Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
{x^2} - 25 \ne 0\\
{x^2} + 2x - 15 \ne 0\\
x + 5 \ne 0\\
3 - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 5\\
x \ne - 5\\
x \ne 3
\end{array} \right.\\
Vậy\,x \ne 5;x \ne - 5;x \ne 3\\
b)\\
A = \left( {\dfrac{{{x^2} - 5x}}{{{x^2} - 25}} - 1} \right):\left( {\dfrac{{25 - {x^2}}}{{{x^2} + 2x - 15}} - \dfrac{{x + 3}}{{x + 5}} - \dfrac{{x - 5}}{{3 - x}}} \right)\\
= \left( {\dfrac{{x\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} - 1} \right)\\
:\dfrac{{25 - {x^2} - \left( {x + 3} \right)\left( {x - 3} \right) + \left( {x - 5} \right)\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 3} \right)}}\\
= \left( {\dfrac{x}{{x + 5}} - 1} \right).\dfrac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{25 - {x^2} - {x^2} + 9 + {x^2} - 25}}\\
= \dfrac{{x - x - 5}}{{x + 5}}.\dfrac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{9 - {x^2}}}\\
= \dfrac{{ - 5}}{1}.\dfrac{{x - 3}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}\\
= \dfrac{5}{{x + 3}}\left( {x \ne - 3} \right)\\
c)Dkxd:x \ne 3;x \ne - 3;x \ne 5;x \ne - 5\\
\left| x \right| = 3\\
\Leftrightarrow x = 3/x = - 3\left( {ktm} \right)\\
Vậy\,A \in \emptyset \\
d)A = \dfrac{2}{7}\\
\Leftrightarrow \dfrac{5}{{x + 3}} = \dfrac{2}{7}\\
\Leftrightarrow 2x + 6 = 35\\
\Leftrightarrow 2x = 29\\
\Leftrightarrow x = \dfrac{{29}}{2}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{29}}{2}\\
e)A > 0\\
\Leftrightarrow \dfrac{5}{{x + 3}} > 0\\
\Leftrightarrow x + 3 > 0\\
\Leftrightarrow x > - 3\\
Vậy\,x > - 3;x \ne 3;x \ne 5\\
f)A = \dfrac{5}{{x + 3}} \in Z\\
\Leftrightarrow \left( {x + 3} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Leftrightarrow x \in \left\{ { - 8; - 4; - 2;2} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ { - 8; - 4; - 2;2} \right\}\\
g)B < 1\\
\Leftrightarrow \dfrac{{x - 3}}{{x + 2}} < 1\\
\Leftrightarrow \dfrac{{x - 3}}{{x + 2}} - 1 < 0\\
\Leftrightarrow \dfrac{{x - 3 - x - 2}}{{x + 2}} < 0\\
\Leftrightarrow \dfrac{{ - 5}}{{x + 2}} < 0\\
\Leftrightarrow x + 2 > 0\\
\Leftrightarrow x > - 2\\
Vậy\,x > - 2\\
h)B = \dfrac{{x - 3}}{{x + 2}} = \dfrac{{x + 2 - 5}}{{x + 2}} = 1 - \dfrac{5}{{x + 2}}\\
B \in Z\\
\Leftrightarrow \dfrac{5}{{x + 2}} \in Z\\
\Leftrightarrow \left( {x + 2} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Leftrightarrow x \in \left\{ { - 7; - 3; - 1;3} \right\}\\
Vậy\,x \in \left\{ { - 7; - 3; - 1;3} \right\}
\end{array}$
