Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
2\sqrt {2x} - 5\sqrt {8x} + 7\sqrt {18x} = 28\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow 2\sqrt {2x} - 5.\sqrt {4.2x} + 7.\sqrt {9.2x} = 28\\
\Leftrightarrow 2\sqrt {2x} - 5.2.\sqrt {2x} + 7.3.\sqrt {2x} = 28\\
\Leftrightarrow 2\sqrt {2x} - 10\sqrt {2x} + 21\sqrt {2x} = 28\\
\Leftrightarrow 13\sqrt {2x} = 28\\
\Leftrightarrow \sqrt {2x} = \dfrac{{28}}{{13}}\\
\Leftrightarrow 2x = \dfrac{{784}}{{169}}\\
\Leftrightarrow x = \dfrac{{392}}{{169}}\\
b,\\
\sqrt {{x^2} - 4x + 4} = 3\\
\Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = 3\\
\Leftrightarrow \left| {x - 2} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.\\
c,\\
\sqrt {{x^2} - 12} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left[ \begin{array}{l}
x \ge 2\sqrt 3 \\
x \le - 2\sqrt 3
\end{array} \right.} \right)\\
\Leftrightarrow {x^2} - 12 = 4\\
\Leftrightarrow {x^2} = 16\\
\Leftrightarrow x = \pm 4\\
d,\\
\sqrt {{x^2} + 2x + 1} = \sqrt {x + 1} \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - 1} \right)\\
\Leftrightarrow {\sqrt {\left( {x + 1} \right)} ^2} - \sqrt {x + 1} = 0\\
\Leftrightarrow \sqrt {x + 1} \left( {\sqrt {x + 1} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 0\\
\sqrt {x + 1} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.\\
e,\\
\sqrt {{x^2} - 9} + \sqrt {{x^2} - 6x + 9} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{x^2} - 9} = 0\\
\sqrt {{x^2} - 6x + 9} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
{\left( {x - 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = 3\\
3,\\
a,\\
\sqrt {\dfrac{{3x - 1}}{{x + 2}}} = 2\,\,\,\,\,\,\,\,\,\,\left( {\left[ \begin{array}{l}
x \ge \dfrac{1}{3}\\
x < - 2
\end{array} \right.} \right)\\
\Leftrightarrow \dfrac{{3x - 1}}{{x + 2}} = 4\\
\Leftrightarrow 3x - 1 = 4.\left( {x + 2} \right)\\
\Leftrightarrow 3x - 1 = 4x + 8\\
\Leftrightarrow x = - 9\\
b,\\
\dfrac{{\sqrt {5x - 7} }}{{\sqrt {2x - 1} }} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{7}{5}} \right)\\
\Leftrightarrow \dfrac{{5x - 7}}{{2x - 1}} = 1\\
\Leftrightarrow 5x - 7 = 2x - 1\\
\Leftrightarrow 3x = 6\\
\Leftrightarrow x = 2\\
c,\\
\dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) = \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
\Leftrightarrow x + \sqrt x - 6 = x - 1\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25
\end{array}\)
