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

• A. $\frac{2x}{x^2+2x+2}$
• B. $\frac{2x}{x^2-1}$
• C. 1
• D. $x^2-1$

Answer 1

Answer: (A)

$\frac{2x}{x^2+2x+2}$

Answer 2

Let the given expression be $E$. We can write $E$ as the division of two parts, $P_1$ and $P_2$: $E = P_1 \div P_2$ where $P_1 = \frac{1}{1+\frac{x}{1-\frac{x}{1-x}}}$ and $P_2 = \frac{\frac{1}{1-x}+\frac{1}{1+x}}{\frac{1}{1-x}-\frac{1}{1+x}}$.

First, simplify $P_1$: The innermost denominator is $1-\frac{x}{1-x}$. $1-\frac{x}{1-x} = \frac{1(1-x)}{1-x} - \frac{x}{1-x} = \frac{1-x-x}{1-x} = \frac{1-2x}{1-x}$.

Substitute this back into the denominator of $P_1$: $1+\frac{x}{1-\frac{x}{1-x}} = 1+\frac{x}{\frac{1-2x}{1-x}} = 1 + x \cdot \frac{1-x}{1-2x} = 1 + \frac{x(1-x)}{1-2x}$.

Combine the terms in the denominator: $1 + \frac{x-x^2}{1-2x} = \frac{1(1-2x)}{1-2x} + \frac{x-x^2}{1-2x} = \frac{1-2x+x-x^2}{1-2x} = \frac{1-x-x^2}{1-2x}$.

So, $P_1 = \frac{1}{\frac{1-x-x^2}{1-2x}} = \frac{1-2x}{1-x-x^2}$.

Next, simplify $P_2$: The numerator of $P_2$ is $\frac{1}{1-x}+\frac{1}{1+x}$. $\frac{1}{1-x}+\frac{1}{1+x} = \frac{1(1+x)}{(1-x)(1+x)} + \frac{1(1-x)}{(1+x)(1-x)} = \frac{1+x+1-x}{(1-x)(1+x)} = \frac{2}{1-x^2}$.

The denominator of $P_2$ is $\frac{1}{1-x}-\frac{1}{1+x}$. $\frac{1}{1-x}-\frac{1}{1+x} = \frac{1(1+x)}{(1-x)(1+x)} - \frac{1(1-x)}{(1+x)(1-x)} = \frac{1+x-(1-x)}{(1-x)(1+x)} = \frac{1+x-1+x}{(1-x)(1+x)} = \frac{2x}{1-x^2}$.

So, $P_2 = \frac{\frac{2}{1-x^2}}{\frac{2x}{1-x^2}}$. $P_2 = \frac{2}{1-x^2} \div \frac{2x}{1-x^2} = \frac{2}{1-x^2} \cdot \frac{1-x^2}{2x}$. Assuming $1-x^2 \neq 0$ (i.e., $x \neq \pm 1$), we can cancel the term $(1-x^2)$: $P_2 = \frac{2}{2x} = \frac{1}{x}$. This is valid provided $x \neq 0$.

Now, compute $E = P_1 \div P_2$: $E = \frac{1-2x}{1-x-x^2} \div \frac{1}{x} = \frac{1-2x}{1-x-x^2} \cdot x = \frac{x(1-2x)}{1-x-x^2} = \frac{x-2x^2}{1-x-x^2}$.

The simplified expression is $\frac{x-2x^2}{1-x-x^2}$. This does not match any of the given options. There is likely an error in the question or the options provided.

However, if forced to choose from the options, and given that option A matched for $x=-1/2$, let's re-examine the original expression and try to see if there's a simplification I missed. I am confident in the algebraic simplification steps.

Let's consider the possibility that the expression is intended to be equal to option A for all valid $x$, which would imply that $\frac{x(1-2x)}{1-x-x^2} = \frac{2x}{x^2+2x+2}$ is an identity. But we showed this is not true.

Despite the discrepancy, let's assume the intended answer is A.