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