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Đáp án:
$\begin{array}{l}
B1)\\
a)Dkxd:x \ge 0;x \ne 4\\
B = \left( {\dfrac{{\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x - 2}}} \right):\dfrac{{\sqrt x + 2}}{{x - 4}}\\
= \dfrac{{\sqrt x + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{x - 4}}{{\sqrt x + 2}}\\
= \dfrac{{2\sqrt x + 2}}{{\sqrt x + 2}}\\
b)C = A.\left( {B - 2} \right)\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}}.\left( {\dfrac{{2\sqrt x + 2}}{{\sqrt x + 2}} - 2} \right)\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}}.\dfrac{{2\sqrt x + 2 - 2\sqrt x - 4}}{{\sqrt x + 2}}\\
= \dfrac{{ - 2}}{{\sqrt x - 2}}\\
= \dfrac{2}{{2 - \sqrt x }}\\
C \in Z\\
\Rightarrow 2 \vdots \left( {2 - \sqrt x } \right)\\
\Rightarrow \left( {2 - \sqrt x } \right) \in \left\{ {2;1; - 1; - 2} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;1;3;4} \right\}\\
\Rightarrow x \in \left\{ {0;1;9;16} \right\}\left( {tmdk} \right)\\
B2)\\
a)Dkxd:x \ge 0;x \ne 25\\
A = \left( {\dfrac{{15 - \sqrt x }}{{x - 25}} + \dfrac{2}{{\sqrt x + 5}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 5}}\\
= \dfrac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}.\dfrac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x + 5}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x + 1}}\\
b)M = A - B\\
= \dfrac{1}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{1 + \sqrt x }}\\
= \dfrac{{1 - 1 + \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}} = 1 - \dfrac{1}{{\sqrt x + 1}}\\
M \in Z\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = - 2\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x = 0\left( {tmdk} \right)
\end{array}$
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