Đáp án:
$\begin{array}{l}
a)x.\left| {x - 1} \right| = - {\left( {x - 1} \right)^2}\left( {x \le 0} \right)\\
\Rightarrow x\left( {1 - x} \right) + {\left( {x - 1} \right)^2} = 0\\
\Rightarrow \left( {x - 1} \right).\left( {x - 1 - x} \right) = 0\\
\Rightarrow \left( {x - 1} \right).\left( { - 1} \right) = 0\\
\Rightarrow x = 1\left( {ktm} \right)\\
\Rightarrow x \in \emptyset \\
b)\frac{{{x^2} - 3x - {m^2}}}{{\sqrt {x - {m^2}} }} = \sqrt {x - {m^2}} \\
\Rightarrow {x^2} - 3x - {m^2} = x - {m^2}\\
\Rightarrow {x^2} - 4x = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
c)Dkxd:{x^2} \le 17\\
x + \sqrt {17 - {x^2}} = a\\
\Rightarrow 17 + 2x\sqrt {17 - {x^2}} = {a^2}\\
\Rightarrow x\sqrt {17 - {x^2}} = \frac{{{a^2} - 17}}{2}\\
pt \Rightarrow a + \frac{{{a^2} - 17}}{2} = 9\\
\Rightarrow {a^2} + 2a - 17 - 18 = 0\\
\Rightarrow {a^2} + 2a - 35 = 0\\
\Rightarrow \left[ \begin{array}{l}
a = - 7\\
a = 5
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x + \sqrt {17 - {x^2}} = - 7\\
x + \sqrt {17 - {x^2}} = 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {17 - {x^2}} = - 7 - x\left( {ktm} \right)\\
\sqrt {17 - {x^2}} = 5 - x\left( {x \le 5} \right)
\end{array} \right.\\
\Rightarrow 17 - {x^2} = {x^2} - 10x + 25\\
\Rightarrow 2{x^2} - 10x + 8 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.
\end{array}$
