Solution 2.1:8c

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m (Lösning 2.1:8c moved to Solution 2.1:8c: Robot: moved page)
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When we come across large and complicated expressions, we have to work step by step;
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<center> [[Image:2_1_8c-1(2).gif]] </center>
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as a first goal, we can multiply the top and bottom of the fraction
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<center> [[Image:2_1_8c-2(2).gif]] </center>
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<math>\frac{1}{1+\frac{1}{1+x}}</math>
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by
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<math>1+x</math>, so as to reduce it to an expression having one fraction sign:
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<math>\begin{align}
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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
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<math>x+2</math>,
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so as to obtain the final answer,
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<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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& \\
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& \frac{x+2}{x+2+x+1}=\frac{x+2}{2x+3} \\
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& \\
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\end{align}</math>

Revision as of 13:52, 16 September 2008

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+\frac{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+\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)}} \\ & \\ & =\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}} \\ \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+\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)} \\ & \\ & \frac{x+2}{x+2+x+1}=\frac{x+2}{2x+3} \\ & \\ \end{align}

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