0 \\ \\ 9 {x}^{2} - 6x + 1= 0 \\ \sqrt{d} = \sqrt{ {6}^{2} - 4 \times 9 \times 1} = 0 \\ x = \frac{6 + - 0}{18} = \frac{1}{3} \\ \\ = > 9 {x}^{2} - 6x + 1 = {(3x - 1)}^{2}" alt="9 {x}^{2} - 6x + 1 > 0 \\ \\ 9 {x}^{2} - 6x + 1= 0 \\ \sqrt{d} = \sqrt{ {6}^{2} - 4 \times 9 \times 1} = 0 \\ x = \frac{6 + - 0}{18} = \frac{1}{3} \\ \\ = > 9 {x}^{2} - 6x + 1 = {(3x - 1)}^{2}" align="absmiddle" class="latex-formula">
По методу интервала получим:
