Đáp án:
$\begin{array}{l}
1){x^2} + 4x + 4 - {y^2}\\
= {\left( {x + 2} \right)^2} - {y^2}\\
= \left( {x + 2 - y} \right)\left( {x + 2 + y} \right)\\
2){x^2} - 10x + 25 - {y^2}\\
= {\left( {x - 5} \right)^2} - {y^2}\\
= \left( {x - 5 - y} \right)\left( {x - 5 + y} \right)\\
3){x^2} + 6a + 9 - {b^2}\\
= {\left( {a + 3} \right)^2} - {b^2}\\
= \left( {a + 3 - b} \right)\left( {a + 3 + b} \right)\\
4)4{a^2} - 4a + 1 - {b^2}\\
= {\left( {2a - 1} \right)^2} - {b^2}\\
= \left( {2a - 1 - b} \right)\left( {2a - 1 + b} \right)\\
5)16{a^2} + 24a + 9 - 4{b^2}\\
= {\left( {4a + 3} \right)^2} - 4{b^2}\\
= \left( {4a + 3 - 2b} \right)\left( {4a + 3 + 2b} \right)\\
6)16{a^2} - 4{b^2} - 8a + 1\\
= {\left( {4a - 1} \right)^2} - 4{b^2}\\
= \left( {4a - 1 - 2b} \right)\left( {4a - 1 + 2b} \right)\\
7){x^2} - {y^2} + 2y - 1\\
= {x^2} - {\left( {y - 1} \right)^2}\\
= \left( {x - y + 1} \right)\left( {x + y - 1} \right)\\
8){x^2} - {y^2} - 4y - 4\\
= {x^2} - {\left( {y + 2} \right)^2}\\
= \left( {x - y - 2} \right)\left( {x + y + 2} \right)\\
9)9{x^2} - 9{y^2} + 6y - 1\\
= 9{x^2} - {\left( {3y - 1} \right)^2}\\
= \left( {3x - 3y + 1} \right)\left( {3x + 3y - 1} \right)
\end{array}$
