Lời giải

a) ĐKXĐ: \(\left\{ \begin{array}{l}x - 4 \ne 0\\x + 4\sqrt x + 4 \ne 0\\x \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 4\\{\left( {\sqrt x + 2} \right)^2} \ne 0\\x \ge 0\end{array} \right.\)

\[ \Leftrightarrow \left\{ \begin{array}{l}x \ne 4\\\sqrt x + 2 \ne 0\,\,\left( {luon\,\,dung} \right)\\x \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 4\\x \ge 0\end{array} \right.\].

b) \(Q = \left( {\frac{1}{{x - 4}} - \frac{1}{{x + 4\sqrt x + 4}}} \right).\frac{{x + 2\sqrt x }}{{\sqrt x }}\)

\( = \left[ {\frac{1}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} - \frac{1}{{{{\left( {\sqrt x + 2} \right)}^2}}}} \right].\frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x }}\)

\( = \frac{{\sqrt x + 2 - \sqrt x + 2}}{{{{\left( {\sqrt x + 2} \right)}^2}\left( {\sqrt x - 2} \right)}}.\left( {\sqrt x + 2} \right)\)

\( = \frac{4}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} = \frac{4}{{x - 4}}\).

Vậy \(Q = \frac{4}{{x - 4}}\).