Đáp án:

\(\begin{array}{l}
6,\\
a,\\
{\left( {x - 5} \right)^2}.\left( {x + 4} \right)\left( {x + 6} \right)\\
b,\\
4.{\left( {2x - 5} \right)^2}.\left( {x + 1} \right)\left( {x + 4} \right)\\
c,\\
 - {\left( {2x - 3} \right)^2}.\left( {6x + 7} \right).\left( {6x + 11} \right)\\
d,\\
{a^2}{\left( {a + 1} \right)^2}\left( {{a^2} - 2a + 2} \right)\\
e,\\
24x\left( {x + 1} \right)\\
7,\\
a,\\
\left( {y - 1} \right)\left( {x - 1} \right).\left( {y + 1} \right)\left( {x + 1} \right)\\
b,\\
2y.\left( {3{x^2} + {y^2}} \right)\\
c,\\
3{y^2}.{\left( {x + 1} \right)^2}\left( {{x^2} - x + 1} \right)\\
d,\\
4.\left( {x - 1 - y - a} \right)\left( {x - 1 + y + a} \right)\\
e,\\
\left( {x + y - 1} \right).\left( {{x^2} - xy + {y^2} + x + y - 1} \right)\\
8,\\
a,\\
\left( {x - 1} \right){\left( {x + 3} \right)^2}\\
b,\\
\left( {a + 1} \right).\left( {{a^2} - a + 1} \right).\left( {{a^2} + a + 1} \right)\\
c,\\
\left( {x - y - 1} \right)\left( {{x^2} + {y^2} + xy - 2x - y + 1} \right)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
6,\\
a,\\
{\left( {{x^2} - 25} \right)^2} - {\left( {x - 5} \right)^2}\\
 = {\left( {{x^2} - {5^2}} \right)^2} - {\left( {x - 5} \right)^2}\\
 = {\left[ {\left( {x - 5} \right)\left( {x + 5} \right)} \right]^2} - {\left( {x - 5} \right)^2}\\
 = {\left( {x - 5} \right)^2}.{\left( {x + 5} \right)^2} - {\left( {x - 5} \right)^2}\\
 = {\left( {x - 5} \right)^2}.\left[ {{{\left( {x + 5} \right)}^2} - 1} \right]\\
 = {\left( {x - 5} \right)^2}\left[ {{{\left( {x + 5} \right)}^2} - {1^2}} \right]\\
 = {\left( {x - 5} \right)^2}.\left[ {\left( {x + 5} \right) - 1} \right].\left[ {\left( {x + 5} \right) + 1} \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[ {{{\left( {2x} \right)}^2} - {5^2}} \right]^2} - 9{\left( {2x - 5} \right)^2}\\
 = {\left[ {\left( {2x - 5} \right)\left( {2x + 5} \right)} \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[ {{{\left( {2x + 5} \right)}^2} - {3^2}} \right]\\
 = {\left( {2x - 5} \right)^2}.\left[ {\left( {2x + 5} \right) - 3} \right]\left[ {\left( {2x + 5} \right) + 3} \right]\\
 = {\left( {2x - 5} \right)^2}.\left( {2x + 2} \right).\left( {2x + 8} \right)\\
 = {\left( {2x - 5} \right)^2}.2.\left( {x + 1} \right).2.\left( {x + 4} \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[ {{{\left( {2x} \right)}^2} - {3^2}} \right]^2}\\
 = 4.{\left( {2x - 3} \right)^2} - 9.{\left[ {\left( {2x - 3} \right)\left( {2x + 3} \right)} \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 - 9{{\left( {2x + 3} \right)}^2}} \right]\\
 = {\left( {2x - 3} \right)^2}.\left[ {{2^2} - {3^2}.{{\left( {2x + 3} \right)}^2}} \right]\\
 = {\left( {2x - 3} \right)^2}.\left[ {2 - 3.\left( {2x + 3} \right)} \right].\left[ {2 + 3.\left( {2x + 3} \right)} \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)\\
