Question

Помогите решить срочнооо!!

Answer 1

$\displaystyle\bf\\\left \{ {{4 {x}^{2} + 5x > 6} \atop {7x \geqslant {x}^{2} }} \right. \\ \\ 1) \: 4 {x}^{2} + 5x > 6 \\ 4 {x}^{2} + 5x - 6 > 0 \\ 4 {x}^{2} + 5x - 6 = 0 \\ a =4 \\ b = 5 \\ c = - 6\\ D = {b}^{2} - 4ac = {5}^{2} - 4 \times 4 \times ( - 6) = 25 + 96 = 121 \\ x_{1} = \frac{ - 5 - 11}{2 \times 4} = - \frac{16}{8} = - 2 \\ x_{2} = \frac{ - 5 + 11}{2 \times 4} = \frac{6}{8} = \frac{3}{4} = 0.75 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 4 {x}^{2} + 5x - 6 = 4(x + 2)(x - 0.75) \\ (x + 2)(x - 0.75) > 6 \\ + + + ( - 2) - - - (0.75) + + + \\ x < - 2 \: \: \: and \: \: \: x > 0.75 \\ \\ 2) \: 7x \geqslant {x}^{2} \\ 7x - {x}^{2} \geqslant 0 \\ {x}^{2} - 7x \leqslant 0 \\ x(x - 7) \leqslant 0 \\ + + + [0] - - - [7] + + + \\ 0 \leqslant x \leqslant 7 \\ \displaystyle\bf\\\left \{ {{x < - 2 \: \: \: and \: \: \: x > 0.75} \atop {0 \leqslant x \leqslant 7 }} \right. \\ \\ x \: \epsilon \: (0.75; \: 7]$