[tex]x\neq - 1 \\ ( \frac{x - 3}{x + 1} ) {}^{2} - ( \frac{x - 3}{x + 1} ) - 2 = 0 \\ \frac{x - 3}{x + 1} = a \\ {a}^{2} - a - 2 = 0[/tex]

По теореме Виета:

[tex] {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c [/tex]

[tex]a_{1} + a_{2} = 1 \\ a_{1}a_{2} = - 2\\ a_{1} = 2 \\ a_{2} = - 1 \\ \\ 1) \: a = 2 \\ \frac{x - 3}{x + 1} = 2 \\ 2(x + 1) = x - 3 \\ 2x + 2 - x = - 3 \\ x = - 3 - 2 \\ x _{1}= - 5 \\ \\ 2) \: a = - 1 \\ \frac{x - 3}{x + 1} = - 1 \\ x - 3 = - (x + 1) \\ x - 3 + x + 1 = 0 \\ 2x - 2 = 0 \\ 2x = 2 \\ x = 2 \div 2 \\ x_{2} = 1[/tex]