Đáp án:

$\begin{array}{l}
1){\left( {2xy + 1} \right)^2} - {\left( {2x + y} \right)^2}\\
 = \left( {2xy + 1 - 2x - y} \right)\left( {2xy + 1 + 2x + y} \right)\\
 = \left( {2x\left( {y - 1} \right) - \left( {y - 1} \right)} \right)\left( {2x\left( {y + 1} \right) + y + 1} \right)\\
 = \left( {y - 1} \right)\left( {2x - 1} \right)\left( {y + 1} \right)\left( {2x + 1} \right)\\
2)\\
{\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2}\\
 = \left( {3x - 2y - 2x + 3y} \right)\left( {3x - 2y + 2x - 3y} \right)\\
 = \left( {x + y} \right)\left( {5x - 5y} \right)\\
 = 5\left( {x + y} \right)\left( {x - y} \right)\\
3)\left( {4{x^2} - 4x + 1} \right) - {\left( {x + 1} \right)^2}\\
 = {\left( {2x - 1} \right)^2} - {\left( {x + 1} \right)^2}\\
 = \left( {2x - 1 - x - 1} \right)\left( {2x - 1 + x + 1} \right)\\
 = \left( {x - 2} \right).3x\\
4){x^2} - 2xy + {y^2} - 1\\
 = {\left( {x - y} \right)^2} - 1\\
 = \left( {x - y - 1} \right)\left( {x - y + 1} \right)\\
5){x^2} - 2xy + {y^2} - 4\\
 = {\left( {x - y} \right)^2} - {2^2}\\
 = \left( {x - y - 2} \right)\left( {x - y + 2} \right)\\
6){x^2} - 2xy + {y^2} - {z^2}\\
 = {\left( {x - y} \right)^2} - {z^2}\\
 = \left( {x - y - z} \right)\left( {x - y + z} \right)\\
7)25 - {x^2} + 4xy - 4{y^2}\\
 = 25 - \left( {{x^2} - 4xy + 4{y^2}} \right)\\
 = {5^2} - {\left( {x - 2y} \right)^2}\\
 = \left( {5 - x + 2y} \right)\left( {5 + x - 2y} \right)\\
8)\\
{x^2} + {y^2} - 2xy - 4{z^2}\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - {\left( {2z} \right)^2}\\
 = {\left( {x - y} \right)^2} - {\left( {2z} \right)^2}\\
 = \left( {x - y - 2z} \right)\left( {x - y + 2z} \right)
\end{array}$