Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{13}}{4}\\
x = \dfrac{5}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 1\\
\sqrt {x - 2\sqrt {x - 1} } = \dfrac{1}{2}\\
\to \sqrt {x - 1 - 2\sqrt {x - 1} .1 + 1} = \dfrac{1}{2}\\
\to \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = \dfrac{1}{2}\\
\to \left| {\sqrt {x - 1} - 1} \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
\sqrt {x - 1} - 1 = \dfrac{1}{2}\\
\sqrt {x - 1} - 1 = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt {x - 1} = \dfrac{3}{2}\\
\sqrt {x - 1} = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x - 1 = \dfrac{9}{4}\\
x - 1 = \dfrac{1}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{13}}{4}\\
x = \dfrac{5}{4}
\end{array} \right.\left( {TM} \right)
\end{array}\)
