Question

Помогитее,алгебра 8 класс​

Answer 1

$\left \{ {{-x^{2}+x+6\leq 0} \atop {5-3(x+1)>x}} \right. \\\\\left \{ {{x^{2}-x-6\geq 0 } \atop {5-3x-3>x}} \right. \\\\\left \{ {{(x-3)(x+2)\geq0 } \atop {-3x-x>-2}} \right. \\\\\left \{ {{(x-3)(x+2)\geq0 } \atop {-4x>-2}} \right.\\\\\left \{ {{(x-3)(x+2)\geq0 } \atop {x<0,5}} \right.$  

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1) _______[- 2]_________[3]______

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2) -___________(0,5)___________

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$Otvet:\boxed{x\in(-\infty \ ; \ -2]}$