Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
x - 9\sqrt x = 0\\
\Leftrightarrow \sqrt x \left( {\sqrt x - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 81\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 0;x = 81\\
b)Dkxd:x \ge 2\\
\sqrt {{x^2} - 2x} - 5\sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt x .\sqrt {x - 2} - 5\sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt {x - 2} \left( {\sqrt x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
\sqrt x = 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = 25\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 2;x = 25
\end{array}$
