Đáp án:
a) \(\dfrac{{x + 2\sqrt x + 2}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{1}{{\sqrt x \left( {\sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x + 1}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{1 + x + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 2}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
b)A > \dfrac{1}{2}\\
\to \dfrac{{x + 2\sqrt x + 2}}{{\sqrt x \left( {\sqrt x + 1} \right)}} > \dfrac{1}{2}\\
\to \dfrac{{2x + 4\sqrt x + 4 - x - \sqrt x }}{{2\sqrt x \left( {\sqrt x + 1} \right)}} > 0\\
\to \dfrac{{x + 3\sqrt x + 4}}{{2\sqrt x \left( {\sqrt x + 1} \right)}} > 0\\
\to x + 3\sqrt x + 4 > 0\left( {do:\left\{ \begin{array}{l}
\sqrt x > 0\forall x > 0\\
\sqrt x + 1 > 0\forall x > 0
\end{array} \right.} \right)\\
\to x + 2.\sqrt x .\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{7}{4} > 0\\
\to {\left( {\sqrt x + \dfrac{3}{2}} \right)^2} + \dfrac{7}{4} > 0\left( {ld} \right)\forall x > 0\\
KL:x > 0
\end{array}\)
