Đáp án:
\(g)\)
Để \(\dfrac{-3x^2-x+4}{x^2+3x+5}>0\)
\(→\) \(\left[ \begin{array}{l}\left\{\begin{matrix} -3x^2-x+5>0 & \\ x^2+3x+5>0 & \end{matrix}\right.\\\left\{\begin{matrix} -3x^2-x+5<0 & \\ x^2+3x+5<0 & \end{matrix}\right.\end{array} \right.\)
\(→ x\in \bigg(-\dfrac 43;1\bigg)\)
\(h)\)
Để \(\dfrac{4x^2+3x-1}{x^2+5x+7}>0\)
\(→4x^2+3x-1>0\)
\(→\) \(\left[ \begin{array}{l}x<-1\\x>\dfrac 14\end{array} \right.\)
\(→ x\in \Big(-∞; -1\Big) ∪\Big(\dfrac{1}{4}; +∞\Big)\)