Giải thích các bước giải:
a) `(2x-3)(x-1)>0`
`=>2x-3;x-1` cùng dấu
`=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}2x-3>0\\x-1>0\end{matrix}\right.\\\left\{\begin{matrix}2x-3<0\\x-1<0\end{matrix}\right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}2x>3\\x>1\end{matrix}\right.\\\left\{\begin{matrix}2x<3\\x<1\end{matrix}\right.\end{array} \right.\)`=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x>\dfrac{3}{2}\\x>1\end{matrix}\right.\\\left\{\begin{matrix}x<\dfrac{3}{2}\\x<1\end{matrix}\right.\end{array} \right.\)`=>`\(\left[ \begin{array}{l}x>\dfrac{3}{2}\\x<1\end{array} \right.\)
Vậy `x>3/2` hoặc `x<1.`
b) `(3x+1).(x-2)<0`
`=>3x+1;x-2` trái dấu
`=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}3x+1>0\\x-2<0\end{matrix}\right.\\\left\{\begin{matrix}3x+1<0\\x-2>0\end{matrix}\right.\end{array} \right.\) `=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}3x>-1\\x<2\end{matrix}\right.\\\left\{\begin{matrix}3x<-1\\x>2\end{matrix}\right.\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}\left\{\begin{matrix}x>-\dfrac{1}{3}\\x<2\end{matrix}\right.\\\left\{\begin{matrix}x<-\dfrac{1}{3}\\x>2\end{matrix}\right.\text{(loại)}\end{array} \right.\)`=>-1/3<x<2`
Vậy `-1/3<x<2.`
