Đáp án:
a. x=2
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:\left\{ \begin{array}{l}
4x - 2 \ge 0\\
2x - 4 \ge 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ge 2
\end{array} \right.\\
\to x \ge 2\\
\sqrt {4 - 2x} = \sqrt {2x - 4} \\
\to {\left( {\sqrt {4 - 2x} } \right)^2} = {\left( {\sqrt {2x - 4} } \right)^2}\\
\to 4 - 2x = 2x - 4\\
\to 4x = 8\\
\to x = 2\left( {TM} \right)\\
b.x + 2 = \sqrt {x + 22} \\
\to {\left( {x + 2} \right)^2} = x + 22\left( {DK:x \ge - 2} \right)\\
\to {x^2} + 4x + 4 = x + 22\\
\to {x^2} + 3x - 18 = 0\\
\to {x^2} - 3x + 6x - 18 = 0\\
\to x\left( {x - 3} \right) + 6\left( {x - 3} \right) = 0\\
\to \left( {x - 3} \right)\left( {x + 6} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\left( {TM} \right)\\
x = - 6\left( l \right)
\end{array} \right.
\end{array}\)
