c. $\dfrac{1}{x-4}<\dfrac{3}{x^2+x-4}$
$\Leftrightarrow \dfrac{1}{x-4}-\dfrac{3}{x^2+x-4}<0$
$\Leftrightarrow \dfrac{x^2+x-4-3x+12}{(x-4)(x^2+x-4)}<0$
$\Leftrightarrow \dfrac{x^2-2x+8}{(x-4)(x^2+x-4)}<0$
Ta có: $x^2-2x+8=0$
$\Rightarrow$ pt vô nghiệm
$x-4=0 \Leftrightarrow x=4$
$x^2+x-4=0$
$\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{-1+\sqrt{17}}{2}\\x=\dfrac{-1-\sqrt{17}}{2}\end{array} \right.$
$*BXD:$ Bạn xem ảnh !
$\Leftrightarrow x∈(-∞,\dfrac{-1+\sqrt{17}}{2})∪(\dfrac{-1+\sqrt{17}}{2},4)$
