Đáp án:
$\begin{array}{l}
a){x^3} - 2{x^2} + 2x - 1 = {x^3} - 2{x^2} + x + x - 1\\
= x\left( {{x^2} - 2x + 1} \right) + \left( {x - 1} \right)\\
= x{\left( {x - 1} \right)^2} + \left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x\left( {x - 1} \right) + 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - x + 1} \right)\\
b)20{x^2} - 5{y^2} + 10y - 5\\
= 5\left( {4{x^2} - {y^2} + 2y - 1} \right)\\
= 5\left[ {{{\left( {2x} \right)}^2} - \left( {{y^2} - 2y + 1} \right)} \right]\\
= 5\left[ {{{\left( {2x} \right)}^2} - {{\left( {y - 1} \right)}^2}} \right]\\
= 5\left( {2x - y + 1} \right)\left( {2x + y - 1} \right)
\end{array}$
c)
\begin{array}{l}
4{y^2} + 2x{y^2} - xy - 1\\
= \left( {4{y^2} - 1} \right) + \left( {2x{y^2} - xy} \right)\\
= \left( {2y - 1} \right)\left( {2y + 1} \right) + xy\left( {2y - 1} \right)\\
= \left( {2y - 1} \right)\left( {2y + 1 + xy} \right)
\end{array}
