Đáp án:

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

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
{x^2} - x - {y^2} - y\\
 = \left( {{x^2} - {y^2}} \right) + \left( { - x - y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) - \left( {x + y} \right)\\
 = \left( {x + y} \right).\left[ {\left( {x - y} \right) - 1} \right]\\
 = \left( {x + y} \right)\left( {x - y - 1} \right)\\
b,\\
{x^2} - 2xy + {y^2} - {z^2}\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - {z^2}\\
 = {\left( {x - y} \right)^2} - {z^2}\\
 = \left[ {\left( {x - y} \right) - z} \right].\left[ {\left( {x - y} \right) + z} \right]\\
 = \left( {x - y - z} \right)\left( {x - y + z} \right)\\
c,\\
5x - 5y + ax - ay\\
 = \left( {5x - 5y} \right) + \left( {ax - ay} \right)\\
 = 5.\left( {x - y} \right) + a.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {5 + a} \right)\\
d,\\
{a^3} - {a^2}x - ay + xy\\
 = \left( {{a^3} - {a^2}x} \right) + \left( { - ay + xy} \right)\\
 = {a^2}.\left( {a - x} \right) - y.\left( {a - x} \right)\\
 = \left( {a - x} \right).\left( {{a^2} - y} \right)\\
e,\\
4{x^2} - {y^2} + 4x + 1\\
 = \left( {4{x^2} + 4x + 1} \right) - {y^2}\\
 = \left[ {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} \right] - {y^2}\\
 = {\left( {2x + 1} \right)^2} - {y^2}\\
 = \left[ {\left( {2x + 1} \right) - y} \right].\left[ {\left( {2x + 1} \right) + y} \right]\\
 = \left( {2x + 1 - y} \right)\left( {2x + 1 + y} \right)\\
f,\\
{x^3} - x + {y^3} - y\\
 = \left( {{x^3} + {y^3}} \right) + \left( { - x - y} \right)\\
 = \left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right) - \left( {x + y} \right)\\
 = \left( {x + y} \right).\left[ {\left( {{x^2} - xy + {y^2}} \right) - 1} \right]\\
 = \left( {x + y} \right).\left( {{x^2} - xy + {y^2} - 1} \right)\\
2,\\
a,\\
10\left( {x - y} \right) - 8y\left( {y - x} \right)\\
 = 10\left( {x - y} \right) - 8y.\left[ { - \left( {x - y} \right)} \right]\\
 = 10\left( {x - y} \right) + 8y\left( {x - y} \right)\\
 = \left( {x - y} \right).\left( {10 + 8y} \right)\\
 = 2.\left( {x - y} \right).\left( {5 + 4y} \right)\\
b,\\
2y + 3z + 6y + y\\
 = \left( {2y + 6y + y} \right) + 3z\\
 = 9y + 3z\\
 = 3.\left( {3y + z} \right)\\
3,\\
a,\\
{x^2} - {y^2} - 2x + 2y\\
 = \left( {{x^2} - {y^2}} \right) + \left( { - 2x + 2y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) - 2.\left( {x - y} \right)\\
 = \left( {x - y} \right).\left[ {\left( {x + y} \right) - 2} \right]\\
 = \left( {x - y} \right)\left( {x + y - 2} \right)\\
b,\\
2x + 2y - {x^2} - xy\\
 = \left( {2x + 2y} \right) - \left( {{x^2} + xy} \right)\\
 = 2.\left( {x + y} \right) - x.\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {2 - x} \right)\\
c,\\
3{a^2} - 6ab + 3{b^2} - 12{c^2}\\
 = 3.\left( {{a^2} - 2ab + {b^2} - 4{c^2}} \right)\\
 = 3.\left[ {\left( {{a^2} - 2ab + {b^2}} \right) - 4{c^2}} \right]\\
 = 3.\left[ {{{\left( {a - b} \right)}^2} - {{\left( {2c} \right)}^2}} \right]\\
 = 3.\left[ {\left( {a - b} \right) - 2c} \right].\left[ {\left( {a - b} \right) + 2c} \right]\\
 = 3.\left( {a - b - 2c} \right).\left( {a - b + 2c} \right)\\
d,\\
{x^2} - 25 + {y^2} + 2xy\\
 = \left( {{x^2} + 2xy + {y^2}} \right) - 25\\
 = {\left( {x + y} \right)^2} - {5^2}\\
 = \left[ {\left( {x + y} \right) - 5} \right].\left[ {\left( {x + y} \right) + 5} \right]\\
 = \left( {x + y - 5} \right)\left( {x + y + 5} \right)\\
e,\\
{a^2} + 2ab + {b^2} - ac - bc\\
 = \left( {{a^2} + 2ab + {b^2}} \right) - \left( {ac + bc} \right)\\
 = {\left( {a + b} \right)^2} - c.\left( {a + b} \right)\\
 = \left( {a + b} \right).\left[ {\left( {a + b} \right) - c} \right]\\
 = \left( {a + b} \right)\left( {a + b - c} \right)\\
f,\\
{x^2} - 2x - 4{y^2} - 4y\\
 = \left( {{x^2} - 4{y^2}} \right) + \left( { - 2x - 4y} \right)\\
 = \left[ {{x^2} - {{\left( {2y} \right)}^2}} \right] - 2.\left( {x + 2y} \right)\\
 = \left( {x - 2y} \right)\left( {x + 2y} \right) - 2.\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right).\left[ {\left( {x - 2y} \right) - 2} \right]\\
 = \left( {x + 2y} \right).\left( {x - 2y - 2} \right)
\end{array}\)