Đáp án:
5) x=5
Giải thích các bước giải:
\(\begin{array}{l}
1)\left[ \begin{array}{l}
5x - 1 = 0\\
2x - \dfrac{1}{3} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x = \dfrac{1}{6}
\end{array} \right.\\
2)\left| {x + 5} \right| = 15\\
\to \left[ \begin{array}{l}
x + 5 = 15\\
x + 5 = - 15
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\\
x = - 20
\end{array} \right.\\
3){\left( {\dfrac{{x - 1}}{2}} \right)^3} = - \dfrac{1}{{27}}\\
\to \dfrac{{x - 1}}{2} = \sqrt[3]{{ - \dfrac{1}{{27}}}}\\
\to \dfrac{{x - 1}}{2} = - \dfrac{1}{3}\\
\to 3x - 3 = - 2\\
\to 3x = 1\\
\to x = \dfrac{1}{3}\\
4){\left( {\dfrac{{x + 1}}{2}} \right)^2} = \dfrac{4}{{25}}\\
\to \left| {\dfrac{{x + 1}}{2}} \right| = \dfrac{2}{5}\\
\to \left[ \begin{array}{l}
\dfrac{{x + 1}}{2} = \dfrac{2}{5}\\
\dfrac{{x + 1}}{2} = - \dfrac{2}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + 1 = \dfrac{4}{5}\\
x + 1 = - \dfrac{4}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{5}\\
x = - \dfrac{9}{5}
\end{array} \right.\\
5){2^{x - 1}} = 16\\
\to \dfrac{1}{2}{.2^x} = 16\\
\to {2^x} = 32\\
\to {2^x} = {2^5}\\
\to x = 5
\end{array}\)
