Đáp án:
a) x=-2
b) \(x = \dfrac{8}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.\\
\sqrt {\left( {x - 2} \right)\left( {x + 2} \right)} + \sqrt {{{\left( {x + 2} \right)}^2}} = 0\\
\to \sqrt {x + 2} \left( {\sqrt {x - 2} + \sqrt {x + 2} } \right) = 0\\
\to x + 2 = 0\left( {do:\sqrt {x - 2} + \sqrt {x + 2} > 0\forall \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.} \right)\\
\to x = - 2\\
b)\sqrt {{{\left( {3 - 2x} \right)}^2}} = 5 - 3x\\
\to \left| {3 - 2x} \right| = 5 - 3x\\
\to \left[ \begin{array}{l}
3 - 2x = 5 - 3x\left( {DK:\dfrac{3}{2} \ge x} \right)\\
3 - 2x = - 5 + 3x\left( {DK:\dfrac{3}{2} < x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\left( l \right)\\
5x = 8
\end{array} \right.\\
\to x = \dfrac{8}{5}
\end{array}\)
