Đáp án:
$\begin{array}{l}
a)Dkxd:17 - {x^2} \ge 0 \Leftrightarrow {x^2} \le 17\\
Đặt:x + \sqrt {17 - {x^2}} = a\\
\Leftrightarrow {x^2} + 2x\sqrt {17 - {x^2}} + 17 - {x^2} = {a^2}\\
\Leftrightarrow x\sqrt {17 - {x^2}} = \dfrac{{{a^2} - 17}}{2}\\
Pt\\
\Leftrightarrow a + \dfrac{{{a^2} - 17}}{2} = 9\\
\Leftrightarrow {a^2} + 2a - 17 = 18\\
\Leftrightarrow {a^2} + 2a - 35 = 0\\
\Leftrightarrow \left( {a - 5} \right)\left( {a + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = 5\\
a = - 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x\sqrt {17 - {x^2}} = 4\\
x\sqrt {17 - {x^2}} = 16
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2}\left( {17 - {x^2}} \right) = 16\\
{x^2}\left( {17 - {x^2}} \right) = 256
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^4} - 17{x^2} + 16 = 0\\
{x^4} - 17{x^2} + 256 = 0\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow \left( {{x^2} - 16} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow {x^2} = 16/{x^2} = 1\left( {tmdk} \right)\\
Vậy\,x \in \left\{ { - 4; - 1;1;4} \right\}\\
c)Dkxd:{x^3} - 1 \ge 0 \Leftrightarrow x \ge 1\\
2{x^2} + 5x - 1 = 7\sqrt {{x^3} - 1} \\
\Leftrightarrow 2{x^2} + 5x - 1 = 7\sqrt {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} \\
Đặt:\sqrt {{x^2} + x + 1} = a;\sqrt {x - 1} = b\left( {a,b \ge 0} \right)\\
\Leftrightarrow 2{a^2} + 3{b^2} = 2{x^2} + 5x - 1\\
Pt \Leftrightarrow 2{a^2} + 3{b^2} = 7a.b\\
\Leftrightarrow 2{a^2} - 7ab + 3{b^2} = 0\\
\Leftrightarrow 2{a^2} - ab - 6ab + 3{b^2} = 0\\
\Leftrightarrow \left( {2a - b} \right)\left( {a - 3b} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2a = b\\
a = 3b
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2\sqrt {{x^2} + x + 1} = \sqrt {x - 1} \\
\sqrt {{x^2} + x + 1} = 3\sqrt {x - 1}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4{x^2} + 4x + 4 = x - 1\\
{x^2} + x + 1 = 9x - 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4{x^2} + 3x + 5 = 0\left( {vn} \right)\\
{x^2} - 8x + 10 = 0
\end{array} \right.\\
\Leftrightarrow {x^2} - 8x + 16 - 6 = 0\\
\Leftrightarrow {\left( {x - 4} \right)^2} = 6\\
\Leftrightarrow x = 4 \pm \sqrt 6 \left( {tmdk} \right)\\
Vậy\,x = 4 \pm \sqrt 6
\end{array}$
