\( \left. \begin{array} { l l } { 1.1 .2 } & { 2 x ^ { 2 } - 4 x = - 1 } \\ { 1.1 .3 } & { ( x - 2 ) ^ { 2 } \geq 1 } \end{array} \right. \)

1. **Solve the equation** The equation is \[ 2x^2 - 4x = -1. \] Add \(1\) to both sides: \[ 2x^2 - 4x + 1 = 0. \] Use the quadratic formula with \(a = 2\), \(b = -4\), and \(c = 1\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4\cdot2\cdot1}}{4} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4}. \] Simplify the square root: \[ \sqrt{8} = 2\sqrt{2}. \] Hence, \[ x = \frac{4 \pm 2\sqrt{2}}{4} = \frac{2 \pm \sqrt{2}}{2} = 1 \pm \frac{\sqrt{2}}{2}. \] This gives two potential solutions: \[ x_1 = 1 + \frac{\sqrt{2}}{2}, \qquad x_2 = 1 - \frac{\sqrt{2}}{2}. \] 2. **Solve the inequality** The inequality is \[ (x-2)^2 \geq 1. \] Taking square roots (or by rewriting as an absolute value), we have: \[ |x-2| \geq 1. \] This can be split into two cases: - \(x-2 \geq 1\) which gives \[ x \geq 3, \] - \(x-2 \leq -1\) which gives \[ x \leq 1. \] So the solution to the inequality is \[ x \leq 1 \quad \text{or} \quad x \geq 3. \] 3. **Combine the results** We now determine which of the solutions from the equation satisfy the inequality: - For \(x_1 = 1 + \frac{\sqrt{2}}{2}\), numerically \[ 1 + \frac{\sqrt{2}}{2} \approx 1 + 0.7071 = 1.7071. \] This value does not satisfy either \(x \leq 1\) or \(x \geq 3\). - For \(x_2 = 1 - \frac{\sqrt{2}}{2}\), numerically \[ 1 - \frac{\sqrt{2}}{2} \approx 1 - 0.7071 = 0.2929. \] This value satisfies \(x \leq 1\). Therefore, the only solution that satisfies both the equation and the inequality is: \[ x = 1 - \frac{\sqrt{2}}{2}. \]