Question

Розв'яжіть систему нерівностей:
ЛЮДИ, ПОМОГИТЕ, СРОЧНО!!!!!!!​

Answer 1

$\displaystyle\bf\\\left \{ {{2 {x}^{2} -9 x + 10 \geqslant 0 } \atop {(3 - x)(1 - 2x) \leqslant 3 }} \right. \\ \\ 1) \: 2 {x}^{2} - 9x + 10 \geqslant 0 \\ 2 {x}^{2} - 9x + 10 = 0 \\ a = 2 \\ b = - 9 \\ c = 10\\ D = {b}^{2} - 4ac = ( - 9) {}^{2} - 4 \times 2 \times 10 = 81 - 80 = 1 \\ x_{1} = \frac{9 - 1}{2 \times 2} = \frac{8}{4} = 2 \\ x_{2} = \frac{9 + 1}{2 \times 2} = \frac{10}{4} = 2.5 \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ 2 {x}^{2} - 9x + 10 = 2(x - 2)(x - 2.5) \\ (x - 2)(x - 2.5) \geqslant 0 \\ + + + + [2] - - - - [2.5] + + + + \\ x \leqslant 2 \: \: \: and \: \: \: x \geqslant 2.5 \\ \\ 2) \: (3 - x)(1 - 2x) \leqslant 3 \\ 3 - 6x - x + 2 {x}^{2} - 3 \leqslant 0 \\ 2 {x}^{2} - 7x \leqslant 0 \\ {x}^{2} - 3.5 \leqslant 0 \\ x(x - 3.5) \leqslant 0 \\ + + + + [0] - - - - [3.5] + + + + \\ 0 \leqslant x \leqslant 3.5 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant 2 \: \: \: and \: \: \: x \geqslant 2.5} \atop {0 \leqslant x \leqslant 3.5 }} \right. \\ \\ x \: \epsilon \: [0; \: 2]U[2.5; \: 3.5]$