[tex]\displaystyle\bf\\1)\\\\\frac{(x-6)(x+2)^{2} }{x-3} \leq 0[/tex]
x = 6 и x = - 2 - нули функции , а x = 3 - число не входящее в область определения .
[tex]\displaystyle\bf\\+ + + + + [-2] + + + + + (3) - - - - - [6] + + + + + \\\\\\Otvet \ : \ x\in \Big(3 \ ; \ 6\Big] \ \cup \ (-2)\\\\\\2)\\\\g(x)=-6-2x\\\\|g(x)|\leq 3\\\\|-6-2x|\leq 3\\\\\\\left \{ {{-6-2x\geq -3} \atop {-6-2x\leq 3}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{-2x\geq 3} \atop {-2x\leq 9}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x\leq -1,5} \atop {x\geq -4,5}} \right. \\\\\\x\in\Big[-4,5 \ ; \ -1,5\Big][/tex]
Наименьшее решение неравенства : x = - 4,5
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[tex]\displaystyle\bf\\1)\\\\\frac{(x-6)(x+2)^{2} }{x-3} \leq 0[/tex]
x = 6 и x = - 2 - нули функции , а x = 3 - число не входящее в область определения .
[tex]\displaystyle\bf\\+ + + + + [-2] + + + + + (3) - - - - - [6] + + + + + \\\\\\Otvet \ : \ x\in \Big(3 \ ; \ 6\Big] \ \cup \ (-2)\\\\\\2)\\\\g(x)=-6-2x\\\\|g(x)|\leq 3\\\\|-6-2x|\leq 3\\\\\\\left \{ {{-6-2x\geq -3} \atop {-6-2x\leq 3}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{-2x\geq 3} \atop {-2x\leq 9}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x\leq -1,5} \atop {x\geq -4,5}} \right. \\\\\\x\in\Big[-4,5 \ ; \ -1,5\Big][/tex]
Наименьшее решение неравенства : x = - 4,5