Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^4} + 2{x^3} + {x^2} - {y^4}\\
= {\left( {{x^2}} \right)^2} + 2.{x^2}.x + {x^2} - {y^4}\\
= {\left( {{x^2} + x} \right)^2} - {\left( {{y^2}} \right)^2}\\
= \left( {{x^2} + x + {y^2}} \right)\left( {{x^2} + x - {y^2}} \right)\\
b,\\
{x^2}.\left( {x - z} \right) + {y^2}\left( {y - z} \right) + xyz\\
= {x^3} - {x^2}z + {y^3} - {y^2}z + xyz\\
= \left( {{x^3} + {y^3}} \right) - \left( {{x^2}z + {y^2}z - xyz} \right)\\
= \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - z.\left( {{x^2} - xy + {y^2}} \right)\\
= \left( {{x^2} - xy + {y^2}} \right).\left( {x + y - z} \right)\\
c,\\
a{c^2} - {a^2}c + {b^2}c - b{c^2} + ab\left( {a - b} \right)\\
= \left( {a{c^2} - b{c^2}} \right) - \left( {{a^2}c - {b^2}c} \right) + ab\left( {a - b} \right)\\
= {c^2}\left( {a - b} \right) - c\left( {{a^2} - {b^2}} \right) + ab\left( {a - b} \right)\\
= {c^2}\left( {a - b} \right) - c\left( {a - b} \right)\left( {a + b} \right) + ab\left( {a - b} \right)\\
= \left( {a - b} \right).\left( {{c^2} - c.\left( {a + b} \right) + ab} \right)\\
= \left( {a - b} \right).\left( {{c^2} - ca - cb + ab} \right)\\
= \left( {a - b} \right).\left[ {c\left( {c - a} \right) - b.\left( {c - a} \right)} \right]\\
= \left( {a - b} \right)\left( {c - a} \right)\left( {c - b} \right)\\
d,\\
{a^3} + 2020{a^2} + 2020a + 2019\\
= \left( {{a^3} + 2019{a^2}} \right) + \left( {{a^2} + 2019a} \right) + \left( {a + 2019} \right)\\
= {a^2}\left( {a + 2019} \right) + a\left( {a + 2019} \right) + \left( {a + 2019} \right)\\
= \left( {a + 2019} \right).\left( {{a^2} + a + 1} \right)
\end{array}\)
