Giải thích các bước giải:
1,
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne 0\\
x \ne - 3
\end{array} \right.\,\,\,\,\,\,\left( * \right)\)
Ta có:
\(\begin{array}{l}
A = \left( {\frac{3}{{x + 2}} + \frac{{x + 2}}{{2 - x}} - \frac{{x - 10}}{{{x^2} - 4}}} \right).\frac{{{x^2} - 4x + 4}}{{{x^2} + 3x}}\\
= \left( {\frac{3}{{x + 2}} - \frac{{x + 2}}{{x - 2}} - \frac{{x - 10}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\frac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
= \left( {\frac{{3\left( {x - 2} \right) - {{\left( {x + 2} \right)}^2} - x + 10}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\frac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
= \frac{{3x - 6 - {x^2} - 4x - 4 - x + 10}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
= \frac{{ - {x^2} - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
= \frac{{ - x\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right).x.\left( {x + 3} \right)}}\\
= - \frac{{x - 2}}{{x + 3}}\\
b,\\
\left| {x + 3} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 1\\
x + 3 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 2\,\,\,\left( L \right)\\
x = - 4\,\,\,\left( {t/m\,\,\,\left( * \right)} \right)
\end{array} \right.\\
x = - 1 \Rightarrow A = - \frac{{\left( { - 4} \right) - 2}}{3} = 2\\
c,\\
A = - \frac{{x - 2}}{{x + 3}} = - \frac{{x + 3 - 5}}{{x + 3}} = - 1 + \frac{5}{{x + 3}}\\
A \in Z \Leftrightarrow \frac{5}{{x + 3}} \in Z \Leftrightarrow x + 3 \in \left\{ { \pm 1; \pm 5} \right\}\\
\Rightarrow x \in \left\{ { - 8;\,\, - 4;\,\, - 2;\,\,2} \right\}\\
\left( * \right) \Rightarrow x \in \left\{ { - 8;\,\, - 4} \right\}
\end{array}\)
Bài 2:
\(\begin{array}{l}
a,\\
9{x^2} - 25 = {\left( {3x} \right)^2} - {5^2} = \left( {3x - 5} \right)\left( {3x + 5} \right)\\
b,\\
2x - 10 - {\left( {x - 5} \right)^2}\\
= 2\left( {x - 5} \right) - {\left( {x - 5} \right)^2}\\
= \left( {x - 5} \right).\left[ {2 - \left( {x - 5} \right)} \right]\\
= \left( {x - 5} \right).\left( {7 - x} \right)\\
c,\\
{x^2} - 13x + 30\\
= \left( {{x^2} - 3x} \right) - \left( {10x - 30} \right)\\
= x\left( {x - 3} \right) - 10\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {x - 10} \right)\\
d,\\
{x^3} - x - 2{x^2} + 2\\
= \left( {{x^3} - 2{x^2}} \right) - \left( {x - 2} \right)\\
= {x^2}\left( {x - 2} \right) - \left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {{x^2} - 1} \right)\\
= \left( {x - 2} \right)\left( {x + 1} \right)\left( {x - 1} \right)
\end{array}\)
