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