Đáp án:

\(\begin{array}{l}
1,\\
\left( {x + 1} \right).\left( {{x^2} + x + 1} \right).{\left( {{x^2} - x + 1} \right)^2}\\
2,\\
\left( {x - 2} \right).\left( {{x^4} + {x^3} + {x^2} + x + 1} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
{x^7} + {x^5} + {x^4} + {x^3} + {x^2} + 1\\
 = \left( {{x^7} - x} \right) + \left( {{x^5} + {x^4} + {x^3}} \right) + \left( {{x^2} + x + 1} \right)\\
 = x\left( {{x^6} - 1} \right) + {x^3}.\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
 = x.\left[ {{{\left( {{x^3}} \right)}^2} - {1^2}} \right] + \left( {{x^2} + x + 1} \right).\left( {{x^3} + 1} \right)\\
 = x.\left( {{x^3} - 1} \right).\left( {{x^3} + 1} \right) + \left( {{x^2} + x + 1} \right).\left( {{x^3} + 1} \right)\\
 = x.\left( {x - 1} \right).\left( {{x^2} + x + 1} \right)\left( {{x^3} + 1} \right) + \left( {{x^2} + x + 1} \right).\left( {{x^3} + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^3} + 1} \right).\left[ {x\left( {x - 1} \right) + 1} \right]\\
 = \left( {{x^2} + x + 1} \right).\left( {x + 1} \right).\left( {{x^2} - x + 1} \right).\left( {{x^2} - x + 1} \right)\\
 = \left( {x + 1} \right).\left( {{x^2} + x + 1} \right).{\left( {{x^2} - x + 1} \right)^2}\\
2,\\
{x^5} - {x^4} - {x^3} - {x^2} - x - 2\\
 = \left( {{x^5} - 2{x^4}} \right) + \left( {{x^4} - 2{x^3}} \right) + \left( {{x^3} - 2{x^2}} \right) + \left( {{x^2} - 2x} \right) + \left( {x - 2} \right)\\
 = {x^4}.\left( {x - 2} \right) + {x^3}.\left( {x - 2} \right) + {x^2}.\left( {x - 2} \right) + x.\left( {x - 2} \right) + \left( {x - 2} \right)\\
 = \left( {x - 2} \right).\left( {{x^4} + {x^3} + {x^2} + x + 1} \right)
\end{array}\)