Đáp án:
c) x>4
Giải thích các bước giải:
\(\begin{array}{l}
a)M = \dfrac{{x - 4\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
b)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to M = \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 2}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 2 + 1 + 2}}{{\sqrt 2 + 1}}\\
= \dfrac{{\sqrt 2 + 3}}{{\sqrt 2 + 1}} = 2\sqrt 2 - 1\\
c)M > 0\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x }} > 0\\
\to \sqrt x - 2 > 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to x > 4
\end{array}\)
