Đáp án:
c. \(25 > x > 4\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 4\\
b.M = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} + \dfrac{{8\sqrt x + 19}}{{x + \sqrt x + 6}} + \dfrac{1}{{2 - \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 8\sqrt x + 19 - \sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 4 + 8\sqrt x + 19 - \sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 7\sqrt x + 12}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 4} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x - 2}}\\
c.M > 3\\
\to \dfrac{{\sqrt x + 4}}{{\sqrt x - 2}} > 3\\
\to \dfrac{{\sqrt x + 4 - 3\sqrt x + 6}}{{\sqrt x - 2}} > 0\\
\to \dfrac{{10 - 2\sqrt x }}{{\sqrt x - 2}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
10 - 2\sqrt x > 0\\
\sqrt x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
10 - 2\sqrt x < 0\\
\sqrt x - 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5 > \sqrt x \\
\sqrt x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt x > 5\\
\sqrt x < 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 25 > x > 4
\end{array}\)
