Complex Fractions
A complex fraction is a fraction with a numerator or denominator that also contains a fraction. As an example, the following is a complex fraction:
\[ \frac{2}{3-\frac{1}{2}}. \]
The easiest way to approach such questions is to first express(if possible) the numerator and the denominator as fractions.
In the above example, it would be like
\[\frac{\frac{2}{1}}{\frac{6-1}{2}}.\]
Now, we can simplify it using division
\[\frac{2}{1} \div \frac{6-1}{2}\]
\[= \frac{2}{1} \times \frac{2}{5}\]
\[= \frac{4}{5}.\]
Another way to put it is
\[\frac{\frac{2}{1}}{\frac{6-1}{2}}\]
\[= \frac{2(2)}{1(6-1)}\]
\[= \frac{2(2)}{1(5)}\]
\[= \frac{4}{5}.\]
Simplify \[ \frac{\frac{4}{3}}{\frac{2}{3}+1} .\]
Multiplying both the numerator and denominator by \(3,\) which is the common denominator of \(\frac{4}{3}\) and \( \frac{2}{3},\) we have
\[ \begin{align} \frac{\frac{4}{3}}{\frac{2}{3}+1} &=\frac{3\cdot \frac{4}{3}}{3\cdot \frac{2}{3} + 3 \cdot 1} \\ &= \frac{4}{2+3} \\ &= \frac{4}{5}. \end{align} \]
Simplify \[ 1+\frac{1+ \frac{1+\frac{1}{x}}{x}}{x}.\]
\[ \begin{align} 1+\frac{1+ \frac{1+\frac{1}{x}}{x}}{x} &= 1+ \frac{1+\frac{\frac{x+1}{x}}{x}}{x} \\ &= 1+\frac{1+\frac{x+1}{x^2}}{x} \\ &= 1+\frac{\frac{x^2+x+1}{x^2}}{x} \\ &= 1+ \frac{x^2+x+1}{x^3} \\ &= \frac{x^3+x^2+x+1}{x^3}. \end{align} \]
Simplify.
\[ \large\frac { \frac { x+1 }{ x-1 } +\frac { x-1 }{ x+1 } }{ \frac { x+1 }{ x-1 } -\frac { x-1 }{ x+1 } }\]
\[\large\frac { \frac { x+1 }{ x-1 } +\frac { x-1 }{ x+1 } }{ \frac { x+1 }{ x-1 } -\frac { x-1 }{ x+1 } } =\frac { \frac { { (x+1) }^{ 2 }+{ (x-1) }^{ 2 } }{ (x-1)(x+1) } }{ \frac { { (x+1) }^{ 2 }-{ (x-1) }^{ 2 } }{ (x-1)(x+1) } } =\frac { \frac { { x }^{ 2 }+2x+1+{ x }^{ 2 }-2x+1 }{ { x }^{ 2 }-1 } }{ \frac { { x }^{ 2 }+2x+1-({ x }^{ 2 }-2x+1) }{ { x }^{ 2 }-1 } } \]
\[\large=\frac { 2{ x }^{ 2 }+2 }{ { x }^{ 2 }-1 } \times \frac { { x }^{ 2 }-1 }{ 4x }\]
\[\large=\frac { { x }^{ 2 }+1 }{ 2x } ,\quad x\neq 0,\quad x\neq \pm 1\\\]
Simplify.
\[ \frac{x-\frac{1}{x}}{1+\frac{x}{1-x}}\div \frac{x^{2}-2x+1}{x} \]
\[ =\frac{\frac{x^{2}-1}{x}}{\frac{1-x+x}{1-x}}\div \frac{(x-1)^{2}}{x} =\frac{x^{2}-1}{x}(1-x)\times \frac{x}{(x-1)^{2}} \]
\[=\frac{(x-1)(x+1)(1-x)}{(x-1)^{2}} = \frac{-(x+1)(x-1)}{x-1}=-x-1, x\neq 0, 1\]