Verified answer

[tex]\dfrac{x^{3}-3x+2 }{6-x} \leq 0\\\\\\\dfrac{(x^{3} -x)-(2x-2)}{x-6} \geq 0\\\\\\\dfrac{x(x^{2} -1)-2(x-1)}{x-6} \geq 0\\\\\\\dfrac{x(x -1)(x+1)-2(x-1)}{x-6} \geq 0\\\\\\\dfrac{(x-1)(x^{2} +x-2)}{x-6} \geq 0\\\\\\\dfrac{(x-1)(x+2)(x-1)}{x-6} \geq 0[/tex]

Нули функции : x = 1  ;  x = - 2

x = 6  -  число не входящее в область определения .

[tex]\displaystyle\bf\\+ + + + + \Big[-2\Big] - - - - -\Big [1\Big] - - - - - \Big(6\Big) + + + + +[/tex]

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[tex]\displaystyle\bf\\Otvet \ : \ x\in\Big(-\infty \ ; \ -2\Big] \ \cup \ \Big(6 \ ; +\infty\Big)[/tex]