Đáp án:
\(\dfrac{{2x + 1}}{{4\sqrt x }} > \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 1\\
Xét:\dfrac{{2x + 1}}{{4\sqrt x }} > \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{4x + 2 - 4\sqrt x }}{{8\sqrt x }} > 0\\
\Leftrightarrow \dfrac{{2x - 2\sqrt x + 1}}{{4\sqrt x }} > 0\\
Do:4\sqrt x > 0\forall x > 0\\
\to 2x - 2\sqrt x + 1 > 0\\
\to {\left( {\sqrt {2x} } \right)^2} - 2.\sqrt {2x} .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2} > 0\\
\to {\left( {\sqrt {2x} - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} > 0\left( {ld} \right)\forall x > 0\\
\to \dfrac{{2x + 1}}{{4\sqrt x }} > \dfrac{1}{2}
\end{array}\)
