ĐK: \(x\ne-5;x\ne3\)
\(\dfrac{x-3}{x+5}+\dfrac{x+5}{x-3}< 2\\ \Rightarrow\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+5\right)}+\dfrac{\left(x+5\right)^2}{\left(x-3\right)\left(x+5\right)}-\dfrac{2\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}< 0\)
\(\Rightarrow\dfrac{x^2-6x+9+x^2+10x+25-2x^2-4x+30}{\left(x-3\right)\left(x+5\right)}< 0\)
\(\Rightarrow\dfrac{64}{\left(x-3\right)\left(x+5\right)}< 0\Rightarrow\left(x-3\right)\left(x+5\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-3< 0\\x+5>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x>-5\end{matrix}\right.\Rightarrow-5< x< 3\) (TM)
TH2: \(\left\{{}\begin{matrix}x-3>0\\x+5< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x< -5\end{matrix}\right.\) (vô lí)
ĐKXĐ : x ≠ -5; x ≠ 3
\(\dfrac{x-3}{x+5}+\dfrac{x+5}{x-3}-2< 0\) ⇔ \(\dfrac{64}{\left(x+5\right)\left(x-3\right)}< 0\) ⇔ (x+5) và (x-3) khác dấu
⇔ \(\left\{{}\begin{matrix}x+5>0\\x-3< 0\end{matrix}\right.\) ⇔ -5 < x < 3. Vì x ϵ \(\left\{-4;-3;-2;-1;0;1;2\right\}\)
