Đáp án: $x = \dfrac{{9 + \sqrt {17} }}{2}$
Giải thích các bước giải:
$\begin{array}{l}
c)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}\\
B = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
A.B = \dfrac{{\sqrt x }}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{\sqrt x }}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x }}{2}\\
\Leftrightarrow 2\sqrt x + 4 = x + \sqrt x \\
\Leftrightarrow x - \sqrt x - 4 = 0\\
\Leftrightarrow x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4} - 4 = 0\\
\Leftrightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} = \dfrac{{17}}{4}\\
\Leftrightarrow \sqrt x = \dfrac{{1 + \sqrt {17} }}{2}\\
\Leftrightarrow x = \dfrac{{18 + 2\sqrt {17} }}{4} = \dfrac{{9 + \sqrt {17} }}{2}\left( {tmdk} \right)\\
Vay\,x = \dfrac{{9 + \sqrt {17} }}{2}
\end{array}$
