Đáp án:
$\begin{array}{l}
1){x^2} - {y^2} - 2x - 2y\\
= \left( {x - y} \right)\left( {x + y} \right) - 2\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - y - 2} \right)\\
2){x^3} - 4{x^2} - 8x + 8\\
= {x^3} + 8 - 4{x^2} - 8x\\
= \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - 4x\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - 2x + 4 - 4x} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - 6x + 4} \right)\\
3)12{x^2} - 3{y^2} + 12x + 3\\
= 3\left( {4{x^2} - {y^2} + 4x + 1} \right)\\
= 3\left( {4x\left( {x + 1} \right) - {y^2} + 1} \right)\\
4){x^2} - 9 - {x^2}\left( {{x^2} - 9} \right)\\
= \left( {{x^2} - 9} \right)\left( {1 - {x^2}} \right)\\
= \left( {x - 3} \right)\left( {x + 3} \right)\left( {1 - x} \right)\left( {1 + x} \right)\\
5){x^4} - 5{x^2} + 4\\
= {x^4} - {x^2} - 4{x^2} + 4\\
= {x^2}\left( {{x^2} - 1} \right) - 4\left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)
\end{array}$
