Đáp án:
\(\begin{array}{l}
a)x = - \dfrac{8}{3}\\
b)\left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{{11}}{{12}}
\end{array} \right.\\
c)x = \dfrac{{15}}{{32}}\\
d)x = 1\\
e)x = 5\\
g)\left[ \begin{array}{l}
x = 7\\
x = - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{1}{5}\left( {x - \dfrac{1}{4}} \right) = \dfrac{2}{3} - \dfrac{3}{2} + \dfrac{1}{4}\\
\to \dfrac{1}{5}\left( {x - \dfrac{1}{4}} \right) = - \dfrac{7}{{12}}\\
\to x - \dfrac{1}{4} = - \dfrac{{35}}{{12}}\\
\to x = - \dfrac{8}{3}\\
b)\left| {3 - 4x} \right| = \dfrac{2}{3}\\
\to \left[ \begin{array}{l}
3 - 4x = \dfrac{2}{3}\\
3 - 4x = - \dfrac{2}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = \dfrac{7}{3}\\
4x = \dfrac{{11}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{{11}}{{12}}
\end{array} \right.\\
c)\dfrac{3}{2} - 3x - 5x + \dfrac{1}{2} = - \dfrac{7}{4}\\
\to - 8x = - \dfrac{{15}}{4}\\
\to x = \dfrac{{15}}{{32}}\\
d){3^{x + 1}} = {3^2}\\
\to x + 1 = 2\\
\to x = 1\\
e){2^{3\left( {x + 1} \right)}} = {2^{2.9}}\\
\to 3\left( {x + 1} \right) = 18\\
\to x + 1 = 6\\
\to x = 5\\
g)2{\left( {x - 3} \right)^2} = 32\\
\to {\left( {x - 3} \right)^2} = 16\\
\to \left| {x - 3} \right| = 4\\
\to \left[ \begin{array}{l}
x - 3 = 4\\
x - 3 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 1
\end{array} \right.
\end{array}\)
