Đáp án:

$\begin{array}{l}
B1)a){x^2} - {y^2} - 2x + 2y\\
 = \left( {x - y} \right)\left( {x + y} \right) - 2\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {x + y - 2} \right)\\
b){x^2}\left( {x - 1} \right) + 16\left( {1 - x} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} - 16} \right)\\
 = \left( {x - 1} \right)\left( {x - 4} \right)\left( {x + 4} \right)\\
c){x^2} + 4x - {y^2} + 4\\
 = {x^2} + 4x + 4 - {y^2}\\
 = {\left( {x + 2} \right)^2} - {y^2}\\
 = \left( {x + 2 + y} \right)\left( {x + 2 - y} \right)\\
d){x^3} - 3{x^2} - 3x + 1\\
 = {x^3} + 1 - 3{x^2} - 3x\\
 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - 3x\left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - x + 1 - 3x} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
e){x^4} + 4{y^4}\\
 = {x^4} + 4{x^2}{y^2} + 4{y^4} - 4{x^2}{y^2}\\
 = {\left( {{x^2} + 2{y^2}} \right)^2} - {\left( {2xy} \right)^2}\\
 = \left( {{x^2} + 2{y^2} - 2xy} \right)\left( {{x^2} + 2{y^2} + 2xy} \right)\\
f){x^4} - 13{x^2} + 36\\
 = {x^4} - 12{x^2} + 36 - {x^2}\\
 = {\left( {{x^2} - 6} \right)^2} - {x^2}\\
 = \left( {{x^2} - x - 6} \right)\left( {{x^2} + x - 6} \right)\\
 = \left( {x - 3} \right)\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 3} \right)\\
g){\left( {{x^2} + x} \right)^2} + 4{x^2} + 4x - 12\\
 = {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) - 12\\
 = {\left( {{x^2} + x} \right)^2} + 6\left( {{x^2} + x} \right) - 2\left( {{x^2} + x} \right) - 12\\
 = \left( {{x^2} + x + 6} \right)\left( {{x^2} + x - 2} \right)\\
 = \left( {{x^2} + x + 6} \right)\left( {x - 1} \right)\left( {x + 2} \right)\\
h){x^6} + 2{x^5} + {x^4} - 2{x^3} - 2{x^2} + 1\\
 = {x^4}\left( {{x^2} + 2x + 1} \right) - 2{x^2}\left( {x + 1} \right) + 1\\
 = {x^4}{\left( {x + 1} \right)^2} - 2{x^2}\left( {x + 1} \right) + 1\\
 = {\left( {{x^2}\left( {x + 1} \right) - 1} \right)^2}\\
 = {\left( {{x^3} + {x^2} - 1} \right)^2}
\end{array}$

2 bài còn lại đã trả lời ở câu khác