Đáp án:
$\begin{array}{l}
a){x^2} + xy - {x^3} + {y^2} + {y^3}\\
= {y^3} - {x^3} + {x^2} + xy + {y^2}\\
= \left( {y - x} \right)\left( {{y^2} + xy + {x^2}} \right) + \left( {{y^2} + xy + {x^2}} \right)\\
= \left( {{y^2} + xy + {x^2}} \right)\left( {y - x + 1} \right)\\
b){x^3} - 2x - {x^2}y + 2y\\
= x\left( {{x^2} - 2} \right) - y\left( {{x^2} - 2} \right)\\
= \left( {{x^2} - 2} \right)\left( {x - y} \right)\\
c){x^3} - {y^3} + {x^2} - x{y^2} + {x^2}y - {y^2}\\
= {x^3} - {y^3} + {x^2} - {y^2} + {x^2}y - x{y^2}\\
= \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) + \left( {x - y} \right)\left( {x + y} \right)\\
+ xy\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {{x^2} + xy + {y^2} + x + y + xy} \right)\\
= \left( {x - y} \right)\left( {{x^2} + 2xy + {y^2} + x + y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right)\left( {x + y + 1} \right)\\
d){x^2}y + 2xy - 2y\\
= y\left( {{x^2} + 2x - 2} \right)\\
e){x^2} - 4{y^2} + 6x + 9\\
= {x^2} + 6x + 9 - 4{y^2}\\
= {\left( {x + 3} \right)^2} - 4{y^2}\\
= \left( {x + 3 + 2y} \right)\left( {x + 3 - 2y} \right)\\
f){\left( {x + 1} \right)^4} - {\left( {2x - 3} \right)^2}\\
= \left[ {{{\left( {x + 1} \right)}^2} - 2x + 3} \right]\left[ {{{\left( {x + 1} \right)}^2} + 2x - 3} \right]\\
= \left( {{x^2} + 4} \right)\left( {{x^2} + 4x - 2} \right)\\
g)2{x^2} - x - 1\\
= 2{x^2} - 2x + x - 1\\
= \left( {x - 1} \right)\left( {2x + 1} \right)\\
h){x^2} - 10x + 5\\
= {x^2} - 10x + 25 - 20\\
= {\left( {x - 5} \right)^2} - 20\\
= \left( {x - 5 - 2\sqrt 5 } \right)\left( {x - 5 + 2\sqrt 5 } \right)
\end{array}$
