Đáp án:
a. \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.B = \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b.P = \dfrac{1}{B} = \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 4}}{{\sqrt x + 1}}\\
= 1 - \dfrac{4}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{4}{{\sqrt x + 1}} \le 4\\
\to - \dfrac{4}{{\sqrt x + 1}} \ge - 4\\
\to 1 - \dfrac{4}{{\sqrt x + 1}} \ge - 3\\
\to Min = - 3\\
\Leftrightarrow x = 0\\
c.Q = \dfrac{A}{B} = \dfrac{{x + 2\sqrt x + 5}}{{\sqrt x - 3}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{x + 2\sqrt x + 5}}{{\sqrt x + 1}} = \dfrac{{x + 2\sqrt x + 1 + 4}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + 4}}{{\sqrt x + 1}} = \left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}}\\
Do:x > 0\\
Co - si:\left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{4}{{\sqrt x + 1}}} \\
\to \left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} \ge 2.2 = 4\\
\to Min = 4\\
\Leftrightarrow \left( {\sqrt x + 1} \right) = \dfrac{4}{{\sqrt x + 1}}\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = - 2\left( l \right)
\end{array} \right.\\
\to x = 1
\end{array}\)
