Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2a,\\
\sqrt {9x - 18} + \sqrt {4x - 8} = 20 + \dfrac{1}{4}\sqrt {16x - 32} \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \sqrt {9\left( {x - 2} \right)} + \sqrt {4\left( {x - 2} \right)} = 20 + \dfrac{1}{4}\sqrt {16\left( {x - 2} \right)} \\
\Leftrightarrow \sqrt {{3^2}.\left( {x - 2} \right)} + \sqrt {{2^2}.\left( {x - 2} \right)} = 20 + \dfrac{1}{4}\sqrt {{4^2}.\left( {x - 2} \right)} \\
\Leftrightarrow 3\sqrt {x - 2} + 2\sqrt {x - 2} = 20 + \dfrac{1}{4}.4\sqrt {x - 2} \\
\Leftrightarrow 5\sqrt {x - 2} = 20 + \sqrt {x - 2} \\
\Leftrightarrow 4\sqrt {x - 2} = 20\\
\Leftrightarrow \sqrt {x - 2} = 5\\
\Leftrightarrow x - 2 = 25\\
\Leftrightarrow x = 27\\
3,\\
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
3 - \sqrt x \ne 0\\
3 + \sqrt x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
A = \left( {\dfrac{{\sqrt x }}{{3 - \sqrt x }} - \dfrac{{\sqrt x }}{{3 + \sqrt x }}} \right):\dfrac{{12}}{{3 + \sqrt x }}\\
= \dfrac{{\sqrt x \left( {3 + \sqrt x } \right) - \sqrt x \left( {3 - \sqrt x } \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{12}}{{3 + \sqrt x }}\\
= \dfrac{{\left( {3\sqrt x + x} \right) - \left( {3\sqrt x - x} \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{3 + \sqrt x }}{{12}}\\
= \dfrac{{2x}}{{\left( {3 - \sqrt x } \right).\left( {3 + \sqrt x } \right)}}.\dfrac{{3 + \sqrt x }}{{12}}\\
= \dfrac{x}{{6.\left( {3 - \sqrt x } \right)}}\\
c,\\
x \ge 0\\
A > 0 \Leftrightarrow \dfrac{x}{{6.\left( {3 - \sqrt x } \right)}} > 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
3 - \sqrt x > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
\sqrt x < 9
\end{array} \right. \Leftrightarrow 0 < x < 9
\end{array}\)