 =  - {\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^2}.\left( {{a^4} - {a^2} + 2a + 2} \right)\\
 = {a^2}.\left[ {\left( {{a^4} + {a^3}} \right) + \left( { - {a^3} - {a^2}} \right) + \left( {2a + 2} \right)} \right]\\
 = {a^2}.\left[ {{a^3}\left( {a + 1} \right) - {a^2}\left( {a + 1} \right) + 2\left( {a + 1} \right)} \right]\\
 = {a^2}.\left( {a + 1} \right).\left( {{a^3} - {a^2} + 2} \right)\\
 = {a^2}\left( {a + 1} \right).\left[ {\left( {{a^3} + {a^2}} \right) + \left( { - 2{a^2} - 2a} \right) + \left( {2a + 2} \right)} \right]\\
 = {a^2}\left( {a + 1} \right).\left[ {{a^2}\left( {a + 1} \right) - 2a\left( {a + 1} \right) + 2\left( {a + 1} \right)} \right]\\
 = {a^2}\left( {a + 1} \right).\left[ {\left( {a + 1} \right).\left( {{a^2} - 2a + 2} \right)} \right]\\
 = {a^2}{\left( {a + 1} \right)^2}\left( {{a^2} - 2a + 2} \right)\\
e,\\
{\left( {3{x^2} + 3x + 2} \right)^2} - {\left( {3{x^2} + 3x - 2} \right)^2}\\
 = \left[ {\left( {3{x^2} + 3x + 2} \right) - \left( {3{x^2} + 3x - 2} \right)} \right].\left[ {\left( {3{x^2} + 3x + 2} \right) + \left( {3{x^2} + 3x - 2} \right)} \right]\\
 = \left( {3{x^2} + 3x + 2 - 3{x^2} - 3x + 2} \right).\left( {3{x^2} + 3x + 2 + 3{x^2} + 3x - 2} \right)\\
 = 4.\left( {6{x^2} + 6x} \right)\\
 = 4.6x\left( {x + 1} \right)\\
 = 24x\left( {x + 1} \right)\\
7,\\
a,\\
{\left( {xy + 1} \right)^2} - {\left( {x + y} \right)^2}\\
 = \left[ {\left( {xy + 1} \right) - \left( {x + y} \right)} \right].\left[ {\left( {xy + 1} \right) + \left( {x + y} \right)} \right]\\
 = \left( {xy + 1 - x - y} \right)\left( {xy + 1 + x + y} \right)\\
 = \left[ {\left( {xy - x} \right) + \left( { - y + 1} \right)} \right].\left[ {\left( {xy + x} \right) + \left( {y + 1} \right)} \right]\\
 = \left[ {x\left( {y - 1} \right) - \left( {y - 1} \right)} \right].\left[ {x\left( {y + 1} \right) + \left( {y + 1} \right)} \right]\\
 = \left( {y - 1} \right)\left( {x - 1} \right).\left( {y + 1} \right)\left( {x + 1} \right)\\
b,\\
{\left( {x + y} \right)^3} - {\left( {x - y} \right)^3}\\
 = \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) - \left( {{x^3} - 3{x^2}y + 3x{y^2} - {y^3}} \right)\\
 = {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^3}\left( {x + 1} \right) + \left( {x + 1} \right)} \right]\\
 = 3{y^2}.\left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
 = 3{y^2}.\left( {x + 1} \right).\left( {x + 1} \right).\left( {{x^2} - x + 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{x^2} - 4{y^2} - 8x + 8ay - 4{a^2} + 4\\
 = 4.\left( {{x^2} - {y^2} - 2x - 2ay - {a^2} + 1} \right)\\
 = 4.\left[ {\left( {{x^2} - 2x + 1} \right) + \left( { - {y^2} - 2ay - {a^2}} \right)} \right]\\
