Đáp án:
$\begin{array}{l}
a){x^2} - 2x - 4{y^2} - 4y\\
= \left( {{x^2} - 4{y^2}} \right) - 2\left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {x - 2y} \right) - 2\left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {x - 2y - 2} \right)\\
b){x^4} + 2{x^3} - 4x - 4\\
= {x^4} - 4 + 2x\left( {{x^2} - 2} \right)\\
= \left( {{x^2} - 2} \right)\left( {{x^2} + 2} \right) + 2x\left( {{x^2} - 2} \right)\\
= \left( {{x^2} - 2} \right)\left( {{x^2} + 2x + 2} \right)\\
c){x^3} + 2{x^2}y - x - 2y\\
= {x^2}\left( {x + 2y} \right) - \left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {{x^2} - 1} \right)\\
= \left( {x + 2y} \right)\left( {x + 1} \right)\left( {x - 1} \right)\\
d)3{x^2} - 3{y^2} - 2{\left( {x - y} \right)^2}\\
= 3\left( {{x^2} - {y^2}} \right) - 2{\left( {x - y} \right)^2}\\
= 3\left( {x - y} \right)\left( {x + y} \right) - 2{\left( {x - y} \right)^2}\\
= \left( {x - y} \right)\left( {3x + 3y - 2x + 2y} \right)\\
= \left( {x - y} \right)\left( {x + 5y} \right)\\
e){x^3} - 4{x^2} - 9x + 36\\
= {x^2}\left( {x - 4} \right) - 9\left( {x - 4} \right)\\
= \left( {x - 4} \right)\left( {{x^2} - 9} \right)\\
= \left( {x - 4} \right)\left( {x - 3} \right)\left( {x + 3} \right)\\
f){x^2} - {y^2} - 2x - 2y\\
= \left( {x - y} \right)\left( {x + y} \right) - 2\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - y - 2} \right)
\end{array}$
