Đáp án:
3) \(0 < x \le \dfrac{{25}}{4};x \ne 4\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = \dfrac{1}{4}\\
\to A = \dfrac{{2 + \sqrt {\dfrac{1}{4}} }}{{\sqrt {\dfrac{1}{4}} }} = 5\\
2)B = \dfrac{{x + \sqrt x + 2 + \sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
3)P = \dfrac{A}{B} = \dfrac{{\sqrt x + 2}}{{\sqrt x }}:\dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
= \dfrac{{x - 4}}{x}\\
P.x \le \dfrac{3}{2}\left( {\sqrt x - 1} \right)\\
\to \dfrac{{x - 4}}{x}.x \le \dfrac{3}{2}\left( {\sqrt x - 1} \right)\\
\to x - 4 \le \dfrac{3}{2}\left( {\sqrt x - 1} \right)\\
\to x - 4 \le \dfrac{3}{2}\sqrt x - \dfrac{3}{2}\\
\to x - \dfrac{3}{2}\sqrt x - \dfrac{5}{2} \le 0\\
\to \left( {2\sqrt x - 5} \right)\left( {\sqrt x + 1} \right) \le 0\\
\to 2\sqrt x - 5 \le 0\left( {do:\sqrt x + 1 > 0\forall x > 0} \right)\\
\to x \le \dfrac{{25}}{4}\\
\to 0 < x \le \dfrac{{25}}{4};x \ne 4
\end{array}\)
