Đáp án:
$\begin{array}{l}
B2)a)\left( {2x - \dfrac{1}{3}} \right)\left( {3x + \dfrac{7}{9}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = 0\\
3x + \dfrac{7}{9} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{6}\\
x = \dfrac{{ - 7}}{{27}}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 7}}{{27}};x = \dfrac{1}{6}\\
b)\left( {\dfrac{1}{3}x + 2} \right)\left( {\dfrac{{ - 2}}{5}x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{3}x + 2 = 0\\
\dfrac{{ - 2}}{5}x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 6\\
x = \dfrac{{ - 15}}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 15}}{2};x = - 6\\
c)\left( {\dfrac{{ - 11}}{9}x + \dfrac{{ - 3}}{{15}}} \right).\left( {\dfrac{{ - 10}}{{12}}x - \dfrac{{12}}{{13}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ - 11}}{9}x + \dfrac{{ - 3}}{{15}} = 0\\
\dfrac{{ - 10}}{{12}}x - \dfrac{{12}}{{13}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{ - 11}}{9}x = \dfrac{1}{5}\\
\dfrac{{ - 5}}{6}x = \dfrac{{12}}{{13}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 9}}{{55}}\\
x = \dfrac{{ - 72}}{{65}}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 9}}{{55}};x = \dfrac{{ - 72}}{{65}}\\
d)\left( {\dfrac{5}{8} - \dfrac{3}{4}x} \right)\left( {\dfrac{{ - 21}}{{17}}x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{5}{8} = \dfrac{3}{4}x\\
x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{6}\\
x = 0
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{5}{6}\\
B3)a)\left| {\dfrac{8}{9} - x} \right| = \dfrac{1}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{8}{9} - x = \dfrac{1}{3}\\
\dfrac{8}{9} - x = - \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{8}{9} - \dfrac{1}{3} = \dfrac{5}{9}\\
x = \dfrac{8}{9} + \dfrac{1}{3} = \dfrac{{11}}{9}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{9};x = \dfrac{{11}}{9}\\
b)\left| { - x + \dfrac{1}{5}} \right| = \dfrac{3}{{10}}\\
\Leftrightarrow \left[ \begin{array}{l}
- x + \dfrac{1}{5} = \dfrac{3}{{10}}\\
- x + \dfrac{1}{5} = - \dfrac{3}{{10}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{5} - \dfrac{3}{{10}} = - \dfrac{1}{{10}}\\
x = \dfrac{1}{5} + \dfrac{3}{{10}} = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 1}}{{10}};x = \dfrac{1}{2}\\
c)\left| { - x - \dfrac{6}{7}} \right| = \dfrac{3}{{21}} = \dfrac{1}{7}\\
\Leftrightarrow \left[ \begin{array}{l}
- x - \dfrac{6}{7} = \dfrac{1}{7}\\
- x - \dfrac{6}{7} = - \dfrac{1}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 6}}{7} - \dfrac{1}{7} = - 1\\
x = \dfrac{{ - 6}}{7} + \dfrac{1}{7} = \dfrac{{ - 5}}{7}
\end{array} \right.\\
Vậy\,x = - 1;x = \dfrac{{ - 5}}{7}\\
d)\left| {x + \dfrac{3}{4}} \right| - \dfrac{1}{2} = \dfrac{3}{8}\\
\Leftrightarrow \left| {x + \dfrac{3}{4}} \right| = \dfrac{3}{8} + \dfrac{1}{2} = \dfrac{7}{8}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{4} = \dfrac{7}{8}\\
x + \dfrac{3}{4} = - \dfrac{7}{8}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{8} - \dfrac{3}{4} = \dfrac{1}{8}\\
x = - \dfrac{7}{8} - \dfrac{3}{4} = \dfrac{{ - 9}}{8}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{8};x = - \dfrac{9}{8}\\
e){\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{4}{{25}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{2}{5}\\
x - \dfrac{1}{2} = - \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{5} + \dfrac{1}{2} = \dfrac{9}{{10}}\\
x = \dfrac{1}{2} - \dfrac{2}{5} = \dfrac{1}{{10}}
\end{array} \right.\\
Vậy\,x = \dfrac{9}{{10}};x = \dfrac{1}{{10}}\\
f){\left( {x + \dfrac{1}{3}} \right)^3} = \dfrac{1}{8}\\
\Leftrightarrow x + \dfrac{1}{3} = \dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}\\
Vậy\,x = \dfrac{1}{6}\\
g){\left( {\dfrac{2}{5}x + \dfrac{3}{4}} \right)^2} = \dfrac{{25}}{{49}}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5}x + \dfrac{3}{4} = \dfrac{5}{7}\\
\dfrac{2}{5}x + \dfrac{3}{4} = - \dfrac{5}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5}x = \dfrac{5}{7} - \dfrac{3}{4} = \dfrac{{ - 1}}{{28}}\\
\dfrac{2}{5}x = - \dfrac{5}{7} - \dfrac{3}{4} = \dfrac{{ - 41}}{{28}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5}}{{56}}\\
x = \dfrac{{ - 205}}{{56}}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 5}}{{56}};x = \dfrac{{ - 205}}{{56}}\\
h){\left( {\dfrac{2}{3}x - \dfrac{1}{4}} \right)^3} = \dfrac{{ - 8}}{{27}}\\
\Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{4} = \dfrac{{ - 2}}{3}\\
\Leftrightarrow \dfrac{2}{3}x = \dfrac{{ - 2}}{3} + \dfrac{1}{4} = \dfrac{{ - 5}}{{12}}\\
\Leftrightarrow x = \dfrac{{ - 5}}{{12}}.\dfrac{3}{2} = \dfrac{{ - 5}}{8}\\
Vậy\,x = \dfrac{{ - 5}}{8}
\end{array}$
