Question

Помогите пожалуйста ​

Answer 1

$- 3 {x}^{2} + 4x + 4 > 0 \\ 3 {x }^{2} - 4x - 4 < 0 \\ \\ 3 {x}^{2} - 4x - 4 = 0 \\ a = 3 \\ b = - 4 \\ c = - 4 \\ D = {b}^{2} - 4ac =( - 4) {}^{2} - 4 \times 3 \times ( - 4) = 16 + 48 = 64 \\ x_{1} = \frac{4 - 8}{2 \times 3} = - \frac{4}{6} = - \frac{2}{3} \\ x_{2} = \frac{4 + 8}{2 \times 3} = \frac{12}{6} = 2 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 3 {x {}^{2} - 4x - x} = 3(x + \frac{2}{3} )(x - 2) \\ \\ (x + \frac{2}{3} )(x - 2) < 0 \\ + + + ( - \frac{2}{3} ) - - - (2) + + + \\ x \: \epsilon \: ( - \frac{2}{3} ; \: 2)$

Ответ: 2)