Đáp án:
d. x=2
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:3x + 8 \ge 0 \to x \ge - \dfrac{8}{3}\\
\sqrt {3x + 8} = 2\\
\to 3x + 8 = 4\\
\to 3x = - 4\\
\to x = - \dfrac{4}{3}\left( {TM} \right)\\
b.\sqrt {{{\left( {2x + 1} \right)}^2}} = 5\\
\to \left| {2x + 1} \right| = 5\\
\to \left[ \begin{array}{l}
2x + 1 = 5\\
2x + 1 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 3
\end{array} \right.\\
c.DK:x \ge \dfrac{1}{4}\\
\sqrt {4x - 1} = \left| { - 2} \right|\\
\to \sqrt {4x - 1} = 2\\
\to 4x - 1 = 4\\
\to x = \dfrac{5}{4}\left( {TM} \right)\\
d.\sqrt {{{\left( {x - 1} \right)}^2}} = \left| {x - 3} \right|\\
\to \left| {x - 1} \right| = \left| {x - 3} \right|\\
\to \left[ \begin{array}{l}
x - 1 = x - 3\\
x - 1 = - x + 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 1 = - 3\left( l \right)\\
2x = 4
\end{array} \right.\\
\to x = 2
\end{array}\)
