Solution 2.1:6d
From Förberedande kurs i matematik 1
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| - | {{ | + | First, we simplify the numerator and denominator for the whole fraction by rewriting as |
| - | + | ||
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | a-b+\frac{b^{2}}{a+b} | ||
| + | &= (a-b)\cdot\frac{a+b}{a+b} + \frac{b^{2}}{a+b} = \frac{(a-b)\cdot (a+b)+b^{2}}{a+b}\\[5pt] | ||
| + | &= \frac{a^{2}-b^{2}+b^{2}}{a+b} = \frac{a^{2}}{a+b}\,,\\[15pt] | ||
| + | 1-\biggl(\frac{a-b}{a+b}\biggr)^{2} | ||
| + | &= 1-\frac{(a-b)^{2}}{(a+b)^{2}} = \frac{(a+b)^{2}}{(a+b)^{2}} - \frac{(a-b)^{2}}{(a+b)^{2}}\\[5pt] | ||
| + | &= \frac{(a+b)^{2}-(a-b)^{2}}{(a+b)^{2}}\\[5pt] | ||
| + | &= \frac{(a^{2}+2ab+b^{2})-(a^{2}-2ab+b^{2})}{(a+b)^{2}} = \frac{4ab}{(a+b)^{2}}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The whole fraction is therefore | ||
| + | |||
| + | {{Displayed math||<math>\frac{a-b+\dfrac{b^{2}}{a+b}}{1-\biggl(\dfrac{a-b}{a+b}\biggr)^{2}} = \frac{\dfrac{a^{2}}{a+b}}{\dfrac{4ab}{(a+b)^{2}}} = \frac{a^{2}}{a+b}\cdot\frac{(a+b)^{2}}{4ab} = \frac{a(a+b)}{4b}\,\textrm{.}</math>}} | ||
Current revision
First, we simplify the numerator and denominator for the whole fraction by rewriting as
| \displaystyle \begin{align}
a-b+\frac{b^{2}}{a+b} &= (a-b)\cdot\frac{a+b}{a+b} + \frac{b^{2}}{a+b} = \frac{(a-b)\cdot (a+b)+b^{2}}{a+b}\\[5pt] &= \frac{a^{2}-b^{2}+b^{2}}{a+b} = \frac{a^{2}}{a+b}\,,\\[15pt] 1-\biggl(\frac{a-b}{a+b}\biggr)^{2} &= 1-\frac{(a-b)^{2}}{(a+b)^{2}} = \frac{(a+b)^{2}}{(a+b)^{2}} - \frac{(a-b)^{2}}{(a+b)^{2}}\\[5pt] &= \frac{(a+b)^{2}-(a-b)^{2}}{(a+b)^{2}}\\[5pt] &= \frac{(a^{2}+2ab+b^{2})-(a^{2}-2ab+b^{2})}{(a+b)^{2}} = \frac{4ab}{(a+b)^{2}}\,\textrm{.} \end{align} |
The whole fraction is therefore
| \displaystyle \frac{a-b+\dfrac{b^{2}}{a+b}}{1-\biggl(\dfrac{a-b}{a+b}\biggr)^{2}} = \frac{\dfrac{a^{2}}{a+b}}{\dfrac{4ab}{(a+b)^{2}}} = \frac{a^{2}}{a+b}\cdot\frac{(a+b)^{2}}{4ab} = \frac{a(a+b)}{4b}\,\textrm{.} |
