Bài $4$
$a$)$\dfrac{x}{5} = \dfrac{2}{3}$
$⇔ 3x = 10$
$⇔ x = \dfrac{10}{3}$
Vậy $x = \dfrac{10}{3}$
$b$) $\dfrac{x}{3} - \dfrac{1}{2}= \dfrac{1}{5}$
$⇔ \dfrac{x}{3} = \dfrac{7}{10}$
$⇔ 10x = 21$
$⇔ x =\dfrac{21}{10}$
Vậy $x =\dfrac{21}{10}$
$c$) $\dfrac{2}{3}.x + \dfrac{1}{2} = \dfrac{1}{10}$
$⇔ \dfrac{2}{3}.x = \dfrac{-2}{5}$
$⇔ x = \dfrac{-3}{5}$
Vậy $x = \dfrac{-3}{5}$
$d$) $\dfrac{x}{5} + \dfrac{1}{2} = \dfrac{6}{10}$
$⇔ \dfrac{x}{5} = \dfrac{1}{10}$
$⇔ \dfrac{2x}{10} = \dfrac{1}{10}$
$⇔ 2x = 1$
$⇔ x = \dfrac{1}{2}$
Vậy $x = \dfrac{1}{2}$
Bài $9$.
$A=\dfrac{3}{2.5} + \dfrac{3}{5.8} + \dfrac{3}{8.11} + ..... + \dfrac{3}{92.95} + \dfrac{3}{95.98}$
$⇔ A= \dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + \dfrac{1}{8} - \dfrac{1}{11} + .... + \dfrac{1}{92} - \dfrac{1}{95} + \dfrac{1}{95} - \dfrac{1}{98}$
$⇔ A = \dfrac{1}{2} - \dfrac{1}{98}$
$⇔ A = \dfrac{48}{98} = \dfrac{24}{49}$
