b, 2x-1>6-2
c, 2x+3>2-1
mk thấy đề sai sai
\(a,\left(x+3\right)^2-3\left(2x-1\right)>x\left(x-4\right)\)
\(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
\(\Leftrightarrow x^2-x^2+4x+6x-6x>-3-9\)
\(\Leftrightarrow4x>-12\)
\(\Leftrightarrow x>-3\)
\(d,\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1}{x-3}-2\le0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)}{x-3}-\dfrac{2\left(x-3\right)}{x-3}\le0\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}\le0\)
\(\Leftrightarrow\dfrac{7}{x-3}\le0\)
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy BPT có nghiệm x ≥ 3
\(e,\dfrac{12-3x}{2x+6}>3\)
\(\Leftrightarrow\dfrac{12-3x}{2x+6}-3>0\)
\(\Leftrightarrow\dfrac{12-3x}{2x+6}-\dfrac{3\left(2x+6\right)}{2x+6}>0\)
\(\Leftrightarrow\dfrac{12-3x-6x-18}{2x+6}>0\)
\(\Leftrightarrow\dfrac{-9x-6}{2x+6}>0\)
\(\Leftrightarrow\dfrac{-3\left(3x+2\right)}{2\left(x+3\right)}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+2>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+2< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-\dfrac{2}{3}\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -\dfrac{2}{3}\\x< -3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{-3;-\dfrac{2}{3}\right\}\)
