Question

помогите с алгеброй
$\frac{x {}^{2} (1 - x)}{x {}^{2} - 6x + 9 } \leqslant 0$
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Answer 1

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