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}\)