Đáp án:
$\begin{array}{l}
a)\sqrt {x - 2\sqrt {x - 1} } = 1 - \sqrt {x - 1} \left( {x \ge 1} \right)\\
\Rightarrow \sqrt {x - 1 - 2\sqrt {x - 1} + 1} = 1 - \sqrt {x - 1} \\
\Rightarrow \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = 1 - \sqrt {x - 1} \\
\Rightarrow \left| {\sqrt {x - 1} - 1} \right| = 1 - \sqrt {x - 1} \\
\Rightarrow \sqrt {x - 1} - 1 \le 0\\
\Rightarrow \sqrt {x - 1} \le 1\\
\Rightarrow x - 1 \le 1\\
\Rightarrow x \le 2\\
Vay\,1 \le x \le 2\\
b)Dkxd:x \ge 0\\
\frac{{\sqrt x - 1}}{{\sqrt x + 3}} = \frac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) = \left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)\\
\Rightarrow x - 1 = x + \sqrt x - 6\\
\Rightarrow \sqrt x = 5\\
\Rightarrow x = 25\left( {tmdk} \right)\\
c)\sqrt {9{{\left( {2 - 3x} \right)}^2}} = 6\\
\Rightarrow 3\left| {2 - 3x} \right| = 6\\
\Rightarrow \left| {2 - 3x} \right| = 2\\
\Rightarrow \left[ \begin{array}{l}
2 - 3x = 2\\
2 - 3x = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{4}{3}
\end{array} \right.\\
d)\sqrt {{x^2} - 4} - \sqrt {x - 2} = 0\left( {dkxd:x \ge 2} \right)\\
\Rightarrow \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)} - \sqrt {x - 2} = 0\\
\Rightarrow \sqrt {x - 2} .\sqrt {x + 2} - \sqrt {x - 2} = 0\\
\Rightarrow \sqrt {x - 2} .\left( {\sqrt {x + 2} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {x - 2} = 0\\
\sqrt {x + 2} = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x + 2 = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 2
\end{array}$