 = 4.\left[ {\left( {{x^2} - 2.x.1 + {1^2}} \right) - \left( {{y^2} + 2ya + {a^2}} \right)} \right]\\
 = 4.\left[ {{{\left( {x - 1} \right)}^2} - {{\left( {y + a} \right)}^2}} \right]\\
 = 4.\left[ {\left( {x - 1} \right) - \left( {y + a} \right)} \right].\left[ {\left( {x - 1} \right) + \left( {y + a} \right)} \right]\\
 = 4.\left( {x - 1 - y - a} \right)\left( {x - 1 + y + a} \right)\\
e,\\
{\left( {x + y} \right)^3} - 1 - 3xy\left( {x + y - 1} \right)\\
 = \left[ {{{\left( {x + y} \right)}^3} - {1^3}} \right] - 3xy\left( {x + y - 1} \right)\\
 = \left[ {\left( {x + y} \right) - 1} \right].\left[ {{{\left( {x + y} \right)}^2} + \left( {x + y} \right).1 + {1^2}} \right] - 3xy\left( {x + y - 1} \right)\\
 = \left( {x + y - 1} \right).\left( {{x^2} + 2xy + {y^2} + x + y + 1} \right) - 3xy\left( {x + y - 1} \right)\\
 = \left( {x + y - 1} \right).\left( {{x^2} + 2xy + {y^2} + x + y + 1 - 3xy} \right)\\
 = \left( {x + y - 1} \right).\left( {{x^2} - xy + {y^2} + x + y - 1} \right)\\
8,\\
a,\\
{x^3} - 1 + 5{x^2} - 5 + 3x - 3\\
 = \left( {{x^3} - 1} \right) + \left( {5{x^2} - 5} \right) + \left( {3x - 3} \right)\\
 = \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} \right) + 5.\left( {x - 1} \right)\left( {x + 1} \right) + 3\left( {x - 1} \right)\\
 = \left( {x - 1} \right).\left[ {\left( {{x^2} + x + 1} \right) + 5.\left( {x + 1} \right) + 3} \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^2} + 2.x.3 + {3^2}} \right)\\
 = \left( {x - 1} \right){\left( {x + 3} \right)^2}\\
b,\\
{a^5} + {a^4} + {a^3} + {a^2} + a + 1\\
 = \left( {{a^5} + {a^4}} \right) + \left( {{a^3} + {a^2}} \right) + \left( {a + 1} \right)\\
 = {a^4}\left( {a + 1} \right) + {a^2}\left( {a + 1} \right) + \left( {a + 1} \right)\\
 = \left( {a + 1} \right).\left( {{a^4} + {a^2} + 1} \right)\\
 = \left( {a + 1} \right).\left[ {\left( {{a^4} + 2{a^2} + 1} \right) - {a^2}} \right]\\
 = \left( {a + 1} \right).\left[ {\left( {{{\left( {{a^2}} \right)}^2} + 2.{a^2}.1 + {1^2}} \right) - {a^2}} \right]\\
 = \left( {a + 1} \right).\left[ {{{\left( {{a^2} + 1} \right)}^2} - {a^2}} \right]\\
 = \left( {a + 1} \right).\left[ {\left( {{a^2} + 1} \right) - a} \right].\left[ {\left( {{a^2} + 1} \right) + a} \right]\\
 = \left( {a + 1} \right).\left( {{a^2} - a + 1} \right).\left( {{a^2} + a + 1} \right)\\
c,\\
{x^3} - 3{x^2} + 3x - 1 - {y^3}\\
 = \left( {{x^3} - 3{x^2} + 3x - 1} \right) - {y^3}\\
 = {\left( {x - 1} \right)^3} - {y^3}\\
 = \left[ {\left( {x - 1} \right) - y} \right].\left[ {{{\left( {x - 1} \right)}^2} + \left( {x - 1} \right).y + {y^2}} \right]\\
 = \left( {x - 1 - y} \right).\left( {{x^2} - 2x + 1 + xy - y + {y^2}} \right)\\
 = \left( {x - y - 1} \right)\left( {{x^2} + {y^2} + xy - 2x - y + 1} \right)
\end{array}\)