Đáp án:
c) \(Min = \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
A = \left[ {\dfrac{{\sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x - 1}}} \right].\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}}\\
= \left[ {\dfrac{{\sqrt x + \sqrt x + 3}}{{\sqrt x - 1}}} \right].\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 3}}{{2\sqrt x - 1}}\\
b)Thay:x = 3 - 2\sqrt 2 \\
= 2 - 2\sqrt 2 .1 + 1 = {\left( {\sqrt 2 - 1} \right)^2}\\
\to A = \dfrac{{2\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + 3}}{{2\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} - 1}}\\
= \dfrac{{2\left( {\sqrt 2 - 1} \right) + 3}}{{2\left( {\sqrt 2 - 1} \right) - 1}} = \dfrac{{2\sqrt 2 + 1}}{{2\sqrt 2 - 3}}\\
= - 11 - 8\sqrt 2 \\
c)P = \dfrac{1}{A} + 1 = \dfrac{{2\sqrt x - 1}}{{2\sqrt x + 3}} + 1\\
= \dfrac{{2\sqrt x - 1 + 2\sqrt x + 3}}{{2\sqrt x + 3}}\\
= \dfrac{{4\sqrt x + 2}}{{2\sqrt x + 3}} = \dfrac{{2\left( {2\sqrt x + 3} \right) - 4}}{{2\sqrt x + 3}}\\
= 2 - \dfrac{4}{{2\sqrt x + 3}}\\
Do:x \ge 0\\
\to 2\sqrt x \ge 0\\
\to 2\sqrt x + 3 \ge 3\\
\to \dfrac{4}{{2\sqrt x + 3}} \le \dfrac{4}{3}\\
\to - \dfrac{4}{{2\sqrt x + 3}} \ge - \dfrac{4}{3}\\
\to 2 - \dfrac{4}{{2\sqrt x + 3}} \ge 2 - \dfrac{4}{3}\\
\to A \ge \dfrac{2}{3}\\
\to Min = \dfrac{2}{3}\\
\Leftrightarrow x = 0
\end{array}\)
