Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
c,\\
\dfrac{3}{4} + \dfrac{1}{4}:x = - 3\\
\Leftrightarrow \dfrac{1}{4}:x = - 3 - \dfrac{3}{4}\\
\Leftrightarrow \dfrac{1}{4}:x = - \dfrac{{15}}{4}\\
\Leftrightarrow x = \dfrac{1}{4}:\left( { - \dfrac{{15}}{4}} \right)\\
\Leftrightarrow x = - \dfrac{1}{{15}}\\
d,\\
\dfrac{3}{7}x + 2\dfrac{3}{8} = 1\dfrac{2}{5}\\
\Leftrightarrow \dfrac{3}{7}x + \dfrac{{19}}{8} = \dfrac{7}{5}\\
\Leftrightarrow \dfrac{3}{7}x = \dfrac{7}{5} - \dfrac{{19}}{8}\\
\Leftrightarrow \dfrac{3}{7}x = - \dfrac{{39}}{{40}}\\
\Leftrightarrow x = \left( { - \dfrac{{39}}{{40}}} \right):\dfrac{3}{7}\\
\Leftrightarrow x = - \dfrac{{91}}{{40}}\\
e,\\
{\left( { - \dfrac{1}{3}} \right)^3} - x = \dfrac{1}{{81}}\\
\Leftrightarrow - \dfrac{1}{{27}} - x = \dfrac{1}{{81}}\\
\Leftrightarrow x = - \dfrac{1}{{27}} - \dfrac{1}{{81}}\\
\Leftrightarrow x = - \dfrac{4}{{81}}\\
h,\\
{\left( { - 3x - 2} \right)^2} = 64\\
\Leftrightarrow \left[ \begin{array}{l}
- 3x - 2 = 8\\
- 3x - 2 = - 8
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{10}}{3}\\
x = 2
\end{array} \right.\\
i,\\
{\left( { - 3} \right)^x}:81 = - 27\\
\Leftrightarrow {\left( { - 3} \right)^x} = 81.\left( { - 27} \right)\\
\Leftrightarrow {\left( { - 3} \right)^x} = {\left( { - 3} \right)^4}.{\left( { - 3} \right)^3}\\
\Leftrightarrow {\left( { - 3} \right)^x} = {\left( { - 3} \right)^7}\\
\Leftrightarrow x = 7
\end{array}\)
