Giải thích các bước giải:
\[\begin{array}{l}
1,\\
\left( {2x + {x^2} + 1} \right):\left( {x + 1} \right)\\
= \left( {{x^2} + 2x + 1} \right):\left( {x + 1} \right) = {\left( {x + 1} \right)^2}:\left( {x + 1} \right) = x + 1\\
2,\\
\left( {3x - 4 + {x^2}} \right):\left( {x - 1} \right)\\
= \left( {{x^2} + 3x - 4} \right):\left( {x - 1} \right) = \left( {{x^2} - x + 4x - 4} \right):\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x + 4} \right):\left( {x - 1} \right) = x + 4\\
3,\\
\left( { - 6x + {x^2} + 5} \right):\left( {x - 5} \right)\\
= \left( {{x^2} - 6x + 5} \right):\left( {x - 5} \right) = \left( {x - 5} \right)\left( {x - 1} \right):\left( {x - 5} \right)\\
= x - 1\\
5,\\
\left( { - x + {x^2} - 12} \right):\left( {x + 3} \right)\\
= \left( {{x^2} - x - 12} \right):\left( {x + 3} \right)\\
= \left( {x - 4} \right)\left( {x + 3} \right):\left( {x + 3} \right) = x - 4
\end{array}\]
