h. \( \frac{2 x}{x^{2}-y^{2}}+\frac{1}{x+y}-\frac{1}{x-y} \)

To solve the expression \( \frac{2 x}{x^{2}-y^{2}}+\frac{1}{x+y}-\frac{1}{x-y} \), we first notice that \( x^2 - y^2 \) can be factored as \( (x+y)(x-y) \). Thus, we can rewrite the first term as: \[ \frac{2x}{(x+y)(x-y)} \] Now, let's find a common denominator for the entire expression. The common denominator will be \( (x+y)(x-y) \). Rewriting the remaining terms gives us: \[ \frac{1}{x+y} = \frac{x-y}{(x+y)(x-y)} \quad \text{and} \quad \frac{1}{x-y} = \frac{x+y}{(x+y)(x-y)} \] Now, substituting these back into our expression, we have: \[ \frac{2x}{(x+y)(x-y)} + \frac{x-y}{(x+y)(x-y)} - \frac{x+y}{(x+y)(x-y)} \] Combining these into a single fraction, we get: \[ \frac{2x + (x - y) - (x + y)}{(x+y)(x-y)} = \frac{2x + x - y - x - y}{(x+y)(x-y)} = \frac{x - 2y}{(x+y)(x-y)} \] Thus, the simplified expression is: \[ \frac{x - 2y}{(x+y)(x-y)} \]