Đáp án:
$\begin{array}{l}
a)M = \dfrac{{\sqrt x }}{{\sqrt x - 2}} - \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x .\sqrt x - 4\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
b)x = 3 + 2\sqrt 2 \\
\Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 2 + 1\\
\Rightarrow M = \dfrac{{\sqrt x + 2}}{{\sqrt x }} = 1 + \dfrac{2}{{\sqrt x }}\\
= 1 + \dfrac{2}{{\sqrt 2 + 1}}\\
= 1 + \dfrac{{2\left( {\sqrt 2 - 1} \right)}}{{2 - 1}}\\
= 1 + 2\sqrt 2 - 2\\
= 2\sqrt 2 - 1\\
c)M > 0\\
\Rightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x }} > 0\\
\Rightarrow 1 + \dfrac{2}{{\sqrt x }} > 0\left( {\text{Luôn đúng}dung} \right)
\end{array}$
Vậy M>0 với mọi x>0;x#4
