$A=\dfrac{2}{x+3}$ ĐK: $x\neq0;\,x\neq-3$
a) Để $A<1$
$⇔\dfrac{2}{x+3}<1$
$⇔\dfrac{2}{x+3}-\dfrac{x+3}{x+3}<0$
$⇔\dfrac{2-x-3}{x+3}<0$
$⇔\dfrac{-x-1}{x+3}<0$
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}-x-1>0\\x+3<0\end{cases}\\\begin{cases}-x-1<0\\x+3>0\end{cases}\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}x<-1\\x<-3\end{cases}\\\begin{cases}x>-1\\x>-3\end{cases}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x<-3\\x>-1\end{array} \right.\)
Vậy với $x<-3$ hoặc $x>-1$ thì $A<1$
b) Để $A>1$
$⇔\dfrac{2}{x+3}>1$
$⇔\dfrac{2}{x+3}-\dfrac{x+3}{x+3}>0$
$⇔\dfrac{2-x-3}{x+3}>0$
$⇔\dfrac{-x-1}{x+3}>0$
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}-x-1>0\\x+3>0\end{cases}\\\begin{cases}-x-1<0\\x+3<0\end{cases}\end{array} \right.\)
\(\Leftrightarrow\left[ \begin{array}{l}\begin{cases}x<-1\\x>-3\end{cases}\\\begin{cases}x>-1\\x<-3\end{cases}\end{array} \right.\)
$⇔-3<x<-1$
Vậy với $-3<x<-1;x\neq0$ thì $A<1$
