Đáp án:
Giải thích các bước giải:
$a)(2x-5).(3x-\dfrac{5}{3})<0$
$⇔$\(\left[ \begin{array}{l}\left \{ {{2x-5>0} \atop {3x-\dfrac{5}{3}<0}} \right.\\\left \{ {{2x-5<0} \atop {3x-\dfrac{5}{3}>0}} \right.\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\left \{ {{2x>5} \atop {3x<\dfrac{5}{3}}} \right.\\\left \{ {{2x<5} \atop {3x>\dfrac{5}{3}}} \right.\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\left \{ {{x>\dfrac{5}{2}} \atop {x<\dfrac{5}{9}}} \right.\\\left \{ {{x<\dfrac{5}{2}} \atop {x>\dfrac{5}{9}}} \right.\end{array} \right.\)
$b)x^2+2020x<0$
$⇔x.(x+2020)<0$
$⇔$\(\left[ \begin{array}{l}\left \{ {{x>0} \atop {x+2020<0}} \right.\\\left \{ {{x<0} \atop {x+2020>0}} \right.\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\left \{ {{x>0} \atop {x<-2020}} \right.\\\left \{ {{x<0} \atop {x>-2020}} \right.\end{array} \right.\)
