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\[ \frac{x + 1}{2} + \frac{2}{x + 1} = \frac{x + 1}{3} + \frac{3}{x + 1} - \frac{5}{6} \] Or, \[ \frac{x + 1}{2} + \frac{2}{x + 1} - \frac{x + 1}{3} - \frac{3}{x + 1} = -\frac{5}{6} \] Or, \[ \left(\frac{x + 1}{2} - \frac{x + 1}{3}\right) + \left(\frac{2}{x + 1} - \frac{3}{x + 1}\right) = -\frac{5}{6} \] Or, \[ \frac{3x + 3 - 2x - 2}{6} + \frac{2 - 3}{x + 1} = -\frac{5}{6} \] Or, \[ \frac{x + 1}{6} - \frac{1}{x + 1} = -\frac{5}{6} \] Or, \[ \frac{(x + 1)^2 - 6}{6(x + 1)} = -\frac{5}{6} \] Or, \[ \frac{x^2 + 2x + 1 - 6}{x + 1} = -5 \Rightarrow x^2 + 2x - 5 = -5(x + 1) \] \[ x^2 + 2x - 5 + 5x + 5 = 0 \Rightarrow x^2 + 7x = 0 \Rightarrow x(x + 7) = 0 \] ∴ Either \(x = 0\) Or \(x + 7 = 0\) ⟹ \(x = -7\) ∴ The required solutions are \(x = 0\) or \(x = -7\)(Answer)