Ta có : \(\dfrac{4x+3}{x^2+1}< 0\)
\(\Leftrightarrow4x+3< 0\) ( Vì : \(x^2+1>0\) )
\(\Leftrightarrow4x< -3\)
\(\Leftrightarrow x< -\dfrac{3}{4}\)
Để \(\dfrac{4x+3}{x^2+1}\) là số nguyên âm
thì \(\dfrac{4x+3}{x^2+1}< 0\)
\(\Leftrightarrow4x+3< 0\left(Vì\text{ }x^2+1>0\forall x\right)\\ \Leftrightarrow4x< -3\\ \Leftrightarrow x< -\dfrac{3}{4}\)