Đáp án:
$a)$
$\dfrac{3}{4}-x=\dfrac{1}{5}$
$⇔$ $x=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}$
Vậy $x=\dfrac{11}{20}$
$b)$ $|x+\dfrac{2}{5}|-\dfrac{3}{7}=\dfrac{4}{7}$
$⇔$ $|x+\dfrac{2}{5}|=1$
$⇔$ \(\left[ \begin{array}{l}x+\dfrac{2}{5}=1\\x+\dfrac{2}{5}=-1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{-7}{5}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{-7}{5}\end{array} \right.\)
$c)$ $(x+\dfrac{1}{3})^3:2=\dfrac{-1}{16}$
$⇔$ $(x+\dfrac{1}{3})^3= $ $\dfrac{-1}{8}=(\dfrac{-1}{2})^3$
$⇔$ $x=\dfrac{-5}{6}$
Vậy $x=\dfrac{-5}{6}$
$d)$ $\dfrac{x+2}{3}=$ $\dfrac{12}{x+2}$
$⇔(x+2)^2=36=(±6)^2$
$⇔$ \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
