Đáp án:

$\begin{array}{l}
a){x^2} - 4 - 3\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x + 2} \right) - 3\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x + 2 - 3} \right)\\
 = \left( {x - 2} \right)\left( {x - 1} \right)\\
b){x^2} - xy + 5y - 25\\
 = {x^2} - 25 - xy + 5y\\
 = \left( {x + 5} \right)\left( {x - 5} \right) - y\left( {x - 5} \right)\\
 = \left( {x - 5} \right)\left( {x + 5 - y} \right)\\
c){x^3} + {x^2} - 2x - 8\\
 = {x^3} - 8 + {x^2} - 2x\\
 = \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + x\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} + 2x + 4 + x} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} + 3x + 4} \right)\\
d){x^3} - 4{x^2} - 8x + 8\\
 = {x^3} + 8 - 4{x^2} - 8x\\
 = \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - 4x\left( {x + 2} \right)\\
 = \left( {x + 2} \right)\left( {{x^2} - 2x + 4 - 4x} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} - 6x + 4} \right)\\
e){x^2} - {y^2} + 4x + 4\\
 = {x^2} + 4x + 4 - {y^2}\\
 = {\left( {x + 2} \right)^2} - {y^2}\\
 = \left( {x + 2 + y} \right)\left( {x + 2 - y} \right)\\
g){x^2} + 2x - 4{y^2} - 4y\\
 = {x^2} - 4{y^2} + 2x - 4y\\
 = \left( {x - 2y} \right)\left( {x + 2y} \right) + 2\left( {x - 2y} \right)\\
 = \left( {x - 2y} \right)\left( {x + 2y + 2} \right)\\
h){x^4} + 5{x^3} + 15x - 9\\
 = {x^4} - 9 + 5{x^3} + 15x\\
 = \left( {{x^2} + 3} \right)\left( {{x^2} - 3} \right) + 5x\left( {{x^2} + 3} \right)\\
 = \left( {{x^2} + 3} \right)\left( {{x^2} - 3 + 5x} \right)\\
i){x^4} - 6{x^3} + 54x - 81\\
 = {x^4} - 81 - 6{x^3} + 54x\\
 = \left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right) - 6x\left( {{x^2} - 9} \right)\\
 = \left( {{x^2} - 9} \right)\left( {{x^2} + 9 - 6x} \right)\\
 = \left( {x - 3} \right)\left( {x + 3} \right){\left( {x - 3} \right)^2}\\
 = {\left( {x - 3} \right)^3}\left( {x + 3} \right)
\end{array}$