Solution 2.1:8c

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

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