Đáp án:

 9) \(\left[ \begin{array}{l}
x = 3\\
x =  - \dfrac{9}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)DK:x \ge 0\\
3x - 4\sqrt x  - 3\sqrt x  + 4 = 0\\
 \to \sqrt x \left( {3\sqrt x  - 4} \right) - \left( {3\sqrt x  - 4} \right) = 0\\
 \to \left( {3\sqrt x  - 4} \right)\left( {\sqrt x  - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
\sqrt x  = \dfrac{4}{3}\\
\sqrt x  = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{16}}{9}\\
x = 1
\end{array} \right.\\
3)DK:\left[ \begin{array}{l}
x \ge \sqrt 6 \\
x \le  - \sqrt 6 
\end{array} \right.\\
{x^2} - 12 + \sqrt {{x^2} - 6}  = 0\\
Đặt:\sqrt {{x^2} - 6}  = t\left( {t \ge 0} \right)\\
 \to {x^2} - 6 = {t^2}\\
 \to {x^2} - 12 = {t^2} - 6\\
Pt \to {t^2} - 6 + t = 0\\
 \to \left( {t - 2} \right)\left( {t + 3} \right) = 0\\
 \to \left[ \begin{array}{l}
t = 2\\
t =  - 3\left( l \right)
\end{array} \right.\\
 \to \sqrt {{x^2} - 6}  = 2\\
 \to {x^2} - 6 = 4\\
 \to {x^2} = 10\\
 \to \left[ \begin{array}{l}
x = \sqrt {10} \\
x =  - \sqrt {10} 
\end{array} \right.\left( {TM} \right)\\
5)DK:x \ge  - 2\\
\sqrt {\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}  = 2\left( {x - 2} \right)\left( {x - 1} \right)\\
 \to \sqrt {\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}  - 2\left( {x - 2} \right)\left( {x - 1} \right) = 0\\
 \to \sqrt {x + 2} \left( {\sqrt {{x^2} - 2x + 4}  - 2\left( {x - 1} \right)\sqrt {x - 2} } \right) = 0\\
 \to \left[ \begin{array}{l}
x + 2 = 0\\
\sqrt {{x^2} - 2x + 4}  = 2\left( {x - 1} \right)\sqrt {x - 2} 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2\\
{x^2} - 2x + 4 = 4\left( {{x^2} - 2x + 1} \right)\left( {x - 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2\\
{x^2} - 2x + 4 = 4\left( {{x^3} - 2{x^2} - 2{x^2} + 4x + x - 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2\\
4{x^3} - 17{x^2} + 22x - 12 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2\\
x = 2,558699778
\end{array} \right.\\
7)DK:{x^2} + 5x + 1 \ge 0 \to \left[ \begin{array}{l}
x \ge \dfrac{{ - 5 + \sqrt {21} }}{2}\\
x \le \dfrac{{ - 5 - \sqrt {21} }}{2}
\end{array} \right.\\
3\left( {{x^2} + 5x} \right) + 2\sqrt {{x^2} + 5x + 1}  = 2\\
Đặt:\sqrt {{x^2} + 5x + 1}  = t\left( {t \ge 0} \right)\\
 \to {x^2} + 5x + 1 = {t^2}\\
 \to {x^2} + 5x = {t^2} - 1\\
 \to 3\left( {{t^2} - 1} \right) + 2t = 2\\
 \to 3{t^2} + 2t - 5 = 0\\
 \to \left[ \begin{array}{l}
t = 1\\
t =  - \dfrac{5}{3}\left( l \right)
\end{array} \right.\\
 \to \sqrt {{x^2} + 5x + 1}  = 1\\
 \to {x^2} + 5x + 1 = 1\\
 \to {x^2} + 5x = 0\\
 \to x\left( {x + 5} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
x =  - 5
\end{array} \right.\left( {TM} \right)\\
9)2{x^2} + 3x + 3 = 5\sqrt {2{x^2} + 3x + 9} \\
Đặt:\sqrt {2{x^2} + 3x + 9}  = t\left( {t \ge 0} \right)\\
 \to 2{x^2} + 3x + 9 = {t^2}\\
 \to 2{x^2} + 3x + 3 = {t^2} - 6\\
Pt \to {t^2} - 6 = 5t\\
 \to \left( {t - 6} \right)\left( {t + 1} \right) = 0\\
 \to \left[ \begin{array}{l}
t = 6\\
t =  - 1\left( l \right)
\end{array} \right.\\
 \to \sqrt {2{x^2} + 3x + 9}  = 6\\
 \to 2{x^2} + 3x + 9 = 36\\
 \to \left( {x - 3} \right)\left( {2x + 9} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  - \dfrac{9}{2}
\end{array} \right.
\end{array}\)