Solution 2.1:8c
From Förberedande kurs i matematik 1
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| - | {{ | + | When we come across large and complicated expressions, we have to work step by step; as a first goal, we can multiply the top and bottom of the fraction |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\frac{1}{1+\dfrac{1}{1+x}}</math>}} |
| + | |||
| + | by <math>1+x</math>, so as to reduce it to an expression having one fraction sign | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}} | ||
| + | &= \frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}\cdot\dfrac{1+x}{1+x}}\\[8pt] | ||
| + | &= \frac{1}{1+\dfrac{1+x}{\Bigl(1+\dfrac{1}{1+x}\Bigr)(1+x)}}\\[8pt] | ||
| + | &= \frac{1}{1+\dfrac{1+x}{1+x+\dfrac{1+x}{1+x}}}\\[8pt] | ||
| + | &= \frac{1}{1+\dfrac{1+x}{1+x+1}}\\[8pt] | ||
| + | &= \frac{1}{1+\dfrac{x+1}{x+2}}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The next step is to multiply the top and bottom of our new expression by | ||
| + | <math>x+2</math>, so as to obtain the final answer, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{1}{1+\dfrac{x+1}{x+2}}\cdot\frac{x+2}{x+2} | ||
| + | &= \frac{x+2}{\Bigl(1+\dfrac{x+1}{x+2}\Bigr)(x+2)}\\[8pt] | ||
| + | &= \frac{x+2}{x+2+\dfrac{x+1}{x+2}(x+2)}\\[8pt] | ||
| + | &= \frac{x+2}{x+2+x+1}\\[8pt] | ||
| + | &= \frac{x+2}{2x+3}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
When we come across large and complicated expressions, we have to work step by step; as a first goal, we can multiply the top and bottom of the fraction
| \displaystyle \frac{1}{1+\dfrac{1}{1+x}} |
by \displaystyle 1+x, so as to reduce it to an expression having one fraction sign
| \displaystyle \begin{align}
\frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}} &= \frac{1}{1+\dfrac{1}{1+\dfrac{1}{1+x}}\cdot\dfrac{1+x}{1+x}}\\[8pt] &= \frac{1}{1+\dfrac{1+x}{\Bigl(1+\dfrac{1}{1+x}\Bigr)(1+x)}}\\[8pt] &= \frac{1}{1+\dfrac{1+x}{1+x+\dfrac{1+x}{1+x}}}\\[8pt] &= \frac{1}{1+\dfrac{1+x}{1+x+1}}\\[8pt] &= \frac{1}{1+\dfrac{x+1}{x+2}}\,\textrm{.} \end{align} |
The next step is to multiply the top and bottom of our new expression by \displaystyle x+2, so as to obtain the final answer,
| \displaystyle \begin{align}
\frac{1}{1+\dfrac{x+1}{x+2}}\cdot\frac{x+2}{x+2} &= \frac{x+2}{\Bigl(1+\dfrac{x+1}{x+2}\Bigr)(x+2)}\\[8pt] &= \frac{x+2}{x+2+\dfrac{x+1}{x+2}(x+2)}\\[8pt] &= \frac{x+2}{x+2+x+1}\\[8pt] &= \frac{x+2}{2x+3}\,\textrm{.} \end{align} |
