b) \( \frac{1}{x+1}+\frac{1}{x-1}=\frac{4}{4 x^{2}-4} \)

To solve the equation \( \frac{1}{x+1}+\frac{1}{x-1}=\frac{4}{4 x^{2}-4} \), we start simplifying both sides. On the left side, we have a common denominator of \((x + 1)(x - 1)\), so we can rewrite it as: \[ \frac{(x-1) + (x+1)}{(x+1)(x-1)} = \frac{2x}{x^2 - 1} \] The right side can be simplified as follows: \[ \frac{4}{4(x^2 - 1)} = \frac{1}{x^2 - 1} \] Now we can set the two expressions equal to each other: \[ \frac{2x}{x^2 - 1} = \frac{1}{x^2 - 1} \] Since the denominators are the same, we can cross-multiply: \[ 2x = 1 \] Now, solving for \(x\): \[ x = \frac{1}{2} \] Lastly, we need to check if \(x = \frac{1}{2}\) is a valid solution by substituting it back into the original equation. It appears that both sides yield a true statement, confirming that the solution is correct. Thus, the solution to the equation is: \[ x = \frac{1}{2} \]