Đáp án:
2) \(\dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 64\\
\to A = \dfrac{{2 + \sqrt {64} }}{{64}} = \dfrac{{10}}{{64}} = \dfrac{5}{{32}}\\
2)B = \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1 + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}} = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
3)\dfrac{A}{B} > \dfrac{3}{2}\\
\to \dfrac{{\sqrt x + 2}}{x}:\dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} > \dfrac{3}{2}\\
\to \dfrac{{\sqrt x + 2}}{x}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} > \dfrac{3}{2}\\
\to \dfrac{{\sqrt x + 1}}{x} > \dfrac{3}{2}\\
\to \dfrac{{2\sqrt x + 2 - 3x}}{{2x}} > 0\\
\to 2\sqrt x + 2 - 3x > 0\left( {do:x > 0} \right)\\
\to 3x - 2\sqrt x - 2 > 0\\
\to 3x - 2.\sqrt {3x} .\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{3} - \dfrac{7}{3} > 0\\
\to {\left( {\sqrt {3x} - \dfrac{1}{{\sqrt 3 }}} \right)^2} > \dfrac{7}{3}\\
\to \left[ \begin{array}{l}
\sqrt {3x} - \dfrac{1}{{\sqrt 3 }} > \sqrt {\dfrac{7}{3}} \\
\sqrt {3x} - \dfrac{1}{{\sqrt 3 }} < - \sqrt {\dfrac{7}{3}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{{1 + \sqrt 7 }}{3}\\
x < \dfrac{{1 - \sqrt 7 }}{3}\left( l \right)
\end{array} \right.\\
\to x > \dfrac{{1 + \sqrt 7 }}{3}
\end{array}\)
