Question

Надо решить це способом введения новой переменной.

Answer 1

$( {x}^{2} + 2x - 3)( {x}^{2} + 2x + 2) = - 4 \\ {x}^{2} + 2x = t \\ (t - 3)(t + 2) = - 4 \\ {t}^{2} - 3t + 2t - 6 + 4 = 0 \\ {t}^{2} - t - 2 = 0 \\ d = {b}^{2} - 4ac = 1 - 4 \times ( - 2) = 9 \\ t1 = \frac{1 + 3}{2} = 2 \\ t2 = \frac{1 - 3}{2} = - 1 \\ 1) \: {x}^{2} + 2x = 2 \\ {x}^{2} + 2x - 2 = 0 \\ d = {b}^{2} - 4ac = 4 - 4 \times ( - 2) = 4 + 8 = 12 \\ x1 = \frac{ - 2 + 2 \sqrt{3} }{2} = - 1 + \sqrt{3} \\ x2 = \frac{ - 2 - 2\sqrt{3} }{2} = - 1 - \sqrt{3} \\ 2) \: {x}^{2} + 2x = - 1 \\ {x}^{2} + 2x + 1 = 0 \\ {(x + 1)}^{2} = 0 \\ x + 1 = 0 \\ x = - 1$