Question

Проверьте, правильно ли я решил?

Answer 1

$1. \ x^2 + 2x - 1 = 0\\\\ D = 4 + 4 = 8\\\\ x_1 = \frac{-2+\sqrt{8}}{2} = \frac{-2 + 2\sqrt{2}}{2} = -1 + \sqrt{2}\\\\ x_2 = \frac{-2-\sqrt{8}}{2} = \frac{-2 - 2\sqrt{2}}{2} = -1 - \sqrt{2}\\\\ (x - x_1)(x - x_2) = (x - ( -1 + \sqrt{2}))(x - (-1 - \sqrt{2})) = \\\\ = (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})$

$2. \ x^2 - 2(\sqrt{2} - 1)x + 3 - 2\sqrt{2} =\\\\ = x^2 - 2(\sqrt{2} - 1)x + 2 - 2\sqrt{2} + 1 =\\\\ = x^2 - 2(\sqrt{2} - 1)x + (\sqrt{2} - 1)^2 = \\\\ = (x - (\sqrt{2} - 1))^2 = (x - \sqrt{2} + 1)^2 = (x + 1 - \sqrt{2})^2$

$3. \ \frac{x^2 + 2x - 1}{x^2 - 2(\sqrt{2} - 1)x + 3 - 2\sqrt{2}} = \frac{(x + 1 - \sqrt{2})(x + 1 + \sqrt{2})}{(x + 1 - \sqrt{2})^2} = \\\\ = \boxed{\frac{x + 1 + \sqrt{2}}{x + 1 - \sqrt{2}}}$