ĐKXĐ:\(x\ne5\)
Ta có:
\(\dfrac{2x+4}{x-5}\ge1\)\(\Leftrightarrow\dfrac{2x+4}{x-5}-1\ge0\Leftrightarrow\dfrac{2x+4}{x-5}-\dfrac{x-5}{x-5}\ge0\)
\(\Leftrightarrow\dfrac{2x+4-x+5}{x-5}\ge0\)
\(\Leftrightarrow\dfrac{x+9}{x-5}\ge0\Leftrightarrow\left[{}\begin{matrix}x+9\ge0;x-5>0\\x+9\le0;x-5< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge-9;x>5\\x\le-9;x< 5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>5\\x\le-9\end{matrix}\right.\)(1)
Ta có:\(\dfrac{2x+4}{x-5}\le3\)\(\Leftrightarrow\dfrac{2x+4}{x-5}-3\le0\Leftrightarrow\dfrac{2x+4}{x-5}-\dfrac{3\left(x-5\right)}{x-5}\le0\)
\(\Leftrightarrow\dfrac{2x+4-3x+15}{x-5}\le0\Leftrightarrow\dfrac{19-x}{x-5}\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}19-x\ge0;x-5< 0\\19-x\le0;x-5>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le19;x< 5\\x\ge19;x>5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x< 5\\x\ge19\end{matrix}\right.\)(2)
Từ (1) và (2) ta có:\(\left[{}\begin{matrix}x\ge19\\x\le-9\end{matrix}\right.\) thoả mãn đề bài
