Đáp án:

\(\begin{array}{l}
3,\\
Tất\,\,cả\,\,các\,\,biểu\,\,thức\,\,đều\,\,không\,\,phụ\,\,thuộc\,\,vào\,\,x\\
4,\\
a,\\
\left( {1 - x + y} \right)\left( {1 + x - y} \right)\\
b,\\
\left( {a - b - c + d} \right).\left( {a - b + c - d} \right)\\
c,\\
\left( {ab - 1} \right).\left( {{a^2}{b^2} + ab + 1} \right)\\
d,\\
\left( {y - z} \right).\left( {x - y} \right)\left( {x - z} \right)\\
e,\\
\left( {x - 3} \right)\left( {x - 12} \right)\\
g,\\
\left( {{x^2} + 20} \right)\left( {x - 6} \right)\left( {x + 6} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
a,\\
{\left( {x - 1} \right)^3} - {\left( {x + 1} \right)^3} + 6.\left( {x + 1} \right)\left( {x - 1} \right)\\
 = \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - \left( {{x^3} + 3.{x^2}.1 + 3.x{{.1}^2} + {1^3}} \right) + 6.\left( {{x^2} - {1^2}} \right)\\
 = \left( {{x^3} - 3{x^2} + 3x - 1} \right) - \left( {{x^3} + 3{x^2} + 3x + 1} \right) + 6.\left( {{x^2} - 1} \right)\\
 = {x^3} - 3{x^2} + 3x - 1 - {x^3} - 3{x^2} - 3x - 1 + 6{x^2} - 6\\
 = \left( {{x^3} - {x^3}} \right) + \left( { - 3{x^2} - 3{x^2} + 6{x^2}} \right) + \left( {3x - 3x} \right) + \left( { - 1 - 1 - 6} \right)\\
 =  - 8\\
b,\\
\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {x + 1} \right).\left( {{x^2} - x.1 + {1^2}} \right) - \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right)\\
 = \left( {{x^3} + {1^3}} \right) - \left( {{x^3} - {1^3}} \right)\\
 = \left( {{x^3} + 1} \right) - \left( {{x^3} - 1} \right)\\
 = {x^3} + 1 - {x^3} + 1\\
 = 2\\
c,\\
{\left( {x - 2} \right)^2} - \left( {x - 3} \right)\left( {x - 1} \right)\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) - \left( {{x^2} - x - 3x + 3} \right)\\
 = \left( {{x^2} - 4x + 4} \right) - \left( {{x^2} - 4x + 3} \right)\\
 = {x^2} - 4x + 4 - {x^2} + 4x - 3\\
 = \left( {{x^2} - {x^2}} \right) + \left( { - 4x + 4x} \right) + \left( {4 - 3} \right)\\
 = 1\\
d,\\
\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {x + 1} \right).\left( {{x^2} - x.1 + {1^2}} \right) - \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right)\\
 = \left( {{x^3} + {1^3}} \right) - \left( {{x^3} - {1^3}} \right)\\
 = \left( {{x^3} + 1} \right) - \left( {{x^3} - 1} \right)\\
 = {x^3} + 1 - {x^3} + 1\\
 = 2\\
 \Rightarrow Tất\,\,cả\,\,các\,\,biểu\,\,thức\,\,đều\,\,không\,\,phụ\,\,thuộc\,\,vào\,\,x\\
4,\\
a,\\
1 + 2xy - {x^2} - {y^2}\\
 = 1 + \left( {2xy - {x^2} - {y^2}} \right)\\
 = 1 - \left( {{x^2} - 2xy + {y^2}} \right)\\
 = 1 - {\left( {x - y} \right)^2}\\
 = {1^2} - {\left( {x - y} \right)^2}\\
 = \left[ {1 - \left( {x - y} \right)} \right].\left[ {1 + \left( {x - y} \right)} \right]\\
 = \left( {1 - x + y} \right)\left( {1 + x - y} \right)\\
b,\\
{a^2} + {b^2} - {c^2} - {d^2} - 2ab + 2cd\\
 = \left( {{a^2} + {b^2} - 2ab} \right) + \left( { - {c^2} - {d^2} + 2cd} \right)\\
 = \left( {{a^2} - 2ab + {b^2}} \right) - \left( {{c^2} - 2cd + {d^2}} \right)\\
 = {\left( {a - b} \right)^2} - {\left( {c - d} \right)^2}\\
 = \left[ {\left( {a - b} \right) - \left( {c - d} \right)} \right].\left[ {\left( {a - b} \right) + \left( {c - d} \right)} \right]\\
 = \left( {a - b - c + d} \right).\left( {a - b + c - d} \right)\\
c,\\
{a^3}{b^3} - 1\\
 = {\left( {ab} \right)^3} - {1^3}\\
 = \left( {ab - 1} \right).\left[ {{{\left( {ab} \right)}^2} + ab.1 + {1^2}} \right]\\
 = \left( {ab - 1} \right).\left( {{a^2}{b^2} + ab + 1} \right)\\
d,\\
{x^2}\left( {y - z} \right) + {y^2}.\left( {z - x} \right) + {z^2}.\left( {x - y} \right)\\
 = {x^2}\left( {y - z} \right) + {y^2}z - {y^2}x + {z^2}x - {z^2}y\\
 = {x^2}\left( {y - z} \right) + \left( {{y^2}z - {z^2}y} \right) + \left( {{z^2}x - {y^2}x} \right)\\
 = {x^2}\left( {y - z} \right) + yz.\left( {y - z} \right) + x\left( {{z^2} - {y^2}} \right)\\
 = {x^2}\left( {y - z} \right) + yz\left( {y - z} \right) - x\left( {{y^2} - {z^2}} \right)\\
 = {x^2}\left( {y - z} \right) + yz\left( {y - z} \right) - x\left( {y - z} \right)\left( {y + z} \right)\\
 = \left( {y - z} \right).\left[ {{x^2} + yz - x.\left( {y + z} \right)} \right]\\
 = \left( {y - z} \right)\left( {{x^2} + yz - xy - xz} \right)\\
 = \left( {y - z} \right)\left[ {\left( {{x^2} - xy} \right) + \left( {yz - xz} \right)} \right]\\
 = \left( {y - z} \right).\left[ {x.\left( {x - y} \right) + z.\left( {y - x} \right)} \right]\\
 = \left( {y - z} \right).\left( {x - y} \right)\left( {x - z} \right)\\
e,\\
{x^2} - 15x + 36\\
 = \left( {{x^2} - 3x} \right) + \left( { - 12x + 36} \right)\\
 = x.\left( {x - 3} \right) - 12.\left( {x - 3} \right)\\
 = \left( {x - 3} \right)\left( {x - 12} \right)\\
g,\\
{\left( {{x^2} - 8} \right)^2} - 784\\
 = {\left( {{x^2} - 8} \right)^2} - {28^2}\\
 = \left[ {\left( {{x^2} - 8} \right) + 28} \right].\left[ {\left( {{x^2} - 8} \right) - 28} \right]\\
 = \left( {{x^2} + 20} \right)\left( {{x^2} - 36} \right)\\
 = \left( {{x^2} + 20} \right)\left( {{x^2} - {6^2}} \right)\\
 = \left( {{x^2} + 20} \right)\left( {x - 6} \right)\left( {x + 6} \right)
\end{array}\)