Đáp án đúng:
Giải chi tiết:Giải:
a) ĐKXĐ: \(x\ge 0;\,\,x\ne 4.\)
\(\begin{array}{l}P = \frac{{\sqrt x }}{{\sqrt x - 2}} - \frac{{\sqrt x + 1}}{{\sqrt x + 2}} + \frac{{\sqrt x + 5}}{{4 - x}}\\\,\,\,\, = \frac{{\sqrt x }}{{\sqrt x - 2}} - \frac{{\sqrt x + 1}}{{\sqrt x + 2}} - \frac{{\sqrt x + 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\\,\,\,\, = \frac{{\sqrt x \left( {\sqrt x + 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \sqrt x - 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\\,\,\,\, = \frac{{x + 2\sqrt x - x + \sqrt x + 2 - \sqrt x - 5}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\\,\,\,\, = \frac{{2\sqrt x - 3}}{{x - 4}}.\\Q = \frac{{3 - \sqrt x }}{{\sqrt x - 2}} + 1 = \frac{{3 - \sqrt x + \sqrt x - 2}}{{\sqrt x - 2}} = \frac{1}{{\sqrt x - 2}}.\end{array}\)
b) Ta có:
\(\frac{P}{Q} - 2 = \frac{{2\sqrt x - 3}}{{x - 4}}:\frac{1}{{\sqrt x - 2}} - 2 = \frac{{2\sqrt x - 3}}{{\sqrt x + 2}} - 2 = \frac{{2\sqrt x - 3 - 2\sqrt x - 4}}{{\sqrt x + 2}} = \frac{{ - 7}}{{\sqrt x + 2}} < 0\)
\( \Rightarrow \frac{P}{Q} < 2.\)
c)
\(\begin{array}{l}\left| P \right| + Q = 0 \Rightarrow \left| P \right| = - Q \Rightarrow Q \le 0\\\Leftrightarrow \frac{1}{{\sqrt x - 2}} \le 0 \Leftrightarrow \sqrt x - 2 < 0 \Leftrightarrow x < 4 \Rightarrow 0 \le x < 4.\end{array}\)
\(\begin{array}{l}\left| P \right| = - Q \Leftrightarrow \left| {\frac{{2\sqrt x - 3}}{{x - 4}}} \right| = \frac{{ - 1}}{{\sqrt x - 2}} \Leftrightarrow \left| {\frac{{2\sqrt x - 3}}{{x - 4}}} \right| = - \frac{{\sqrt x + 2}}{{x - 4}}\\\Leftrightarrow \left[ \begin{array}{l}2\sqrt x - 3 = \sqrt x + 2\\2\sqrt x - 3 = - \sqrt x - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt x = 5\\\sqrt x = \frac{1}{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 25\,\,(l)\\x = \frac{1}{9}(tm)\end{array} \right.\end{array}\)
Vậy \(x=\frac{1}{9}\)
Chú ý:Chú ý: \(\left| A \right|=\left\{ \begin{align} & A\,\,\,\,khi\,\,\,\,\,A\ge 0 \\ & -A\,\,\,\,khi\,\,\,\,A<0 \\ \end{align} \right..\)
