Đáp án:
$ a) (2x-3)(\dfrac{3}{4}x +1)=0$
⇔\(\left[ \begin{array}{l}2x-3=0\\\dfrac{3}{4}x +1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{array} \right.\)
Vậy ....
$b) \dfrac{2}{3}x +\dfrac{5}{7} = \dfrac{3}{10} $
$⇔ \dfrac{2}{3}x = \dfrac{3}{10} - \dfrac{5}{7}$
$⇔ \dfrac{2}{3}x = -\dfrac{29}{70}$
$⇔ x = -\dfrac{29}{70} : \dfrac{2}{3}$
$⇔ x = -\dfrac{87}{140}$
Vậy ....
$c - \dfrac{21}{13}x + \dfrac{1}{3} = \dfrac{2}{3}$
$⇔ -\dfrac{21}{13}x = \dfrac{2}{3} - \dfrac{1}{3}$
$⇔ -\dfrac{21}{13}x = \dfrac{1}{3}$
$⇔ x= -\dfrac{1}{3} : \dfrac{21}{13}$
$⇔ x= -\dfrac{13}{63}$
Vậy ...
$d) \dfrac{3}{7}x + 2\dfrac{3}{8} = 1\dfrac{2}{5}$
$⇔\dfrac{3}{7}x + 2 = \dfrac{7}{5}$
$⇔ \dfrac{3}{7}x = \dfrac{7}{5} - 2$
$⇔ \dfrac{3}{7}x = \dfrac{-3}{5}$
$⇔ x = -\dfrac{3}{5} : \dfrac{3}{7}$
$⇔ x = -\dfrac{7}{5}$
Vậy ...
$e) (5x-1)(2x-\dfrac{1}{3}) = 0$
⇔\(\left[ \begin{array}{l}5x-1=0\\2x-\dfrac{1}{3}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x= \dfrac{1}{5}\\x=\dfrac{1}{6}\end{array} \right.\)
Vậy....
$g) \dfrac{3}{7} + \dfrac{1}{7} : x = \dfrac{3}{14}$
$⇔ \dfrac{1}{7} : x = \dfrac{3}{14} - \dfrac{3}{7}$
$⇔ \dfrac{1}{7} : x = -\dfrac{3}{14}$
$⇔ x = \dfrac{1}{7} : (-\dfrac{3}{14})$
$⇔ x = -\dfrac{2}{3}$
Vậy ....
