$(\frac{x + 1}{2x} + \frac{4}{x + 3} - 2) : \frac{x + 1}{x + 3} - \frac{x^2 - 5x + 3}{2x} = \frac{(x + 1)(x + 3) + 8x - 4x(x + 3)}{2x(x + 3)} * \frac{x + 3}{x + 1} - \frac{x^2 - 5x + 3}{2x} = \frac{x^2 + 4x + 3 + 8x - 4x^2 - 12x}{2x(x + 1)} - \frac{x^2 - 5x + 3}{2x} = \frac{3 - 3x}{2x(x + 1)} - \frac{x^2 - 5x + 3}{2x} = \frac{3(1 - x)(1 + x)}{2x(x + 1)} - \frac{x^2 - 5x + 3}{2x} = \frac{3(1 - x)}{2x} - \frac{x^2 - 5x + 3}{2x} = \frac{3 - 3x - x^2 + 5x - 3}{2x} = \frac{-x^2 + 2x}{2x} = \frac{x(2 - x)}{2x} = \frac{2 - x}{2} = 1 - \frac{x}{2}$
По условию x > 2, тогда $\frac{x}{2} > 1$ и $1 - \frac{x}{2} < 0$