
Current revision
An expression which consists of several fraction signs can be rewritten in terms of one fraction sign by systematically eliminating all partial fractions.
In our expression, we multiply the top and bottom of the main fraction by \displaystyle x+1 (so as to get rid of \displaystyle x+1 from the numerator),
\displaystyle \frac{\,\dfrac{x}{x+1}\,}{3+x} = \frac{\,\dfrac{x}{x+1}\,}{3+x}\cdot\frac{x+1}{x+1} = \frac{\,\dfrac{x}{x+1}\cdot (x+1)\,}{\,(3+x)(x+1)\,} = \frac{x}{(3+x)(x+1)}\,\textrm{.} |