1.A=\(\frac{x^4-2x^2+1}{x^3-3x-2}\)
A có nghĩa \(\Leftrightarrow x^3-3x-2\ne0\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
2 .A = \(\frac{x^4-2x^2+1}{x^3-3x-2}\)=\(\frac{\left(x^2-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x-1\right)^2}{x-2}\)
A<1\(\Rightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Rightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Rightarrow\frac{x^2-3x+3}{x-2}< 0\Rightarrow x-2< 0\)vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy x<2 thỏa mãn yêu cầu A<1
