Đáp án:
\(x>1\)
Giải thích các bước giải:
\(\begin{array}{l}
Q = \left( {\frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \frac{{\sqrt x - 2}}{{x - 1}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x }}\,\,\left( {x > 0;\,\,x \ne 1} \right)\\
Q = \left( {\frac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} - \frac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x }}\\
Q = \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{\sqrt x + 1}}{{\sqrt x }}\\
Q = \frac{{x + \sqrt x - 2 - x + \sqrt x + 2}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{\sqrt x + 1}}{{\sqrt x }}\\
Q = \frac{{2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\frac{1}{{\sqrt x }} = \frac{2}{{x - 1}}\\
TH1:\,\,\left| Q \right| = Q \Leftrightarrow Q \ge 0 \Leftrightarrow \frac{2}{{x - 1}} \ge 0 \Leftrightarrow x - 1 > 0 \Leftrightarrow x > 1\\
\Rightarrow Q > - Q \Leftrightarrow 2Q > 0 \Leftrightarrow \frac{2}{{x - 1}} > 0 \Leftrightarrow x > 1\,\,\left( {tm} \right).\\
TH2:\,\,\left| Q \right| = - Q \Leftrightarrow Q < 0 \Leftrightarrow x < 1\\
\Rightarrow - Q > - Q\,\,\left( {Vo\,\,li} \right)\\
Vay\,\,x > 1\,\,thi\,\,\,\left| Q \right| > - Q.
\end{array}\)
