Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
\frac{{x + 2}}{{x\sqrt x - 1}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\
\Rightarrow \frac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \sqrt x = \frac{1}{{x + \sqrt x + 1}}\\
\Rightarrow \frac{{x + 2 + \sqrt x \left( {x\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
\Rightarrow x + 2 + {x^2} - \sqrt x = \sqrt x - 1\\
\Rightarrow {x^2} + x - 2\sqrt x + 3 = 0\\
\Rightarrow {x^2} + \left( {x - 2\sqrt x + 1} \right) + 2 = 0\\
\Rightarrow {x^2} + {\left( {\sqrt x - 1} \right)^2} + 2 = 0\left( {vô\,nghiệm} \right)
\end{array}$
Vậy phương trình vô nghiệm.
