Question

4. Решите систему неравенств: 1-2x² + 2x + 4 ≤0 4 – 3 (x+2) >5x​

Answer 1

$\displaystyle\bf\\\left \{ {{ - 2 {x}^{2} + 2x + 4 \leqslant 0 } \atop {4 - 3(x + 2) > 5x }} \right. \\ \\ 1) \: - 2 {x}^{2} + 2x + 4 \leqslant 0 \: \: | \div ( - 2) \\ {x}^{2} - x - 2 \geqslant 0 \\ {x}^{2} + x - 2x - 2 \geqslant 0 \\x (x + 1) - 2(x + 1) \geqslant 0 \\ (x + 1)(x - 2) \geqslant 0 \\ + + + [ - 1] - - - [2] + + + \\ x \leqslant - 1 \: \: \: and \: \: \: x \geqslant 2 \\ \\ 2) \: 4 - 3(x + 2) > 5x \\ 4 - 3x - 6 - 5x >0 \\ - 8x > 2 \: \: | \div ( - 8) \\ x < - 0.25 \\ \displaystyle\bf\\3) \: \left \{ {{x \leqslant - 1 \: \: \: and \: \: \: x \geqslant 2} \atop {x < - 0.25 }} \right. \\ \\ x \:\epsilon \: ( - \propto; \: - 1]$