Question

The expression $\frac{1}{1+\frac{x}{1-\frac{x}{1-x}}} \div \frac{\frac{1}{1-x}+\frac{1}{1+x}}{\frac{1}{1-x}-\frac{1}{1+x}}$ when simplified reduces to

Answer 1

$\frac{x(1-2x)}{1-x-x^2}$

Answer 2

Let the given expression be $E$. We can write $E$ as $E = A \div B$, where

$A = \frac{1}{1+\frac{x}{1-\frac{x}{1-x}}}$ and $B = \frac{\frac{1}{1-x}+\frac{1}{1+x}}{\frac{1}{1-x}-\frac{1}{1+x}}$.

First, simplify the expression in the denominator of $A$:

$1 - \frac{x}{1-x} = \frac{1(1-x) - x}{1-x} = \frac{1-x-x}{1-x} = \frac{1-2x}{1-x}$.

Now substitute this back into the denominator of $A$:

$1+\frac{x}{1-\frac{x}{1-x}} = 1+\frac{x}{\frac{1-2x}{1-x}} = 1+\frac{x(1-x)}{1-2x}$.

Combine the terms in the denominator:

$1+\frac{x(1-x)}{1-2x} = \frac{1(1-2x) + x(1-x)}{1-2x} = \frac{1-2x + x-x^2}{1-2x} = \frac{1-x-x^2}{1-2x}$.

So, $A = \frac{1}{\frac{1-x-x^2}{1-2x}} = \frac{1-2x}{1-x-x^2}$.

Next, simplify the expression $B$.

The numerator of $B$ is $\frac{1}{1-x}+\frac{1}{1+x}$. Find a common denominator $(1-x)(1+x) = 1-x^2$:

$\frac{1}{1-x}+\frac{1}{1+x} = \frac{1(1+x) + 1(1-x)}{(1-x)(1+x)} = \frac{1+x+1-x}{1-x^2} = \frac{2}{1-x^2}$.

The denominator of $B$ is $\frac{1}{1-x}-\frac{1}{1+x}$. Find a common denominator $(1-x)(1+x) = 1-x^2$:

$\frac{1}{1-x}-\frac{1}{1+x} = \frac{1(1+x) - 1(1-x)}{(1-x)(1+x)} = \frac{1+x-1+x}{1-x^2} = \frac{2x}{1-x^2}$.

Now, substitute these back into $B$:

$B = \frac{\frac{2}{1-x^2}}{\frac{2x}{1-x^2}}$. Assuming $1-x^2 \neq 0$, we can cancel the denominator $(1-x^2)$:

$B = \frac{2}{2x} = \frac{1}{x}$.

Finally, perform the division $E = A \div B$:

$E = \frac{1-2x}{1-x-x^2} \div \frac{1}{x} = \frac{1-2x}{1-x-x^2} \times x = \frac{x(1-2x)}{1-x-x^2}$.

The simplified expression is $\frac{x(1-2x)}{1-x-x^2}$.