Đáp án:
$1. \frac{1}{3} < x < 1$
$2. \frac{-2}{3} < x < \frac{1}{2}$
$3.$ \(\left[ \begin{array}{l}x>\frac{2}{3}\\x<\frac{-1}{4}\end{array} \right.\)
Giải thích các bước giải:
$1. ( 3x - 1 )( x - 1 ) < 0$
TH1 : $\left \{ {{3x-1>0} \atop {x-1<0}} \right.$
⇔ $\frac{1}{3} < x < 1$
TH2 : $\left \{ {{3x-1<0} \atop {x-1>0}} \right.$
⇔ $x < \frac{1}{3} , x > 1$ (vô lí)
Kết hợp 2 TH ⇒ $\frac{1}{3} < x < 1$
$2. ( -2x + 1 )( -3x - 2 ) < 0$
⇔ $( 2x - 1 )( 3x + 2 ) < 0$
TH1 : $\left \{ {{2x-1>0} \atop {3x+2<0}} \right.$
⇔ $x > \frac{1}{2} , x < \frac{-2}{3}$ (vô lí)
TH2 : $\left \{ {{2x-1<0} \atop {3x+2>0}} \right.$
⇔ $\frac{-2}{3} < x < \frac{1}{2}$
Kết hợp 2TH ⇒ $\frac{-2}{3} < x < \frac{1}{2}$
$3. ( 1 + 4x )( -3x + 2 ) < 0$
⇔ $( 4x + 1 )( 3x - 2 ) > 0$
TH1 : $\left \{ {{4x+1>0} \atop {3x-2>0}} \right.$
⇔ $\left \{ {{x>\frac{-1}{4}} \atop {x>\frac{2}{3}}} \right.$
⇔ $x > \frac{2}{3}$
TH2 : $\left \{ {{4x+1<0} \atop {3x-2<0}} \right.$
⇔ $\left \{ {{x<\frac{-1}{4}} \atop {x<\frac{2}{3}}} \right.$
⇔ $x < \frac{-1}{4}$
Kết hợp 2TH ⇒ \(\left[ \begin{array}{l}x>\frac{2}{3}\\x<\frac{-1}{4}\end{array} \right.\)
