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