Đáp án:
$\begin{array}{l}
3){x^2} - 16\\
= {x^2} - {4^2}\\
= \left( {x - 4} \right)\left( {x + 4} \right)\\
4)9{x^2} - 25{y^2}\\
= {\left( {3x} \right)^2} - {\left( {5y} \right)^2}\\
= \left( {3x - 5y} \right)\left( {3x + 5y} \right)\\
5){x^2} - \frac{1}{4}\\
= {x^2} - {\left( {\frac{1}{2}} \right)^2}\\
= \left( {x - \frac{1}{2}} \right)\left( {x + \frac{1}{2}} \right)\\
6)4{x^2} - 1\\
= {\left( {2x} \right)^2} - 1\\
= \left( {2x - 1} \right)\left( {2x + 1} \right)\\
7)9{x^2} - 24x + 16\\
= {\left( {3x} \right)^2} - 2.3x.4 + {4^2}\\
= {\left( {3x - 4} \right)^2}\\
8){\left( {2x + 5} \right)^2} - {\left( {x + 1} \right)^2}\\
= \left( {2x + 5 - x - 1} \right)\left( {2x + 5 + x + 1} \right)\\
= \left( {x + 4} \right)\left( {3x + 6} \right)\\
= 3.\left( {x + 4} \right)\left( {x + 2} \right)\\
9){\left( {3x + 2y} \right)^2} - {\left( {x + y} \right)^2}\\
= \left( {3x + 2y - x - y} \right)\left( {3x + 2y + x + y} \right)\\
= \left( {2x + y} \right)\left( {4x + 3y} \right)\\
10)16 - {\left( {x + 3} \right)^2}\\
= \left( {4 - x - 3} \right)\left( {4 + x + 3} \right)\\
= \left( {1 - x} \right)\left( {x + 7} \right)\\
11)\\
{\left( {2x + 7} \right)^2} - {y^2}\\
= \left( {2x + 7 + y} \right)\left( {2x + 7 - y} \right)\\
12)4{x^2}{y^2} - {\left( {{x^2} + {y^2} - {z^2}} \right)^2}\\
= \left( {4{x^2}{y^2} + {x^2} + {y^2} - {z^2}} \right).\left( {4{x^2}{y^2} - {x^2} - {y^2} + {z^2}} \right)
\end{array}$
