$1)\\ a)f(x)=(9-x^2)(x-2)=(3-x)(3+x)(x-2)\\ BXD:\\ \begin{array}{|c|cccccccccc|} \hline x&-\infty&&-3&&2&&3&&+\infty\\\hline f(x)&&+&0&-&0&+&0&-\\\hline\end{array}$
Vậy $f(x) \ge 0 \Leftrightarrow x \in (-\infty;-3] \cup [2;3]$
$f(x) < 0 \Leftrightarrow x \in (-3;2) \cup (3;+\infty)$
$b)\dfrac{1}{x-1} \ge \dfrac{2}{x} \\ ĐKXĐ: x\ne 0, x\ne 1\\ \dfrac{1}{x-1} \ge \dfrac{2}{x} \\ \Leftrightarrow \dfrac{1}{x-1} - \dfrac{2}{x} \ge 0\\ \Leftrightarrow \dfrac{x}{(x-1)x} - \dfrac{2(x-1)}{x(x-1)} \ge 0\\ \Leftrightarrow \dfrac{x-2(x-1)}{(x-1)x} \ge 0\\ \Leftrightarrow \dfrac{-x+2}{(x-1)x} \ge 0\\ \begin{array}{|c|cccccccccc|} \hline x&-\infty&&0&&1&&2&&+\infty\\\hline f(x)&&+&||&-&||&+&0&-\\\hline\end{array}$
Vậy $f(x) \ge 0 \Leftrightarrow x \in (-\infty;0) \cup (1;2]$
