Question

Помогите, пожалуйста! Решите данные системы неравенств:

Answer 1

$1)\; \; \left \{ {{x^2<9} \atop {(3x-5)^2<1}} \right. \; \left \{ {{x^2-9<0} \atop {(3x-5)^2-1<0}} \right. \; \left \{ {{(x-3)(x+3)<0} \atop {(3x-5-1)(3x-5+1)<0}} \right. \; \left \{ {{x\in (-3,3)} \atop {(3x-6)(3x-4)<0}} \right. \\\\\left \{ {{x\in (-3,3)} \atop {x\in (1\frac{1}{3};2)}} \right. \; \; \to \; \; x\in (1\frac{1}{3}\, ;2)$

$2)\; \; \left \{ {{x^2\geq 4} \atop {(2-5x)^2\geq 16}} \right. \; \left \{ {{x^2-4\geq 0} \atop {(2-5x)^2-16\geq 0}} \right. \; \left \{ {{(x-2)(x+2)\geq 0} \atop {(2-5x-4)(2-5x+4)\geq 0}} \right. \\\\\left \{ {{x\in (-\infty ,-2\, ]\cup [\, 2,+\infty )} \atop {-(2+5x)(6-5x)\geq 0}} \right. \; \left \{ {{x\in (_\infty ,-2\, ]\cup [\, 2,+\infty )} \atop {x\in (-\infty ,\frac{2}{5}\, ]\cup \, [\frac{6}{5},+\infty )}} \right. \; \to \; x\in (-\infty ,\frac{2}{5}\, ]\cup \, [1\frac{1}{5},+\infty )$

$3)\; \; \left \{ {{x^2\leq 16} \atop {(x-2)^2\geq 9}} \right. \; \left \{ {{x^2-16\leq 0} \atop {(x-2)^2-9\geq 0}} \right. \; \left \{ {{(x-4)(x+4)\leq 0} \atop {(x-2-3)(x-2+3)\geq 0}} \right. \; \left \{ {{(x-4)(x+4)\leq 0} \atop {(x-5)(x+1)\geq 0}} \right. \\\\\left \{ {{x\in [-4,4\, ]} \atop {x\in (-\infty , 1\, ]\cup [\, 5,+\infty )}} \right. \; \; \to \; \; x\in [-4,1\, ]$