0, x \neq 3\\x-1 \geq 0\\\left \{ {{x\geq1 } \atop {x=0}} \right. => x \in [0] \cup [1; 3) \cup (3;+\infty)" alt="\frac{x^2(1-x)}{x^2-6x+9} \leq 0\\\frac{x^2(x-1)}{(x-3)^2} \geq 0 | :x^2 \geq 0\\\frac{x-1}{(x-3)^2}\geq 0 | *(x-3)^2> 0, x \neq 3\\x-1 \geq 0\\\left \{ {{x\geq1 } \atop {x=0}} \right. => x \in [0] \cup [1; 3) \cup (3;+\infty)" align="absmiddle" class="latex-formula">