Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
10,\\
{x^4} - {x^2} + 2x - 1 = {x^4} - \left( {{x^2} - 2x + 1} \right)\\
 = {\left( {{x^2}} \right)^2} - {\left( {x - 1} \right)^2} = \left[ {{x^2} - \left( {x - 1} \right)} \right].\left[ {{x^2} + \left( {x - 1} \right)} \right]\\
 = \left( {{x^2} - x + 1} \right)\left( {{x^2} + x - 1} \right)\\
11,\\
3a - 3b + {a^2} - 2ab + {b^2} = \left( {3a - 3b} \right) + \left( {{a^2} - 2ab + {b^2}} \right)\\
 = 3\left( {a - b} \right) + {\left( {a - b} \right)^2} = \left( {a - b} \right)\left( {3 + a - b} \right)\\
12,\\
{a^2} + 2ab + {b^2} - 2a - 2b + 1 = \left( {{a^2} + 2ab + {b^2}} \right) - \left( {2a + 2b} \right) + 1\\
 = {\left( {a + b} \right)^2} - 2.\left( {a + b} \right).1 + {1^2} = {\left( {a + b - 1} \right)^2}\\
13,\\
{a^2} - {b^2} - 4a + 4b = \left( {{a^2} - {b^2}} \right) - \left( {4a - 4b} \right)\\
 = \left( {a - b} \right)\left( {a + b} \right) - 4\left( {a - b} \right) = \left( {a - b} \right)\left( {a + b - 4} \right)\\
14,\\
{a^3} - {b^3} - 3a + 3b = \left( {{a^3} - {b^3}} \right) - \left( {3a - 3b} \right)\\
 = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) - 3\left( {a - b} \right)\\
 = \left( {a - b} \right)\left( {{a^2} + ab + {b^2} - 3} \right)\\
15,\\
{x^3} + 3{x^2} - 3x - 1 = \left( {{x^3} - {x^2}} \right) + \left( {4{x^2} - 4x} \right) + \left( {x - 1} \right)\\
 = {x^2}\left( {x - 1} \right) + 4x\left( {x - 1} \right) + \left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} + 4x + 1} \right)\\
16,\\
{x^3} - 3{x^2} - 3x + 1 = \left( {{x^3} + {x^2}} \right) - \left( {4{x^2} + 4x} \right) + \left( {x + 1} \right)\\
 = {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
17,\\
{x^3} - 4{x^2} + 4x - 1 = \left( {{x^3} - {x^2}} \right) - \left( {3{x^2} - 3x} \right) + \left( {x - 1} \right)\\
 = {x^2}\left( {x - 1} \right) - 3x\left( {x - 1} \right) + \left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} - 3x + 1} \right)\\
18,\\
4{a^2}{b^2} - {\left( {{a^2} + {b^2} - 1} \right)^2} = {\left( {2ab} \right)^2} - {\left( {{a^2} + {b^2} - 1} \right)^2}\\
 = \left[ {2ab - \left( {{a^2} + {b^2} - 1} \right)} \right].\left[ {2ab + \left( {{a^2} + {b^2} - 1} \right)} \right]\\
 = \left[ {1 - \left( {{a^2} - 2ab + {b^2}} \right)} \right].\left[ {\left( {{a^2} + 2ab + {b^2}} \right) - 1} \right]\\
 = \left[ {1 - {{\left( {a - b} \right)}^2}} \right].\left[ {{{\left( {a + b} \right)}^2} - 1} \right]\\
 = \left( {1 - a + b} \right)\left( {1 + a - b} \right)\left( {a + b - 1} \right)\left( {a + b + 1} \right)\\
19,\\
{\left( {xy + 4} \right)^2} - {\left( {2x + 2y} \right)^2}\\
 = \left[ {\left( {xy + 4} \right) - \left( {2x + 2y} \right)} \right].\left[ {\left( {xy + 4} \right) + \left( {2x + 2y} \right)} \right]\\
 = \left( {xy - 2x - 2y + 4} \right).\left( {xy + 2x + 2y + 4} \right)\\
 = \left[ {x\left( {y - 2} \right) - 2.\left( {y - 2} \right)} \right].\left[ {x\left( {y + 2} \right) + 2\left( {y + 2} \right)} \right]\\
 = \left( {x - 2} \right)\left( {y - 2} \right)\left( {x + 2} \right)\left( {y + 2} \right)\\
20,\\
{\left( {{a^2} + {b^2} + ab} \right)^2} - {a^2}{b^2} - {b^2}{c^2} - {c^2}{a^2}\\
 = \left[ {{{\left( {{a^2} + {b^2} + ab} \right)}^2} - {a^2}{b^2}} \right] - {b^2}{c^2} - {c^2}{a^2}\\
 = \left[ {\left( {{a^2} + {b^2} + ab} \right) - ab} \right].\left[ {\left( {{a^2} + {b^2} + ab} \right) + ab} \right] - {c^2}\left( {{a^2} + {b^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right).\left( {{a^2} + 2ab + {b^2}} \right) - {c^2}\left( {{a^2} + {b^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right).\left( {{a^2} + 2ab + {b^2} - {c^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right).\left[ {{{\left( {a + b} \right)}^2} - {c^2}} \right]\\
 = \left( {{a^2} + {b^2}} \right)\left( {a + b - c} \right)\left( {a + b + c} \right)
\end{array}\)