Question

Допоможіть будь ласка

Answer 1

$\displaystyle\bf\\\left \{ {{(2x - 1)(2x + 1) + 6x \leqslant (2x + 1) {}^{2} } \atop {3x(x - 2) + 5 \geqslant x(3x - 7) }} \right. \\ \displaystyle\bf\\\left \{ {{4 {x}^{2} - 1 + 6x \leqslant 4 {x}^{2} + 4x + 1 } \atop {3 {x}^{2} - 6x + 5 \geqslant 3 {x}^{2} - 7x }} \right. \\ \displaystyle\bf\\\left \{ {{4 {x}^{2} + 6x - 4 {x}^{2} - 4x \leqslant 1 + 1} \atop {3 {x}^{2} - 6x - 3 {x}^{2} + 7x \geqslant - 5}} \right. \\ \displaystyle\bf\\\left \{ {{2x \leqslant 2 \: \: | \div 2} \atop {x \geqslant - 5 }} \right. \\ \displaystyle\bf\\\left \{ {{x \leqslant 1} \atop {x \geqslant - 5 }} \right. \\ \\ otvet \: \: \: x \:\epsilon \: [ - 5; \: 1]$