$(2x+\dfrac{1}{5}).(-\dfrac{3}{5}x+\dfrac{4}{7})=0$
⇒ TH1: $2x+\dfrac{1}{5}=0$
$2x=-\dfrac{1}{5}$
$x=-\dfrac{1}{5}.\dfrac{1}{2}=-\dfrac{1}{10}$
⇒ TH2: $-\dfrac{3}{5}x+\dfrac{4}{7}=0$
$-\dfrac{3}{5}x=-\dfrac{4}{7}$
$x=-\dfrac{4}{7}.(-\dfrac{5}{3})=\dfrac{20}{21}$
⇒ $x=(-\dfrac{1}{10},\dfrac{20}{21})$
⇒ $C$ là đáp án chính xác