Đáp án:

$\begin{array}{l}
1){x^2} - 2xy + {y^2} - 9\\
 = {\left( {x - y} \right)^2} - {3^2}\\
 = \left( {x - y - 3} \right)\left( {x - y + 3} \right)\\
2){x^2} + 2xy + {y^2} - xz - yz\\
 = {\left( {x + y} \right)^2} - z\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {x + y - z} \right)\\
3){x^4} + {x^3} + x + 1\\
 = {x^3}\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
 = {\left( {x + 1} \right)^2}\left( {{x^2} - x + 1} \right)\\
4)3{x^2} - 6xy + 3{y^2}\\
 = 3\left( {{x^2} - 2xy + {y^2}} \right)\\
 = 3{\left( {x - y} \right)^2}\\
5){x^2} + 2xy + {y^2} - 4\\
 = {\left( {x + y} \right)^2} - {2^2}\\
 = \left( {x + y + 2} \right)\left( {x + y - 2} \right)\\
6)x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + ??\\
7){\left( {x - y} \right)^2} + 4\left( {x - y} \right) - 12\\
 = {\left( {x - y} \right)^2} + 6\left( {x - y} \right) - 2\left( {x - y} \right) - 12\\
 = \left( {x - y} \right)\left( {x - y + 6} \right) - 2\left( {x - y + 6} \right)\\
 = \left( {x - y + 6} \right)\left( {x - y - 2} \right)\\
8){x^5} + x + 1\\
 = {x^5} + {x^4} + {x^3} - {x^4} - {x^3} - {x^2}\\
 + {x^2} + x + 1\\
 = {x^3}\left( {{x^2} + x + 1} \right) - {x^2}\left( {{x^2} + x + 1} \right)\\
 + \left( {{x^2} + x + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^3} - {x^2} + 1} \right)\\
9)81{x^4} + 1\\
 = 81{x^4} + 18{x^2} + 1 - 18{x^2}\\
 = {\left( {9{x^2} + 1} \right)^2} - {\left( {3\sqrt 2 x} \right)^2}\\
 = \left( {9{x^2} + 3\sqrt 2 x + 1} \right)\left( {9{x^2} - 3\sqrt 2 x + 1} \right)\\
10){x^3} - 3{x^2} + 2\\
 = {x^3} - {x^2} - 2{x^2} + 2\\
 = {x^2}\left( {x - 1} \right) - 2\left( {{x^2} - 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} - 2x - 2} \right)
\end{array}$