ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{1-3x}{x-3}\ge1\)
<=> \(\dfrac{1-3x}{x-3}-1\ge0\)
<=> \(\dfrac{1-3x-x+3}{x-3}\ge0\)
<=> \(\dfrac{4-4x}{x-3}\ge0\)
TH1: \(\dfrac{4-4x}{x-3}=0\) <=> \(4-4x=0\)
<=> \(-4x=-4\) <=> \(x=1\) (thỏa mãn)
TH2: \(\dfrac{4-4x}{x-3}>0\)
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}4-4x>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-4x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 1\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\1< x< 3\end{matrix}\right.\)( thỏa mãn)
Vậy x=1 hoặc 1<x<3 là nghiệm của bất phương trình
ĐKXĐ: x \(\ne\) 3
\(\dfrac{1-3x}{x-3}\ge1\)
\(\Leftrightarrow\dfrac{1-3x}{x-3}\ge\dfrac{x-1}{x-1}\)
\(\Rightarrow1-3x\ge x-1\)
\(\Leftrightarrow-3x-x\ge-1-1\)
\(\Leftrightarrow-4x\ge-2\Leftrightarrow x\le\dfrac{1}{2}\)
Vậy nghiệm của BPT là \(x\le\dfrac{1}{2}\)
