Question

Tìm x

(3x+2)(x-1)(2x-5)<0

Answer 1

\(\left(3x+2\right)\left(x-1\right)\left(2x-5\right)< 0\)

+TH1: \(\left\{{}\begin{matrix}\left(3x+2\right)< 0\\\left(x-1\right)< 0\\\left(2x-5\right)< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< -\frac{2}{3}\\x< 1\\x< \frac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow x< -\frac{2}{3}\)

+TH2: \(\left\{{}\begin{matrix}\left(3x+2\right)< 0\\\left(x-1\right)>0\\\left(2x-5\right)>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< -\frac{2}{3}\\x>1\\x>\frac{5}{2}\end{matrix}\right.\) (loại)

+TH3: \(\left\{{}\begin{matrix}3x+2>0\\x-1< 0\\2x-5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>-\frac{2}{3}\\x< 1\\x>\frac{5}{2}\end{matrix}\right.\) (loại)

+TH4: \(\left\{{}\begin{matrix}3x+2>0\\x-1>0\\2x-5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-\frac{2}{3}\\x>1\\x< \frac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow1< x< \frac{5}{2}\)

Answer 2

\(\left(3x+2\right).\left(x-1\right).\left(2x-5\right)< 0\)

TH1 : \(\left\{{}\begin{matrix}3x+2>0\\x-1>0\\2x-5< 0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x>-\frac{2}{3}\\x>1\\x< \frac{5}{2}\end{matrix}\right.\)