Ответ:

x належить [7/5; +∞]

Объяснение:

див. фото

[tex]\displaystyle\bf\\\left \{ {{\dfrac{5+2x}{3}-\dfrac{5x-1}{6} < 2} \atop {(x-3)^{2} \leq (x-1)(x+5)}} \right. \\\\\\\left \{ {{\dfrac{5+2x}{3}\cdot 6-\dfrac{5x-1}{6} \cdot 6 < 2\cdot 6} \atop {x^{2} -6x+9 \leq x^{2} +5x-x-5}} \right.\\\\\\\left \{ {{(5+2x)\cdot 2-(5x-1) < 12} \atop {-6x+9\leq 4x-5}} \right.\\\\\\\left \{ {{10+4x-5x+1 < 12} \atop {-6x-4x\leq -5-9}} \right.\\\\\\\left \{ {{-x < 1} \atop {-10x\leq -14}} \right.[/tex]

[tex]\displaystyle\bf\\\left \{ {{x > -1} \atop {x\geq 1,4}} \right. \ \ \ \Rightarrow \ \ \ x\geq 1,4\\\\\\Otvet \ : \ x\in\Big[1,4 \ ; \ +\infty\Big)[/tex]