Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
Q = \left( {\frac{{\sqrt {4x} }}{{\sqrt {4x} - 4}} + \frac{{\sqrt x }}{{\sqrt x + 2}}} \right):\frac{{\sqrt {4x} }}{{x - 4}}\\
\Leftrightarrow Q = \left( {\frac{{2\sqrt x }}{{2\sqrt x - 4}} + \frac{{\sqrt x }}{{\sqrt x + 2}}} \right):\frac{{2\sqrt x }}{{x - 4}}\\
\Leftrightarrow Q = \left( {\frac{{\sqrt x }}{{\sqrt x - 2}} + \frac{{\sqrt x }}{{\sqrt x + 2}}} \right):\frac{{2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
\Leftrightarrow Q = \frac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{2\sqrt x }}\\
\Leftrightarrow Q = \frac{{2x.\left( {\sqrt {x - 2} } \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right).2\sqrt x }} = \sqrt x
\end{array}\)
b,
Khi \(x = \sqrt {24} + 5\) thì
\(Q = \sqrt {\sqrt {24} + 5} = \sqrt {2 + 2\sqrt 2 .\sqrt 3 + 3} = \sqrt {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}} = \sqrt 2 + \sqrt 3 \)
c,
\(\begin{array}{l}
Q < 3 \Leftrightarrow \sqrt x < 3 \Leftrightarrow x < 9\\
\Rightarrow \left\{ \begin{array}{l}
0 < x < 9\\
x \ne 4
\end{array} \right.
\end{array}\)
