Đáp án:
$\begin{array}{l}
a)\dfrac{x}{5} = \dfrac{2}{3}\\
\Leftrightarrow x = 5.\dfrac{2}{3}\\
\Leftrightarrow x = \dfrac{{10}}{3}\\
Vậy\,x = \dfrac{{10}}{3}\\
b)\dfrac{x}{3} - \dfrac{1}{2} = \dfrac{1}{5}\\
\dfrac{x}{3} = \dfrac{1}{5} + \dfrac{1}{2}\\
\dfrac{x}{3} = \dfrac{7}{{10}}\\
x = \dfrac{7}{{10}}.3\\
x = \dfrac{{21}}{{10}}\\
Vậy\,x = \dfrac{{21}}{{10}}\\
c)\dfrac{x}{5} + \dfrac{1}{2} = \dfrac{6}{{10}}\\
\dfrac{x}{5} = \dfrac{6}{{10}} - \dfrac{1}{2}\\
\dfrac{x}{5} = \dfrac{1}{{10}}\\
x = \dfrac{1}{{10}}.5\\
x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}\\
e)\dfrac{{x + 3}}{{15}} = \dfrac{1}{3}\\
x + 3 = \dfrac{1}{3}.15\\
x + 3 = 5\\
x = 5 - 3\\
x = 2\\
Vậy\,x = 2\\
f)\dfrac{{x - 12}}{4} = \dfrac{1}{2}\\
x - 12 = \dfrac{1}{2}.4 = 2\\
x = 2 + 12\\
x = 14\\
Vậy\,x = 14\\
g)\dfrac{4}{{5 + x}} = \dfrac{2}{3}\\
5 + x = \dfrac{{4.3}}{2}\\
5 + x = 6\\
x = 6 - 5\\
x = 1\\
Vậy\,x = 1\\
h)\dfrac{3}{{4 - x}} = \dfrac{1}{3}\\
4 - x = 3.3 = 9\\
x = 4 - 9\\
x = - 5\\
Vậy\,x = - 5\\
i)\dfrac{{ - 5}}{6} - x = \dfrac{2}{3}\\
x = \dfrac{{ - 5}}{6} - \dfrac{2}{3}\\
x = \dfrac{{ - 5 - 4}}{6}\\
x = \dfrac{{ - 3}}{2}\\
Vậy\,x = \dfrac{{ - 3}}{2}\\
k)\dfrac{{ - 2}}{3} - \dfrac{1}{3}\left( {2x - 5} \right) = \dfrac{3}{2}\\
\dfrac{1}{3}\left( {2x - 5} \right) = - \dfrac{2}{3} - \dfrac{3}{2} = \dfrac{{ - 13}}{6}\\
2x - 5 = - \dfrac{{13}}{6}.3\\
2x - 5 = - \dfrac{{13}}{2}\\
2x = \dfrac{{ - 13}}{2} + 5 = \dfrac{{ - 3}}{2}\\
x = \dfrac{{ - 3}}{4}\\
Vậy\,x = - \dfrac{3}{4}\\
l)\left( {3x - 1} \right)\left( { - \dfrac{1}{2}x + 5} \right) = 0\\
\Leftrightarrow 3x - 1 = 0\,hoặc\, - \dfrac{1}{2}x + 5 = 0\\
\Leftrightarrow 3x = 1\,hoặc\,\dfrac{1}{2}.x = 5\\
\Leftrightarrow x = \dfrac{1}{3}\,hoặc\,x = 10\\
Vậy\,x = \dfrac{1}{3};x = 10
\end{array}$
