Question

\(\dfrac{-3x-3}{2x+1}\ge0\)

Answer 1

Đặt \(\dfrac{-3x-3}{2x+1}=f\left(x\right)\)

Có : \(f\left(x\right)=0\Leftrightarrow x=-1\)

2x + 1 = 0 \(\Leftrightarrow x=-\dfrac{1}{2}\)

 

- Lập bảng xét dấu :

- Từ bảng xét dấu

=> Hệ có nghiệm \(x=[-1;-\dfrac{1}{2})\)

 

 

 

Answer 2

ĐKXĐ: \(x\ne-\dfrac{1}{2}\)

Ta có: \(\dfrac{-3x-3}{2x+1}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x-3=0\\\left\{{}\begin{matrix}-3x-3>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}-3x-3< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=3\\\left\{{}\begin{matrix}-3x>3\\2x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}-3x< 3\\2x< -1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\\left\{{}\begin{matrix}x< -1\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< \dfrac{-1}{2}\end{matrix}\right.\left(loại\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\-\dfrac{1}{2}< x< -1\end{matrix}\right.\)

hay \(-\dfrac{1}{2}< x\le-1\)

Vậy: S={x|\(-\dfrac{1}{2}< x\le-1\)}