Đáp án:

$\begin{array}{l}
a)\left( {{x^2} - 2x + 1} \right):\left( {x - 1} \right)\\
 = {\left( {x - 1} \right)^2}:\left( {x - 1} \right)\\
 = x - 1\\
b)\left( {{x^3} + 1} \right):\left( {{x^2} - x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right):\left( {{x^2} - x + 1} \right)\\
 = x + 1\\
c)\left( {{x^3} - {x^2} - 5x - 3} \right):\left( {x - 3} \right)\\
 = \left( {{x^3} - 3{x^2} + 2{x^2} - 6x + x - 3} \right):\left( {x - 3} \right)\\
 = \left( {{x^2} + 2x + 1} \right)\left( {x - 3} \right):\left( {x - 3} \right)\\
 = {x^2} + 2x + 1\\
 = {\left( {x + 1} \right)^2}\\
d)\left( {{x^4} + {x^3} - 6{x^2} - 5x + 5} \right):\left( {{x^2} + x - 1} \right)\\
 = \left( {{x^4} + {x^3} - {x^2} - 5{x^2} - 5x + 5} \right):\left( {{x^2} + x - 1} \right)\\
 = \left( {{x^2} - 5} \right)\left( {{x^2} + x - 1} \right):\left( {{x^2} + x - 1} \right)\\
 = {x^2} - 5
\end{array}$