Giải thích các bước giải:
a)
$\begin{array}{l} P = (\sqrt x - \frac{{x + 2}}{{\sqrt x + 1}}):(\frac{{\sqrt x }}{{\sqrt x + 1}} - \frac{{\sqrt x - 4}}{{1 - x}})\\ = \frac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}:\frac{{\sqrt x - x + \sqrt x - 4}}{{(\sqrt x + 1)(\sqrt x - 1)}}\\ = \frac{{\sqrt x - 2}}{{\sqrt x + 1}}:\frac{{ - x + 2\sqrt x - 4}}{{(\sqrt x + 1)(\sqrt x - 1)}}\\ = \frac{{\sqrt x - 2}}{{\sqrt x + 1}}.\frac{{(\sqrt x + 1)(\sqrt x - 1)}}{{ - x + 2\sqrt x - 4}}\\ = \frac{{(\sqrt x - 2)(\sqrt x - 1)}}{{ - x + 2\sqrt x - 4}} \end{array}$
b) P<0 thì $\frac{{(\sqrt x - 2)(\sqrt x - 1)}}{{ - x + 2\sqrt x - 4}}$<0
<=> (${\sqrt x - 2)(\sqrt x - 1) > 0}$ (do ${ - x + 2\sqrt x - 4 = - {{(\sqrt x - 1)}^2} - 3 < 0\forall x}$)
$\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l} \sqrt x - 2 > 0\\ \sqrt x - 1 < 0 \end{array} \right.(do\,\sqrt x - 2 < \sqrt x - 1)\\ \Leftrightarrow \left[ \begin{array}{l} x > 4\\ x < 1 \end{array} \right. \end{array}$
Vậy $\left[ \begin{array}{l} x > 4\\ 0 \le x < 1 \end{array} \right.$
