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

We start with the expression \[ \frac{\frac{x-2}{x+1} - \frac{x}{x-2}}{x+3}. \] **Step 1. Combine the fractions in the numerator** The two fractions in the numerator are \[ \frac{x-2}{x+1} \quad \text{and} \quad \frac{x}{x-2}. \] To combine them, we need a common denominator, which is \((x+1)(x-2)\): \[ \frac{x-2}{x+1} = \frac{(x-2)(x-2)}{(x+1)(x-2)} = \frac{(x-2)^2}{(x+1)(x-2)}, \] \[ \frac{x}{x-2} = \frac{x(x+1)}{(x-2)(x+1)}. \] Thus, the numerator becomes \[ \frac{(x-2)^2 - x(x+1)}{(x+1)(x-2)}. \] **Step 2. Simplify the numerator's numerator** Expand the expressions in the numerator: \[ (x-2)^2 = x^2 - 4x + 4, \] \[ x(x+1) = x^2 + x. \] Subtract the two: \[ (x-2)^2 - x(x+1) = (x^2 - 4x + 4) - (x^2 + x) = -5x + 4. \] Thus, the combined fraction in the numerator is \[ \frac{-5x+4}{(x+1)(x-2)}. \] **Step 3. Divide by \((x+3)\)** The full expression is now \[ \frac{\frac{-5x+4}{(x+1)(x-2)}}{x+3}. \] Dividing by \(x+3\) is equivalent to multiplying by \(\frac{1}{x+3}\): \[ \frac{-5x+4}{(x+1)(x-2)(x+3)}. \] **Final Answer:** \[ \frac{-5x+4}{(x+1)(x-2)(x+3)}. \]