ĐKXĐ: \(x^3-3x-2\ne0\)
\(\Leftrightarrow x^3-x-2x-2\ne0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-2\left(x+1\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-2\right)\ne0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\ne0\)
hay \(x\notin\left\{2;-1\right\}\)
\(A=\dfrac{x^4-2x^2+1}{x^3-3x-2}=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x-2\right)\cdot\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x-2}\)
Để A<1 thì \(A-1< 0\)
\(\Leftrightarrow\dfrac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Leftrightarrow\dfrac{x^2-3x+3}{x-2}< 0\)
=>x-2<0
hay x<2
Vậy: \(\left\{{}\begin{matrix}x< 2\\x< >-1\end{matrix}\right.\)