Đáp án:
a. \(S = \frac{{\sqrt x }}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \frac{{x + 2}}{{x\sqrt x - 1}} + \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \frac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \frac{{x + 2 + x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \frac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
B = \frac{1}{{\sqrt x - 1}}\\
1)S = A - B = \frac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{1}{{\sqrt x - 1}}\\
= \frac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \frac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \frac{{\sqrt x }}{{x + \sqrt x + 1}}\\
2)Xét:\frac{{\sqrt x }}{{x + \sqrt x + 1}} > \frac{1}{3}\\
\to \frac{{3\sqrt x - x - \sqrt x - 1}}{{3\left( {x + \sqrt x + 1} \right)}} > 0\\
\to - x + 2\sqrt x - 1 > 0\left( {do:x + \sqrt x + 1 > 0\forall x > 0} \right)\\
\to - {\left( {\sqrt x - 1} \right)^2} > 0\\
\to {\left( {\sqrt x - 1} \right)^2} < 0\left( {vôlý} \right)\left( {do:{{\left( {\sqrt x - 1} \right)}^2} \ge 0\forall x > 0} \right)\\
Xét:\frac{{\sqrt x }}{{x + \sqrt x + 1}} \le \frac{1}{3}\\
\to - {\left( {\sqrt x - 1} \right)^2} \le 0\\
\to {\left( {\sqrt x - 1} \right)^2} \ge 0\left( {ld} \right)
\end{array}\)
