Đáp án:
a. P=-8
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = \dfrac{1}{4}\\
P = \dfrac{{4.\dfrac{1}{4} + 3}}{{\sqrt {\dfrac{1}{4}} - 1}} = \dfrac{{1 + 3}}{{\dfrac{1}{2} - 1}}\\
= 4:\left( { - \dfrac{1}{2}} \right) = 4.\left( { - 2} \right) = - 8\\
2)Q = \dfrac{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x - 1} \right) + 5\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x - 3\sqrt x + 1 + 5\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x + 2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x - 1}}\\
3)S = \dfrac{P}{Q} = \dfrac{{4x + 3}}{{\sqrt x - 1}}:\dfrac{{2\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{4x + 3}}{{2\sqrt x }} = 2\sqrt x + \dfrac{3}{{2\sqrt x }}\\
Do:x > 0\\
Cô - si:2\sqrt x + \dfrac{3}{{2\sqrt x }} \ge 2\sqrt {2\sqrt x .\dfrac{3}{{2\sqrt x }}} \\
\to 2\sqrt x + \dfrac{3}{{2\sqrt x }} \ge 2\sqrt 3 \\
\to Min = 2\sqrt 3 \\
\Leftrightarrow 2\sqrt x = \dfrac{3}{{2\sqrt x }}\\
\Leftrightarrow 4x = 3\\
\Leftrightarrow x = \dfrac{3}{4}
\end{array}\)
