Đáp án:

 $1)
\dfrac{8}{15}\\
2)
\dfrac{370}{117}\\
3)
\dfrac{5}{2}\\
4)
\dfrac{3}{2}\\
6)
 x=\dfrac{1}{6}\\
7)
 x=\dfrac{-1}{6}\\
8)
 {\left[\begin{aligned}x =\dfrac{13}{20}\\x =\dfrac{7}{20}\end{aligned}\right.}\\
9)
 {\left[\begin{aligned}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{aligned}\right.}\\$

Giải thích các bước giải:

 $1)
\dfrac{4}{5}.\dfrac{7}{6}+\dfrac{-6}{15}=\dfrac{2.7}{5.3}-\dfrac{6}{15}=\dfrac{14-6}{15}=\dfrac{8}{15}\\
2)
\dfrac{-6}{9}.\dfrac{7}{13}+\dfrac{6}{13}.\dfrac{-5}{9}+3\dfrac{7}{9}\\
=\dfrac{-2}{3}.\dfrac{7}{13}+\dfrac{2}{13}.\dfrac{-5}{3}+\dfrac{34}{9}\\
=\dfrac{-2}{3}.\left ( \dfrac{7}{13}+\dfrac{5}{13} \right )+\dfrac{34}{9}\\
=\dfrac{-2}{3}.\dfrac{12}{13}+\dfrac{34}{9}\\
=\dfrac{-24}{39}+\dfrac{34}{9}\\
=\dfrac{(-24).3}{117}+\dfrac{34.13}{117}\\
=\dfrac{370}{117}\\
3)
6\dfrac{3}{8}-\left ( 4\dfrac{3}{8}-\dfrac{1}{2} \right )\\
=6\dfrac{3}{8}- 4\dfrac{3}{8}+\dfrac{1}{2} \\
=(6-4)+\left ( \dfrac{3}{8}-\dfrac{3}{8} \right )+\dfrac{1}{2}\\
=2+\dfrac{1}{2}=\dfrac{4}{2}+\dfrac{1}{2}=\dfrac{5}{2}\\
4)
2\dfrac{1}{3}.1,5-\left ( \dfrac{11}{10}+\dfrac{50}{100} \right ):\dfrac{4}{5}\\
=\dfrac{7}{3}.\dfrac{3}{2}-\left ( \dfrac{11}{10}+\dfrac{50}{100} \right ).\dfrac{5}{4}\\
=\dfrac{7}{2}-\left ( \dfrac{11}{10}+\dfrac{5}{10} \right ).\dfrac{5}{4}\\
=\dfrac{7}{2}-\dfrac{16}{10}.\dfrac{5}{4}\\
=\dfrac{7}{2}-\dfrac{4}{2}\\
=\dfrac{3}{2}\\
6)
\dfrac{3}{2}+x=\dfrac{5}{3}\\
\Leftrightarrow x=\dfrac{5}{3}-\dfrac{3}{2}\\
\Leftrightarrow x=\dfrac{10}{6}-\dfrac{9}{6}=\dfrac{1}{6}\\
7)
5\dfrac{2}{3}-\left ( \dfrac{3}{2}+x \right )=4\dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{2}+x=\dfrac{17}{3}-\dfrac{13}{3}\\
\Leftrightarrow \dfrac{3}{2}+x=\dfrac{4}{3}\\
\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{2}\\
\Leftrightarrow x=\dfrac{8}{6}-\dfrac{9}{6}=\dfrac{-1}{6}\\
8)
\left |x-\dfrac{1}{2}  \right |+\dfrac{75}{100}=\dfrac{9}{10}\\
\Leftrightarrow \left |x-\dfrac{1}{2}  \right |=\dfrac{9}{10}-\dfrac{75}{100}\\
\Leftrightarrow \left |x-\dfrac{1}{2}  \right |=\dfrac{9}{10}-\dfrac{3}{4}\\
\Leftrightarrow \left |x-\dfrac{1}{2}  \right |=\dfrac{36}{40}-\dfrac{30}{40}\\
\Leftrightarrow \left |x-\dfrac{1}{2}  \right |=\dfrac{6}{40}=\dfrac{3}{20}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{1}{2} =\dfrac{3}{20}\\x-\dfrac{1}{2} =\dfrac{-3}{20}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{3}{20}+\dfrac{1}{2}\\x =\dfrac{-3}{20}+\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{3}{20}+\dfrac{10}{20}\\x =\dfrac{-3}{20}+\dfrac{10}{20}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{13}{20}\\x =\dfrac{7}{20}\end{aligned}\right.}\\
9)
\left ( x-\dfrac{1}{2} \right )^2-\dfrac{1}{9}=0\\
\Leftrightarrow \left ( x-\dfrac{1}{2} \right )^2=\dfrac{1}{9}\\
\Leftrightarrow x-\dfrac{1}{2}=\pm \dfrac{1}{3}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{3}+\dfrac{1}{2}\\x=\dfrac{-1}{3}+\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{2}{6}+\dfrac{3}{6}\\x=\dfrac{-2}{6}+\dfrac{3}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{aligned}\right.}\\$