Đáp án:

$\begin{array}{l}
3)a)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} - \dfrac{4}{{1 - {x^2}}}\\
 = \dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} + \dfrac{4}{{{x^2} - 1}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2} + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1 + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{4x + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{4}{{x - 1}}\\
b)\left( {\dfrac{x}{{xy - {y^2}}} + \dfrac{{2x - y}}{{xy - {x^2}}}} \right):\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right)\\
 = \left( {\dfrac{x}{{y\left( {x - y} \right)}} + \dfrac{{2x - y}}{{x\left( {y - x} \right)}}} \right):\dfrac{{x + y}}{{xy}}\\
 = \dfrac{{x.x - y.\left( {2x - y} \right)}}{{xy\left( {x - y} \right)}}.\dfrac{{xy}}{{x + y}}\\
 = \dfrac{{{x^2} - 2xy + {y^2}}}{{x - y}}.\dfrac{1}{{x + y}}\\
 = \dfrac{{x - y}}{{x + y}}\\
c)\dfrac{{{x^2}y - {y^2}x}}{{{x^2} - {y^2}}}.\left( {\dfrac{y}{{{x^2} - xy}} - \dfrac{x}{{xy - {y^2}}}} \right)\\
 = \dfrac{{xy\left( {x - y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}}.\left( {\dfrac{y}{{x\left( {x - y} \right)}} - \dfrac{x}{{y\left( {x - y} \right)}}} \right)\\
 = \dfrac{{xy}}{{x + y}}.\dfrac{{{y^2} - {x^2}}}{{xy\left( {x - y} \right)}}\\
 =  - 1\\
1)a){x^2} - 64\\
 = \left( {x - 8} \right)\left( {x + 8} \right)\\
b){x^2} - 2x - 4{y^2} - 4y\\
 = \left( {x - 2y} \right)\left( {x + 2y} \right) - 2\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right)\left( {x - 2y - 2} \right)\\
c){x^2} + 4x - {y^2} + 4\\
 = {\left( {x + 2} \right)^2} - {y^2}\\
 = \left( {x + 2 - y} \right)\left( {x + 2 + y} \right)\\
d){x^2} - 7x + 12\\
 = \left( {x - 3} \right)\left( {x - 4} \right)\\
2)a)5x\left( {x - 3} \right) - 2x + 6 = 0\\
 \Rightarrow 5x\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0\\
 \Rightarrow \left( {x - 3} \right)\left( {5x - 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{2}{5}
\end{array} \right.\\
Vậy\,x = 3;x = \dfrac{2}{5}\\
b){x^3} - 8 = {\left( {x - 2} \right)^3}\\
 \Rightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) = \left( {x - 2} \right)\left( {{x^2} - 4x + 4} \right)\\
 \Rightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4 - {x^2} + 4x - 4} \right) = 0\\
 \Rightarrow \left( {x - 2} \right).8x = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
Vậy\,x = 2;x = 0\\
c){x^3} + 5{x^2} - 4x - 20 = 0\\
 \Rightarrow {x^2}\left( {x + 5} \right) - 4\left( {x + 5} \right) = 0\\
 \Rightarrow \left( {x + 5} \right)\left( {{x^2} - 4} \right) = 0\\
 \Rightarrow x =  - 5;x =  \pm 2
\end{array}$