Question

572. Решите неравенство:
г) x²-1/2x+1/16≤0​

Answer 1

${x}^{2} - \frac{1}{2} x + \frac{1}{16} \leqslant 0 \\ {16x}^{2} - 8x + 1 \leqslant 0 \\ {(4x - 1)}^{2} \leqslant 0 \\ {(4x - 1)}^{2} = 0 \\ x = \frac{1}{4} ,0.25$