Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2x{y^2} + 4xy + 2x - 2x{z^2} + 4xzt - 2x{t^2}\\
= 2x.\left( {{y^2} + 2y + 1 - {z^2} + 2zt - {t^2}} \right)\\
= 2x.\left[ {\left( {{y^2} + 2y + 1} \right) - \left( {{z^2} - 2zt + {t^2}} \right)} \right]\\
= 2x.\left[ {{{\left( {y + 1} \right)}^2} - {{\left( {z - t} \right)}^2}} \right]\\
= 2x.\left( {y + 1 - z + t} \right)\left( {y + 1 + z - t} \right)\\
b,\\
{x^4} + {x^3} + 3{x^2} + 2x + 2\\
= \left( {{x^4} + {x^3} + {x^2}} \right) + \left( {2{x^2} + 2x + 2} \right)\\
= {x^2}\left( {{x^2} + x + 1} \right) + 2.\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^2} + 2} \right)\\
c,\\
{x^5} - x{y^4} + {x^4}y - {y^5}\\
= \left( {{x^5} + {x^4}y} \right) - \left( {x{y^4} + {y^5}} \right)\\
= {x^4}\left( {x + y} \right) - {y^4}\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {{x^4} - {y^4}} \right)\\
= \left( {x + y} \right)\left( {{x^2} - {y^2}} \right)\left( {{x^2} + {y^2}} \right)\\
= \left( {x + y} \right)\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)\\
= \left( {x - y} \right){\left( {x + y} \right)^2}\left( {{x^2} + {y^2}} \right)
\end{array}\)
