Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a)\\
DKXD:\,\,\,\,x \ge 0\\
\sqrt {9x}  = 15\\
 \Leftrightarrow {\sqrt {9x} ^2} = {15^2}\\
 \Leftrightarrow 9x = 225\\
 \Leftrightarrow x = 225:9\\
 \Leftrightarrow x = 25\,\,\,\,\,\left( {t/m} \right)\\
b)\\
DKXD:\,\,\,\,\forall x\\
\sqrt {4{x^2}}  = 8\\
 \Leftrightarrow {\sqrt {4{x^2}} ^2} = {8^2}\\
 \Leftrightarrow 4{x^2} = 64\\
 \Leftrightarrow {x^2} = 16\\
 \Leftrightarrow x =  \pm 4\\
c)\\
DKXD:\,\,\,x \ge  - 1\\
\sqrt {4\left( {x + 1} \right)}  = \sqrt 8 \\
 \Leftrightarrow 4\left( {x + 1} \right) = 8\\
 \Leftrightarrow x + 1 = 2\\
 \Leftrightarrow x = 1\,\,\,\,\,\left( {t/m} \right)\\
d)\\
DKXD:\,\,\,\forall x\\
\sqrt {9{{\left( {2 - 3x} \right)}^2}}  = 6\\
 \Leftrightarrow 9{\left( {2 - 3x} \right)^2} = 36\\
 \Leftrightarrow {\left( {2 - 3x} \right)^2} = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
2 - 3x = 2\\
2 - 3x =  - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = 0\\
3x = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{4}{3}
\end{array} \right.\\
e)\\
DKXD:\,\,\,\,x \ge 2\\
\sqrt {{x^2} - 4}  - \sqrt {x - 2}  = 0\\
 \Leftrightarrow \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)}  - \sqrt {x - 2}  = 0\\
 \Leftrightarrow \sqrt {x - 2} .\left( {\sqrt {x + 2}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2}  = 0\\
\sqrt {x + 2}  - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
\sqrt {x + 2}  = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x + 2 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
x \ge 2 \Rightarrow x = 2\\
f,\\
DKXD:\,\,\,\forall x\\
\sqrt {5{x^2}}  = 2x + 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x + 1 \ge 0\\
5{x^2} = {\left( {2x + 1} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
5{x^2} = 4{x^2} + 4x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
{x^2} - 4x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
x = 2 \pm \sqrt 5 
\end{array} \right.\\
 \Leftrightarrow x = 2 \pm \sqrt 5 \\
g,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
{x^2} - 2x \ge 0\\
2 - 3x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 2} \right) \ge 0\\
x \le \dfrac{2}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
x \le \dfrac{2}{3}
\end{array} \right. \Leftrightarrow x \le 0\\
\sqrt {{x^2} - 2x}  = \sqrt {2 - 3x} \\
 \Leftrightarrow {x^2} - 2x = 2 - 3x\\
 \Leftrightarrow {x^2} - 2x - 2 + 3x = 0\\
 \Leftrightarrow {x^2} + x - 2 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.\\
h,\\
DKXD:\,\,\,\,x \ge  - \dfrac{1}{3}\\
1 + \sqrt {3x + 1}  = 3x\\
 \Leftrightarrow \sqrt {3x + 1}  = 3x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
3x - 1 \ge 0\\
3x + 1 = {\left( {3x - 1} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
3x + 1 = 9{x^2} - 6x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
9{x^2} - 9x = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
9x\left( {x - 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
\left\{ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 1\\
i,\\
DKXD:\,\,\,x \ge 5\\
\sqrt {4x - 20}  + 3\sqrt {\dfrac{{x - 5}}{9}}  - \dfrac{1}{3}.\sqrt {9x - 45}  = 4\\
 \Leftrightarrow \sqrt {4.\left( {x - 5} \right)}  + 3.\sqrt {\dfrac{1}{9}.\left( {x - 5} \right)}  - \dfrac{1}{3}.\sqrt {9.\left( {x - 5} \right)}  = 4\\
 \Leftrightarrow 2\sqrt {x - 5}  + 3.\dfrac{1}{3}\sqrt {x - 5}  - \dfrac{1}{3}.3\sqrt {x - 5}  = 4\\
 \Leftrightarrow 2\sqrt {x - 5}  + \sqrt {x - 5}  - \sqrt {x - 5}  = 4\\
 \Leftrightarrow 2\sqrt {x - 5}  = 4\\
 \Leftrightarrow \sqrt {x - 5}  = 2\\
 \Leftrightarrow x - 5 = 4\\
 \Leftrightarrow x = 9
\end{array}\)