Ta có : \(\dfrac{3-7x}{1+x}\ge\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3-7x}{1+x}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{2\left(3-7x\right)-\left(x+1\right)}{2\left(x+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{5-15x}{2\left(x+1\right)}=\dfrac{5\left(3-x\right)}{2\left(x+1\right)}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x\ge0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x\le0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le3\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge3\\x< -1\end{matrix}\right.\end{matrix}\right.\)
Vậy suy ra tập nghiệm
b, (x+4)(5x+9)-x>4
\(\Leftrightarrow\)5x2+29x+36-x>4
\(\Leftrightarrow\)5x2+28x+36>4
\(\Leftrightarrow\)5x2+28x+32>0
\(\Leftrightarrow\)5(x2+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\))>0
\(\Leftrightarrow\)x2+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\)>0
\(\Leftrightarrow\)x2+2.\(\dfrac{14}{5}\)x+\(\dfrac{206}{25}\)+\(\dfrac{32}{5}\)-\(\dfrac{206}{25}\)>0
\(\Leftrightarrow\)(x+\(\dfrac{14}{5}\))2-\(\dfrac{46}{25}\)>0
\(\Leftrightarrow\)(x+\(\dfrac{14-\sqrt{46}}{5}\))(x+\(\dfrac{14+\sqrt{46}}{5}\))>0
\(\Leftrightarrow\)2 trường hợp
