Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0;a \ne 1\\
M = \dfrac{{a + 1}}{{\sqrt a }} - \dfrac{{a\sqrt a + 1}}{{\sqrt a + a}}\\
= \dfrac{{a + 1}}{{\sqrt a }} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}\\
= \dfrac{{a + 1}}{{\sqrt a }} - \dfrac{{a - \sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{a + 1 - a + \sqrt a - 1}}{{\sqrt a }}\\
= \dfrac{{\sqrt a }}{{\sqrt a }} = 1\\
b)N = \dfrac{7}{2} = \dfrac{{a\sqrt a - 1}}{{a - \sqrt a }}\\
\Rightarrow \dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right).\sqrt a }} = \dfrac{7}{2}\\
\Rightarrow \dfrac{{a + \sqrt a + 1}}{{\sqrt a }} = \dfrac{7}{2}\\
\Rightarrow 2a + 2\sqrt a + 2 = 7\sqrt a \\
\Rightarrow 2a - 5\sqrt a + 2 = 0\\
\Rightarrow \left( {2\sqrt a - 1} \right)\left( {\sqrt a - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt a = \dfrac{1}{2}\\
\sqrt a = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = \dfrac{1}{4}\left( {tm} \right)\\
a = 1\left( {ktm} \right)
\end{array} \right.\\
Vay\,a = \dfrac{1}{4}\\
c)M + N - 4\\
= 1 + \dfrac{{a + \sqrt a + 1}}{{\sqrt a }} - 4\\
= \dfrac{{a + \sqrt a + 1 - 3\sqrt a }}{{\sqrt a }}\\
= \dfrac{{a - 2\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a }} > 0\left( {khi:a > 0;a \ne 1} \right)\\
\Rightarrow M + N - 4 > 0\\
\Rightarrow M + N > 4
\end{array}$
