Giải thích các bước giải:
a) Ta có:
$M = \left( {\dfrac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\dfrac{{25 - x}}{{x + 2\sqrt x - 15}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)$
Để M có nghĩa
$ \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 25\\
x \ne 9
\end{array} \right.$
Vậy $x \ge 0;x \ne \left\{ {25;9} \right\}$ thì $M$ có nghĩa.
b) ĐK: $x \ge 0;x \ne \left\{ {25;9} \right\}$
Ta có:
$\begin{array}{l}
M = \left( {\dfrac{{x - 5\sqrt x }}{{x - 25}} - 1} \right):\left( {\dfrac{{25 - x}}{{x + 2\sqrt x - 15}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)\\
= \left( {\dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} - 1} \right):\left( {\dfrac{{25 - x}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x + 5}} + \dfrac{{\sqrt x - 5}}{{\sqrt x - 3}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x + 5}} - 1} \right):\left( {\dfrac{{25 - x - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}}} \right)\\
= \left( {\dfrac{{\sqrt x - \left( {\sqrt x + 5} \right)}}{{\sqrt x + 5}}} \right):\left( {\dfrac{{25 - x - x + 9 + x - 25}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}}} \right)\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}:\dfrac{{ - x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{ - 5}}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 5} \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}\\
= \dfrac{5}{{\sqrt x + 3}}
\end{array}$
Vậy $M = \dfrac{5}{{\sqrt x + 3}}$ với $x \ge 0;x \ne \left\{ {25;9} \right\}$
c) Ta có:
$M = \dfrac{5}{{\sqrt x + 3}}$ với $x \ge 0;x \ne \left\{ {25;9} \right\}$
Để $M \in Z$
$ \Leftrightarrow \dfrac{5}{{\sqrt x + 3}} \in Z$
Mà $x \in Z$ $ \Rightarrow \left( {\sqrt x + 3} \right) \in U\left( 5 \right) = \left\{ { - 5; - 1;1;5} \right\}$
Mặt khác: $\sqrt x + 3 \ge 3,\forall x \ge 0;x \ne \left\{ {25;9} \right\}$
$\begin{array}{l}
\Rightarrow \sqrt x + 3 = 5\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4
\end{array}$
Vậy $x=4$ thỏa mãn đề.
