1) \( \frac{x}{(x+1)(x-1)}+\frac{1}{3 x}-\frac{1}{x^{2}} \)

To simplify the expression \[ \frac{x}{(x+1)(x-1)} + \frac{1}{3x} - \frac{1}{x^2}, \] we first find a common denominator. The denominators in the expression are \((x+1)(x-1)\), \(3x\), and \(x^2\). The least common multiple (LCM) of these denominators is \[ 3x(x+1)(x-1). \] Now, we will rewrite each term with this common denominator. 1. The first term: \[ \frac{x}{(x+1)(x-1)} = \frac{x \cdot 3x}{3x(x+1)(x-1)} = \frac{3x^2}{3x(x+1)(x-1)}. \] 2. The second term: \[ \frac{1}{3x} = \frac{(x+1)(x-1)}{3x(x+1)(x-1)} = \frac{(x^2 - 1)}{3x(x+1)(x-1)}. \] 3. The third term: \[ -\frac{1}{x^2} = -\frac{3(x+1)(x-1)}{3x^2(x+1)(x-1)} = -\frac{3(x^2 - 1)}{3x^2(x+1)(x-1)}. \] Now we can combine all the terms: \[ \frac{3x^2 + (x^2 - 1) - 3(x^2 - 1)}{3x(x+1)(x-1)}. \] Now, simplify the numerator: \[ 3x^2 + (x^2 - 1) - 3(x^2 - 1) = 3x^2 + x^2 - 1 - 3x^2 + 3 = x^2 + 2. \] Thus, the expression simplifies to: \[ \frac{x^2 + 2}{3x(x+1)(x-1)}. \] So, the final simplified form of the expression is: \[ \frac{x^2 + 2}{3x(x+1)(x-1)}. \]