Đáp án:
d) x=4
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)DK:x \ne 0\\
- \dfrac{2}{{5x}} = \dfrac{1}{3} + \dfrac{3}{5}\\
\to - \dfrac{2}{{5x}} = \dfrac{4}{5}\\
\to x = - \dfrac{2}{5}:\dfrac{4}{5}\\
\to x = - \dfrac{1}{2}\\
b)\dfrac{1}{5} + \left| {x - \dfrac{{13}}{{10}}} \right| = \dfrac{3}{2}\\
\to \left| {x - \dfrac{{13}}{{10}}} \right| = \dfrac{{13}}{{10}}\\
\to \left[ \begin{array}{l}
x - \dfrac{{13}}{{10}} = \dfrac{{13}}{{10}}\\
x - \dfrac{{13}}{{10}} = - \dfrac{{13}}{{10}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = 0
\end{array} \right.\\
c)\left| {\dfrac{3}{7} - 2x} \right| = \dfrac{2}{3}\\
\to \left[ \begin{array}{l}
\dfrac{3}{7} - 2x = \dfrac{2}{3}\\
\dfrac{3}{7} - 2x = - \dfrac{2}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - \dfrac{5}{{21}}\\
2x = \dfrac{{23}}{{21}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{5}{{42}}\\
x = \dfrac{{23}}{{42}}
\end{array} \right.\\
d){2^x} + {2^3}{.2^x} = 144\\
\to {2^x}\left( {1 + 8} \right) = 144\\
\to {2^x} = 16\\
\to {2^x} = {2^4}\\
\to x = 4
\end{array}\)
