Đáp án:
$\begin{array}{l}
d)Dkxd:a > 0;a \ne 1\\
N = 2.\dfrac{{a + \sqrt a + 1}}{{\sqrt a }}\\
\Rightarrow N - \sqrt a \\
= 2.\dfrac{{a + \sqrt a + 1}}{{\sqrt a }} - \sqrt a \\
= \dfrac{{2a + 2\sqrt a + 2 - a}}{{\sqrt a }}\\
= \dfrac{{a + 2\sqrt a + 2}}{{\sqrt a }}\\
= \sqrt a + 2 + \dfrac{2}{{\sqrt a }}\\
Do:\sqrt a ;\dfrac{2}{{\sqrt a }} > 0\\
Theo\,Cô - si:\\
\sqrt a + \dfrac{2}{{\sqrt a }} \ge 2.\sqrt {\sqrt a .\dfrac{2}{{\sqrt a }}} = 2.\sqrt 2 \\
\Rightarrow \sqrt a + 2 + \dfrac{2}{{\sqrt a }} \ge 2 + 2\sqrt 2 \\
\Rightarrow N - \sqrt a \ge 2 + 2\sqrt 2 \\
\Rightarrow GTNN:N - \sqrt a = 2 + 2\sqrt 2 \\
Khi:\sqrt a = \dfrac{2}{{\sqrt a }}\\
\Rightarrow a = 2\left( {tmdk} \right)
\end{array}$
