Question

Помогите даю 60 балов​

Answer 1

$- 3 {x}^{2} - x + 2 > 0 \\ 3 {x}^{2} + x - 2 < 0 \\ \\ 3 {x}^{2} + x - 2 = 0 \\ a = 3\\ b = 1 \\ c = - 2 \\ D = {b}^{2} - 4ac = 1 {}^{2} - 4 \times 3 \times ( - 2) = 1+ 24 = 25 \\ x_{1} = \frac{ - 1 - 5}{2 \times 3} = - \frac{6}{6} = - 1 \\ x_{2} = \frac{ - 1 + 5}{2 \times 3} = \frac{4}{6} = \frac{2}{3} \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 3 {x}^{2} + x - 2 = 3(x + 1)(x - \frac{2}{3} ) \\ \\ (x + 1)(x - \frac{2}{3} ) < 0 \\ + + + ( - 1) - - - ( \frac{2}{3} ) + + + \\ x \:\epsilon \: ( - 1; \: \frac{2}{3} )$

Ответ: 0