Verified answer

а)

[tex]\displaystyle\bf\\\left \{ {{8x + {x}^{2} \leqslant 0} \atop {(x - 2) {}^{2} > 4 }} \right. \\ \\ 1) \: {x}^{2} + 8x \leqslant 0 \\ x(x + 8) \leqslant 0 \\ + + + [ - 8] - - - [0] + + + \\ - 8 \leqslant x \leqslant 0 \\ \\ 2) \: (x - 2) {}^{2} > 4 \\ (x - 2) {}^{2} - {2}^{2} > 0 \\ (x - 2 - 2)(x - 2 + 2) > 0 \\ x(x - 4) > 0 \\ + + + (0) - - - (4) + + + \\ x < 0 \: \: \: and \: \: \: x > 4 \\ \displaystyle\bf\\3) \: \left \{ {{ - 8 \leqslant x \leqslant 0} \atop {x < 0 \: \: \: and \: \: \: x > 4 }} \right. \\ \\ otvet \: \: \: x \:\epsilon \: [ - 8; \: 0)[/tex]

б)

[tex]\displaystyle\bf\\\left \{ {{4 {x}^{2} > 64 } \atop {(x - 2) {}^{2} > 9 }} \right. \\ \\ 1) \: 4 {x}^{2} > 64 \\ {x}^{2} - 16 > 0 \\ (x - 4)(x + 4) > 0 \\ + + + ( - 4) - - - (4) + + + \\ x < - 4 \: \: \: and \: \: \: x > 4 \\ \\ 2) \: (x - 2) {}^{2} > 9 \\ (x - 2) {}^{2} - {3}^{2} > 0 \\ (x - 2 - 3)(x - 2 + 3) > 0 \\ ( x- 5)(x + 1) > 0 \\ + + + ( - 1) - - - (5) + + + \\ x < - 1 \: \: \: and \: \: \: x > 5 \\ \displaystyle\bf\\3) \: \left \{ {{x < - 4 \: \: \: and \: \: \: x > 4} \atop {x < - 1 \: \: \: and \: \: \: x > 5 }} \right. \\ \\ otvet \: \: \: x \: \epsilon\: ( - \propto; \: - 4)U(5; \: + \propto)[/tex]