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Đáp án:

$\begin{array}{l}
B1)a)\dfrac{{1 - x}}{{{x^2} - 2x + 1}} + \dfrac{{x + 1}}{{x - 1}}\\
 = \dfrac{{1 - x + \left( {x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{1 - x + {x^2} - 1}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{x\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{x}{{x - 1}}\\
b)\dfrac{{2x}}{{3{y^4}z}}.\left( { - \dfrac{{4{y^2}z}}{{5x}}} \right).\left( { - \dfrac{{15{y^3}}}{{8xz}}} \right)\\
 = \dfrac{{2x.4{y^2}z.15{y^3}}}{{3{y^4}z.5x.8xz}}\\
 = \dfrac{y}{{xz}}\\
B2)a)Dkxd:{x^2} - 1 \ne 0\\
 \Leftrightarrow {x^2} \ne 1\\
 \Leftrightarrow x \ne 1;x \ne  - 1\\
Vậy\,x \ne 1;x \ne  - 1\\
b)\dfrac{{3x + 3}}{{{x^2} - 1}} = \dfrac{{3\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{3}{{x - 1}} =  - 2\\
 \Leftrightarrow x - 1 =  - \dfrac{3}{2}\\
 \Leftrightarrow x =  - \dfrac{3}{2} + 1\\
 \Leftrightarrow x =  - \dfrac{1}{2}\\
Vậy\,x =  - \dfrac{1}{2}\\
c)\dfrac{{3x + 3}}{{{x^2} - 1}} = \dfrac{3}{{x - 1}} \in Z\\
 \Leftrightarrow \left( {x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Leftrightarrow x \in \left\{ { - 2;0;2;4} \right\}\left( {tm} \right)
\end{array}$