Đáp án:
c) \(Min = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{x + 2\sqrt x + 1 + x - 2\sqrt x + 1 - 3\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{2x - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x + \sqrt x - 2}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {2\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x + 2}}\\
b)A = \dfrac{{2\left( {\sqrt x + 2} \right) - 5}}{{\sqrt x + 2}} = 2 - \dfrac{5}{{\sqrt x + 2}}\\
A \in Z \to \dfrac{5}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 5\\
\sqrt x + 2 = 1\left( l \right)
\end{array} \right.\\
\to \sqrt x = 3\\
\to x = 9\\
c)A = 2 - \dfrac{5}{{\sqrt x + 2}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{5}{{\sqrt x + 2}} \le \dfrac{5}{2}\\
\to - \dfrac{5}{{\sqrt x + 2}} \ge - \dfrac{5}{2}\\
\to 2 - \dfrac{5}{{\sqrt x + 2}} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow x = 0
\end{array}\)
