Đáp án:

\(\begin{array}{l}
a)A = \dfrac{{13}}{8}\\
b)\dfrac{{\sqrt x  - 5}}{{\sqrt x  + 8}}\\
c)\left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.\\
d)0 \le x < 1
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)A = \dfrac{{\sqrt x \left( {\sqrt x  + 3} \right) + 2\sqrt x  - 24}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + 3\sqrt x  + 2\sqrt x  - 24}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{x + 5\sqrt x  - 24}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 8} \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{\sqrt x  + 8}}{{\sqrt x  + 3}}\\
Thay:x = 25\\
 \to A = \dfrac{{\sqrt {25}  + 8}}{{\sqrt {25}  + 3}} = \dfrac{{5 + 8}}{{5 + 3}} = \dfrac{{13}}{8}\\
b)B = \dfrac{{\sqrt x  - 5}}{{\sqrt x  + 8}}\\
c)Q = A.B = \dfrac{{\sqrt x  + 8}}{{\sqrt x  + 3}}.\dfrac{{\sqrt x  - 5}}{{\sqrt x  + 8}}\\
 = \dfrac{{\sqrt x  - 5}}{{\sqrt x  + 3}} = \dfrac{{\sqrt x  + 3 - 8}}{{\sqrt x  + 3}} = 1 - \dfrac{8}{{\sqrt x  + 3}}\\
Q \in Z \to \dfrac{8}{{\sqrt x  + 3}} \in Z \to \sqrt x  + 3 \in U\left( 8 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 3 = 8\\
\sqrt x  + 3 = 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 5\\
\sqrt x  = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.\\
d)Q <  - 1\\
 \to \dfrac{{\sqrt x  - 5}}{{\sqrt x  + 3}} + 1 < 0\\
 \to \dfrac{{\sqrt x  - 5 + \sqrt x  + 3}}{{\sqrt x  + 3}} < 0\\
 \to \dfrac{{2\sqrt x  - 2}}{{\sqrt x  + 3}} < 0\\
 \to \sqrt x  - 1 < 0\left( {do:\sqrt x  + 3 > 0\forall x} \right)\\
 \to 0 \le x < 1
\end{array}\)