Đáp án:
d. \(x = \dfrac{1}{{18}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x + 2} \right| = \left| {3x - 5} \right|\\
\to \left[ \begin{array}{l}
x + 2 = 3x - 5\\
x + 2 = - 3x + 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 7\\
4x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{3}{4}
\end{array} \right.\\
b.x:\dfrac{3}{4} + \dfrac{1}{4} = - \dfrac{2}{3}\\
\to \dfrac{{4x}}{3} = - \dfrac{2}{3} - \dfrac{1}{4}\\
\to \dfrac{{4x}}{3} = - \dfrac{{11}}{{12}}\\
\to x = - \dfrac{{11}}{{12}}:\dfrac{4}{3}\\
\to x = - \dfrac{{11}}{{16}}\\
c.{\left( {2x + \dfrac{3}{5}} \right)^2} = \dfrac{9}{{25}}\\
\to \left| {2x + \dfrac{3}{5}} \right| = \dfrac{3}{5}\\
\to \left[ \begin{array}{l}
2x + \dfrac{3}{5} = \dfrac{3}{5}\\
2x + \dfrac{3}{5} = - \dfrac{3}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 0\\
2x = - \dfrac{6}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{3}{5}
\end{array} \right.\\
d.3{\left( {3x - \dfrac{1}{2}} \right)^3} = - \dfrac{1}{9}\\
\to {\left( {3x - \dfrac{1}{2}} \right)^3} = - \dfrac{1}{{27}}\\
\to 3x - \dfrac{1}{2} = - \dfrac{1}{3}\\
\to 3x = \dfrac{1}{6}\\
\to x = \dfrac{1}{{18}}
\end{array}\)
