Đáp án:

 $1)
a) 0\\
b)
0\\
c)
\dfrac{29}{72}\\
2)
a)x=\dfrac{-13}{12}\\
b)
x=\dfrac{2}{3}\\
c)
{\left[\begin{aligned}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{aligned}\right.}\\
3)
A=\dfrac{25}{17}$

Giải thích các bước giải:

 $1)
a)\left ( \dfrac{1}{2}+\dfrac{16}{30} \right )-\left ( 1+\left | -\dfrac{1}{30} \right | \right )\\
=\dfrac{1}{2}+\dfrac{16}{30}-1-\dfrac{1}{30}\\
=\left ( \dfrac{1}{2}-1 \right )+\left ( \dfrac{16}{30}-\dfrac{1}{30} \right )\\
=\dfrac{-1}{2}+\dfrac{15}{30}\\
=\dfrac{-1}{2}+\dfrac{1}{2}=0\\
b)
\dfrac{3}{2}-\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{4}\\
=\dfrac{3}{2}-\dfrac{2}{3}.\left ( 1+\dfrac{5}{4} \right )\\
=\dfrac{3}{2}-\dfrac{2}{3}.\left ( \dfrac{4}{4}+\dfrac{5}{4} \right )\\
=\dfrac{3}{2}-\dfrac{2}{3}.\dfrac{9}{4} \\
=\dfrac{3}{2}-\dfrac{3}{2}=0\\
c)
\left ( \dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3} \right )^2-\left ( \dfrac{1}{3}-\dfrac{5}{8} \right )\\
=\left [ \left (\dfrac{2}{9}+\dfrac{1}{3}  \right ):\dfrac{5}{3} \right ]^2-\left (\dfrac{8}{24}-\dfrac{15}{24}  \right )\\
=\left [ \left (\dfrac{2}{9}+\dfrac{3}{9}  \right ).\dfrac{3}{5} \right ]^2-\dfrac{-7}{24}\\
=\left ( \dfrac{5}{9}.\dfrac{3}{5} \right )^2+\dfrac{7}{24}\\
=\left ( \dfrac{1}{3} \right )^2+\dfrac{7}{24}\\
=\dfrac{1}{9}+\dfrac{7}{24}\\
=\dfrac{8}{72}+\dfrac{21}{72}\\
=\dfrac{29}{72}\\
2)
a)-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\
\Leftrightarrow x=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{1}{3}\\
\Leftrightarrow x=\dfrac{-20}{24}-\dfrac{14}{24}+\dfrac{8}{24}\\
\Leftrightarrow x=\dfrac{-26}{24}=\dfrac{-13}{12}\\
b)
2\left ( x-\dfrac{1}{3} \right )=\left ( \dfrac{1}{3} \right )^2+\dfrac{5}{9}\\
\Leftrightarrow 2\left (x-\dfrac{1}{3}  \right )=\dfrac{1}{9}+\dfrac{5}{9}=\dfrac{6}{9}=\dfrac{2}{3}\\
\Leftrightarrow x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{1}{2}=\dfrac{1}{3}\\
\Leftrightarrow x=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\\
c)
\left | 2x-\dfrac{3}{4} \right |-\dfrac{3}{8}=\dfrac{1}{8}\\
\Leftrightarrow \left | 2x-\dfrac{3}{4} \right |=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{4}{8}=\dfrac{1}{2}\\
\Leftrightarrow {\left[\begin{aligned}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{1}{2}+\dfrac{3}{4}\\2x=\dfrac{-1}{2}+\dfrac{3}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{2}{4}+\dfrac{3}{4}\\2x=\dfrac{-2}{4}+\dfrac{3}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{5}{4}\\2x=\dfrac{1}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{aligned}\right.}\\
3)
A=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\\
=3.\left (\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}  \right )\\
=\dfrac{3}{2}.\left (\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}  \right )\\
=\dfrac{3}{2}.\left ( \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51} \right )\\
=\dfrac{3}{2}.\left ( 1-\dfrac{1}{51} \right )\\
=\dfrac{3}{2}.\left ( \dfrac{51}{51}-\dfrac{1}{51} \right )\\
=\dfrac{3}{2}.\dfrac{50}{51}\\
=\dfrac{25}{17}$