Đáp án:
\(\begin{array}{l}
b)\left[ \begin{array}{l}
x = \dfrac{{11}}{3}\\
x = - 3
\end{array} \right.\\
c)x = 5\\
d)x = \dfrac{5}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
b){\left( {3x - 1} \right)^2} = 2.27 + 46\\
\to {\left( {3x - 1} \right)^2} = 100\\
\to \left| {3x - 1} \right| = 10\\
\to \left[ \begin{array}{l}
3x - 1 = 10\\
3x - 1 = - 10
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{11}}{3}\\
x = - 3
\end{array} \right.\\
c){2^x} + {2^x}{.2^2} = 160\\
\to {5.2^x} = 160\\
\to {2^x} = 32\\
\to {2^x} = {2^5}\\
\to x = 5\\
d)\left| {x + \dfrac{1}{2}} \right| + \left| {x + 2} \right| = 3x\\
\to \left[ \begin{array}{l}
x + \dfrac{1}{2} + x + 2 = 3x\left( {DK:x \ge - \dfrac{1}{2}} \right)\\
- x - \dfrac{1}{2} + x + 2 = 3x\left( {DK: - \dfrac{1}{2} > x \ge - 2} \right)\\
- x - \dfrac{1}{2} - x - 2 = 3x\left( {DK: - 2 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( {TM} \right)\\
3x = \dfrac{3}{2}\\
5x = - \dfrac{5}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{1}{2}\left( l \right)\\
x = - \dfrac{1}{2}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{5}{2}
\end{array}\)
