ĐKXĐ:
$\dfrac{(x-1)(2-x)}{x+3}≥0$
Ta có:
$(x-1)(2-x)=0 ↔ \left[ \begin{array}{l}x=1\\x=2\end{array} \right.$
$(x-1)(2-x)≤0 ↔ \left[ \begin{array}{l}x≤1\\x≥2\end{array} \right.$
$(x-1)(2-x)≥0 ↔ 1≤x≤2$
+) TH1:
$\left\{ \begin{array}{l}(x-1)(2-x)≥0\\x+3>0\end{array} \right.$
$↔ \left[ \begin{array}{l}1≤x≤2\\x>-3\end{array} \right.$
$↔ 1≤x≤2$
+) TH2:
$\left[ \begin{array}{l}(x-1)(2-x)≤0\\x+3<0\end{array} \right.$
$↔ \left[ \begin{array}{l}\left[ \begin{array}{l}x≤1\\x≥2\end{array} \right.\\x<-3\end{array} \right.$
$↔ \left[ \begin{array}{l}x<-3\\x≥2\end{array} \right.$
Vậy để $A$ có nghĩa thì $\left[ \begin{array}{l}x<-3\\1≤x≤2\end{array} \right.$
