Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {{{\left( {3x - 1} \right)}^2}} = \sqrt {3 + 2\sqrt 3 .1 + 1} - \sqrt {3 - 2\sqrt 3 .1 + 1} \\
\to \left| {3x - 1} \right| = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
\to \left| {3x - 1} \right| = \sqrt 3 + 1 - \sqrt 3 + 1\\
\to \left| {3x - 1} \right| = 2\\
\to \left[ \begin{array}{l}
3x - 1 = 2\\
3x - 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
b)\sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt {4 - 2.2.\sqrt 3 + 3} + \sqrt {4 + 2.2.\sqrt 3 + 3} \\
\to \left| {x - 2} \right| = \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \\
\to \left| {x - 2} \right| = 2 - \sqrt 3 + 2 + \sqrt 3 \\
\to \left| {x - 2} \right| = 4\\
\to \left[ \begin{array}{l}
x - 2 = 4\\
x - 2 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - 2
\end{array} \right.
\end{array}\)
