736. 1) \frac{1}{9}x² + \frac{1}{2}x + \frac{9}{16} = 0; 2) \frac{5}{4}x² – x + \frac{1}{9} = 0;

Решим уравнения.

1. $$\frac{1}{9}x^2 + \frac{1}{2}x + \frac{9}{16} = 0$$
   $$D = b^2 - 4ac = (\frac{1}{2})^2 - 4 \cdot \frac{1}{9} \cdot \frac{9}{16} = \frac{1}{4} - \frac{36}{144} = \frac{1}{4} - \frac{1}{4} = 0$$
   $$x = \frac{-b}{2a} = \frac{-\frac{1}{2}}{2 \cdot \frac{1}{9}} = \frac{-\frac{1}{2}}{\frac{2}{9}} = -\frac{1}{2} \cdot \frac{9}{2} = -\frac{9}{4} = -2.25$$

   Ответ: x = -2.25

2. $$\frac{5}{4}x^2 - x + \frac{1}{9} = 0$$
   $$D = b^2 - 4ac = (-1)^2 - 4 \cdot \frac{5}{4} \cdot \frac{1}{9} = 1 - \frac{5}{9} = \frac{9 - 5}{9} = \frac{4}{9}$$
   $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{\frac{4}{9}}}{2 \cdot \frac{5}{4}} = \frac{1 + \frac{2}{3}}{\frac{5}{2}} = \frac{\frac{5}{3}}{\frac{5}{2}} = \frac{5}{3} \cdot \frac{2}{5} = \frac{2}{3}$$
   $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{\frac{4}{9}}}{2 \cdot \frac{5}{4}} = \frac{1 - \frac{2}{3}}{\frac{5}{2}} = \frac{\frac{1}{3}}{\frac{5}{2}} = \frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}$$

   Ответ: x₁ = $$\frac{2}{3}$$, x₂ = $$\frac{2}{15}$$