Đáp án:
1) \(\dfrac{{2x + 1}}{{4\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \left[ {\dfrac{{x + 2\sqrt x + 1 + x - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]:\left[ {\dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \dfrac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{4\sqrt x }}\\
= \dfrac{{2x + 1}}{{4\sqrt x }}\\
2)Thay:x = 4\\
\to A = \dfrac{{2.4 + 1}}{{4\sqrt 4 }} = \dfrac{9}{4}\\
3)Xét:A > \dfrac{1}{2}\\
\to \dfrac{{2x + 1}}{{4\sqrt x }} > \dfrac{1}{2}\\
\to \dfrac{{4x + 2 - 4\sqrt x }}{{8\sqrt x }} > 0\\
\to \dfrac{{2x - 2\sqrt x + 1}}{{4\sqrt x }} > 0\\
Do:\left\{ \begin{array}{l}
2x - 2\sqrt x + 1 > 0\forall x > 0\\
4\sqrt x > 0\forall x > 0
\end{array} \right.\\
\to \dfrac{{2x - 2\sqrt x + 1}}{{4\sqrt x }} > 0\left( {ld} \right)\\
\to A > \dfrac{1}{2}
\end{array}\)
