d)
\( \dfrac{x(x-1)(-x^{2}+2)(x+3)}{(x^{2}-5x+6)(x-3)}\geqslant 0 \)
\[
\frac{\overbrace{x}^{A}(\overbrace{x-1}^{B})(\overbrace{-x^{2}+2}^{C})(\overbrace{x+3}^{D})}{(\underbrace{x^{2}-5x+6}_{E})(\underbrace{x-3}_{F})}\geqslant 0
\]
\[
x=0
\]
\[
\begin{gathered}
x-1=0\\[5pt]
x=1
\end{gathered}
\]
\[
\begin{gathered}
-x^{2}+2=0\\
x^{2}=2\\
x=\sqrt{2}\\[5pt]
x=-\sqrt{2}\qquad \text{ou}\qquad x=+\sqrt{2}
\end{gathered}
\]
Observação:
\( \sqrt{2\;}\simeq 1,41 \)
\[
\begin{gathered}
x+3=0\\[5pt]
x=-3
\end{gathered}
\]
\[
\begin{gathered}
x^{2}-5x+6=0\\[5pt]
\Delta=b^{2}-4ac=(-5)^{2}-4.1.6=25-24=1\\[5pt]
t=\frac{-b\pm\sqrt{\Delta \;}}{2a}=\frac{-(-5)\pm \sqrt{1\;}}{2.1}=\frac{5\pm1}{2}\\[5pt]
x=2\qquad \text{ou}\qquad x=3
\end{gathered}
\]
\[
\begin{gathered}
x-3=0\\[5pt]
x=3
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{V=\left[\phantom{{}}\right.\;-3;-\sqrt{2\;}\;\left.\phantom{{}}\right]\;\cup\;\left[\phantom{{}}\right.\;0;1\;\left.\phantom{{}}\right]\;\cup\;\left[\phantom{{}}\right.\;+\sqrt{2\;};2\;\left[\phantom{{}}\right.}
\]