Đáp án:
$\begin{array}{l}
1)a){\left( {{x^2} - 25} \right)^2} - {\left( {x - 5} \right)^2}\\
= \left( {{x^2} - 25 - x + 5} \right)\left( {{x^2} - 25 + x - 5} \right)\\
= \left( {{x^2} - x - 20} \right)\left( {{x^2} + x - 30} \right)\\
= \left( {x - 5} \right)\left( {x + 4} \right)\left( {x - 5} \right)\left( {x + 6} \right)\\
= {\left( {x - 5} \right)^2}\left( {x + 4} \right)\left( {x + 6} \right)\\
b){\left( {4{x^2} - 25} \right)^2} - 9{\left( {2x - 5} \right)^2}\\
= {\left( {2x - 5} \right)^2}{\left( {2x + 5} \right)^2} - 9{\left( {2x - 5} \right)^2}\\
= {\left( {2x - 5} \right)^2}.\left[ {{{\left( {2x + 5} \right)}^2} - 9} \right]\\
= {\left( {2x - 5} \right)^2}\left( {2x + 5 - 3} \right)\left( {2x + 5 + 3} \right)\\
= {\left( {2x - 5} \right)^2}\left( {2x + 2} \right)\left( {2x + 8} \right)\\
= 4.{\left( {2x - 5} \right)^2}.\left( {x + 1} \right)\left( {x + 4} \right)\\
c)4{\left( {2x - 3} \right)^2} - 9{\left( {4{x^2} - 9} \right)^2}\\
= 4.{\left( {2x - 3} \right)^2} - 9.{\left( {2x - 3} \right)^2}{\left( {2x + 3} \right)^2}\\
= {\left( {2x - 3} \right)^2}\left[ {4 - {{\left( {6x + 9} \right)}^2}} \right]\\
= {\left( {2x - 3} \right)^2}.\left( {2 - 6x - 9} \right)\left( {2 + 6x + 9} \right)\\
= {\left( {2x - 3} \right)^2}.\left( { - 6x - 7} \right).\left( {6x + 11} \right)\\
d){a^6} - {a^4} + 2{a^3} + 2{a^2}\\
= {a^4}\left( {{a^2} - 1} \right) + 2{a^2}\left( {a + 1} \right)\\
= {a^4}\left( {a - 1} \right)\left( {a + 1} \right) + 2{a^2}\left( {a + 1} \right)\\
= {a^2}\left( {a + 1} \right).\left( {{a^2}\left( {a - 1} \right) + 2} \right)\\
= {a^2}\left( {a + 1} \right)\left( {{a^3} - {a^2} + 2} \right)\\
e){\left( {3{x^2} + 3x + 2} \right)^2} - {\left( {3{x^2} + 3x - 2} \right)^2}\\
= \left( {3{x^2} + 3x + 2 + \left( {3{x^2} + 3x - 2} \right)} \right)\\
.\left( {3{x^2} + 3x + 2 - 3{x^2} - 3x + 2} \right)\\
= \left( {6{x^2} + 6x} \right).4\\
= 24x\left( {x + 1} \right)\\
2)a){\left( {xy + 1} \right)^2} - {\left( {x + y} \right)^2}\\
= \left( {xy + 1 - x - y} \right)\left( {xy + 1 + x + y} \right)\\
= \left( {x - 1} \right)\left( {y - 1} \right).\left( {x + 1} \right)\left( {y + 1} \right)\\
b){\left( {x + y} \right)^3} - {\left( {x - y} \right)^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
- {x^3} + 3{x^2}y - 3x{y^2} + {y^3}\\
= 6{x^2}y + 2{y^3}\\
= 2y\left( {3{x^2} + {y^2}} \right)\\
c)3{x^4}{y^2} + 3{x^3}{y^2} + 3x{y^2} + 3{y^2}\\
= 3{y^2}\left( {{x^4} + {x^3} + x + 1} \right)\\
= 3{y^2}\left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
= 3{y^2}{\left( {x + 1} \right)^2}\left( {{x^2} - x + 1} \right)\\
d)4.\left( {{x^2} - {y^2}} \right) - 8\left( {x - ay} \right) - 4\left( {{a^2} - 1} \right)\\
= 4.\left( {{x^2} - {y^2} - 2x + 2ay - {a^2} + 1} \right)\\
= 4.\left[ {\left( {{x^2} - 2x + 1} \right) - \left( {{a^2} - 2ay + {y^2}} \right)} \right]\\
= 4.\left[ {{{\left( {x - 1} \right)}^2} - {{\left( {a - y} \right)}^2}} \right]\\
= 4.\left( {x - 1 - a + y} \right)\left( {x - 1 + a - y} \right)\\
e){\left( {x + y} \right)^3} - 1 - 3xy\left( {x + y - 1} \right)\\
= \left( {x + y - 1} \right)\left( {{{\left( {x + y} \right)}^2} + x + y + 1} \right)\\
- 3xy\left( {x + y - 1} \right)\\
= \left( {x + y - 1} \right).\left[ {{{\left( {x + y} \right)}^2} + x + y + 1 - 3xy} \right]\\
= \left( {x + y - 1} \right)\left( {{x^2} + {y^2} + x + y + 1 - xy} \right)\\
3)a){x^3} - 1 + 5{x^2} - 5 + 3x - 3\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + 5\left( {{x^2} - 1} \right) + 3\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1 + 5x + 5 + 3} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + 6x + 9} \right)\\
= \left( {x - 1} \right){\left( {x + 3} \right)^2}
\end{array}$
