`{x^2-3x+1}/{x^2-1}`
Ta có : `x^2-3x+1=0` $\Leftrightarrow$ $\left[\begin{matrix} x=\dfrac{3+\sqrt{5}}{2}\\ x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.$
`x^2-1=0` $\Leftrightarrow$ $\left[\begin{matrix} x=1\\ x=-1\end{matrix}\right.$
Trục số :
`+` (-1) `-` (`{3-\sqrt{5}}/2`) `+` (1) `-` (`{3+\sqrt{5}}/2`) `+`
Vậy `f(x)` > 0 `AA x in (-oo;-1) uu ({3-\sqrt{5}}/2;1) uu ({3+\sqrt{5}}/2;+oo)`
`f(x)` < 0 `AA x in (-1;{3-\sqrt{5}}/2) uu (1;{3+\sqrt{5}}/2)`
