Đáp án:

 $3)
a) x=\dfrac{-1}{44}\\
c)x=\dfrac{3}{22}\\
d)
x=\dfrac{1}{81}\\
g)
 {\left[\begin{aligned}x= \dfrac{-1}{4}\\x=- \dfrac{3}{4}\end{aligned}\right.}\\
4)
 x=\{-1;0\}$

Giải thích các bước giải:

 $3)
a) -x-\dfrac{3}{4}=\dfrac{-8}{11}\\
\Leftrightarrow x=\dfrac{-3}{4}+\dfrac{8}{11}\\
\Leftrightarrow x=\dfrac{-33}{44}+\dfrac{32}{44}\\
\Leftrightarrow x=\dfrac{-1}{44}\\
c) \dfrac{1}{3}+\dfrac{1}{2}:x=4\\
\Leftrightarrow \dfrac{1}{2}:x=4-\dfrac{1}{3}\\
\Leftrightarrow \dfrac{1}{2}:x=\dfrac{12}{3}-\dfrac{1}{3}\\
\Leftrightarrow \dfrac{1}{2}:x=\dfrac{11}{3}\\
\Leftrightarrow x=\dfrac{1}{2}:\dfrac{11}{3}\\
\Leftrightarrow x=\dfrac{1}{2}.\dfrac{3}{11}=\dfrac{3}{22}\\
d)
x:\left ( -\dfrac{1}{3} \right )^3=\dfrac{-1}{3}\\
\Leftrightarrow =\left ( -\dfrac{1}{3} \right ).\left ( -\dfrac{1}{3} \right )^3
\Leftrightarrow x=\left (\dfrac{-1}{3}  \right )^4=\dfrac{1}{81}\\
g)
\left ( x+\dfrac{1}{2} \right )^2=\dfrac{1}{16}\\
\Leftrightarrow x+\dfrac{1}{2}=\pm \dfrac{1}{4}\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{2}= \dfrac{1}{4}\\x+\dfrac{1}{2}=- \dfrac{1}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{1}{4}-\dfrac{1}{2}\\x=- \dfrac{1}{4}-\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{1}{4}-\dfrac{2}{4}\\x=- \dfrac{1}{4}-\dfrac{2}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{-1}{4}\\x=- \dfrac{3}{4}\end{aligned}\right.}\\
4)
-4\dfrac{1}{3}.\left ( \dfrac{1}{2}-\dfrac{1}{6} \right )\leq x\leq -\dfrac{2}{3}.\left ( \dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4} \right )\\
\Leftrightarrow \dfrac{-13}{3}.\left ( \dfrac{3}{6}-\dfrac{1}{6} \right )\leq x\leq -\dfrac{2}{3}.\left ( \dfrac{4}{12}-\dfrac{6}{12}-\dfrac{9}{12} \right )\\
\Leftrightarrow \dfrac{-13}{3}.\dfrac{2}{6}\leq x\leq \dfrac{-2}{3}.\dfrac{-11}{12}\\
\Leftrightarrow \dfrac{-13}{9}\leq x\leq \dfrac{11}{18}\\
\Rightarrow x=\{-1;0\}$