Giải bất phương trình sau
Lời giải:
ĐKXĐ: $x\neq \frac{1}{2}; x\neq \frac{-5}{4}$
\(\frac{-3x^2-x+4}{(2x-1)(16x^2+4x+25)}>0\Leftrightarrow \frac{-3x^2-x+4}{(2x-1)(4x+5)^2}>0\)
\(\Leftrightarrow \frac{-3x^2-x+4}{2x-1}>0\)
\(\Rightarrow \left[\begin{matrix} \left\{\begin{matrix} -3x^2-x+4>0\\ 2x-1>0\end{matrix}\right.\\ \left\{\begin{matrix} -3x^2-x+4<0\\ 2x-1<0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} (1-x)(3x+4)>0\\ 2x-1>0\end{matrix}\right.\\ \left\{\begin{matrix} (1-x)(3x+4)<0\\ 2x-1<0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} -\frac{4}{3}< x< 1\\ x>\frac{1}{2}\end{matrix}\right.\\ \left\{\begin{matrix} x>1 \text{hoặc} x< \frac{-4}{3}\\ x< \frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} \frac{1}{2}< x< 1\\ x< \frac{-4}{3}\end{matrix}\right.\)
Vậy ..........
