Đáp án:
\(\begin{array}{l}
a.P = \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
b.x = 16
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.P = \left( {\sqrt x - \frac{{x + 2}}{{\sqrt x + 1}}} \right):\left( {\frac{{\sqrt x }}{{\sqrt x + 1}} - \frac{{\sqrt x - 4}}{{1 - x}}} \right)\\
DKXD:\left\{ \begin{array}{l}
\sqrt x + 1 \ne 0\\
x \ge 0\\
1 - x \ne 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
P = \left[ {\frac{{\sqrt x (\sqrt x + 1)}}{{\sqrt x + 1}} - \frac{{x + 2}}{{\sqrt x + 1}}} \right]:\left[ {\frac{{\sqrt x (\sqrt x - 1)}}{{(\sqrt x + 1)(\sqrt x - 1)}} + \frac{{\sqrt x - 4}}{{(\sqrt x + 1)(\sqrt x - 1)}}} \right]\\
= \frac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}:\frac{{x - \sqrt x + \sqrt x - 4}}{{(\sqrt x + 1)(\sqrt x - 1)}}\\
= \frac{{\sqrt x - 2}}{{\sqrt x + 1}}.\frac{{(\sqrt x + 1)(\sqrt x - 1)}}{{(\sqrt x - 2)(\sqrt x + 2)}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
b.P = \frac{1}{2}\\
\leftrightarrow \frac{{\sqrt x - 1}}{{\sqrt x + 2}} = \frac{1}{2}\\
\leftrightarrow 2\sqrt x - 2 = \sqrt x + 2\\
\leftrightarrow \sqrt x = 4\\
\leftrightarrow x = 16(tm)
\end{array}\)
