\(\Leftrightarrow\dfrac{1}{x-1}>\dfrac{1}{x-2}-\dfrac{1}{x+2}=\dfrac{\left(x+2\right)-\left(x-2\right)}{x^2-4}=\dfrac{4}{x^2-4}\)\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{4}{x^2-4}>0\Leftrightarrow\dfrac{x^2-4-4x+4}{\left(x-2\right)\left(x-1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x+2\right)}>0\)
Điều kiện tồn tại A
\(\left\{{}\begin{matrix}x\ne2\\x\ne1\\x\ne-2\end{matrix}\right.\) \(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+2\right)}\)
\(\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x>1\)(1)
\(\left\{{}\begin{matrix}x< 0\\-2< x< 1\end{matrix}\right.\) \(\Rightarrow-2< x< 0\)(2)
từ (1)&(2)kết luận\(\Rightarrow\left[{}\begin{matrix}-2< x< 0\\x>1\end{matrix}\right.\)
