Đáp án:
$\begin{array}{l}
a)\dfrac{x}{{3,15}} = \dfrac{{0,15}}{{7,2}}\\
\Leftrightarrow x = \dfrac{{0,15.3,15}}{{7,2}}\\
\Leftrightarrow x = \dfrac{{21}}{{320}}\\
Vậy\,x = \dfrac{{21}}{{320}}\\
b)0,2:1\dfrac{1}{5} = \dfrac{2}{3}:\left( {6x + 7} \right)\\
\Leftrightarrow \dfrac{1}{5}:\dfrac{6}{5} = \dfrac{2}{3}:\left( {6x + 7} \right)\\
\Leftrightarrow \dfrac{1}{5}.\dfrac{5}{6} = \dfrac{2}{3}:\left( {6x + 7} \right)\\
\Leftrightarrow \dfrac{1}{6} = \dfrac{2}{3}:\left( {6x + 7} \right)\\
\Leftrightarrow \left( {6x + 7} \right) = \dfrac{2}{3}:\dfrac{1}{6}\\
\Leftrightarrow \left( {6x + 7} \right) = \dfrac{2}{3}.6\\
\Leftrightarrow 6x + 7 = 4\\
\Leftrightarrow 6x = 4 - 7\\
\Leftrightarrow 6x = - 3\\
\Leftrightarrow x = - \dfrac{1}{2}\\
Vậy\,x = - \dfrac{1}{2}
\end{array}$
