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