Đáp án:
$\begin{array}{l}
a)3a + 3b = 3\left( {a + b} \right)\\
b){a^2} - 6ab + 9{b^2}\\
= {\left( {a - 3b} \right)^2}\\
c)ab - ac - 4b + 4c\\
= a\left( {b - c} \right) - 4\left( {b - c} \right)\\
= \left( {b - c} \right)\left( {a - 4} \right)\\
d){a^2} + 5a + 4\\
= {a^2} + a + 4a + 4\\
= a\left( {a + 1} \right) + 4\left( {a + 1} \right)\\
= \left( {a + 1} \right)\left( {a + 4} \right)\\
e)x - xy + y - {y^2}\\
= x.\left( {1 - y} \right) + y.\left( {1 - y} \right)\\
= \left( {1 - y} \right)\left( {x + y} \right)\\
f){x^2} - 4x - {y^2} + 4\\
= {x^2} - 4x + 4 - {y^2}\\
= {\left( {x - 2} \right)^2} - {y^2}\\
= \left( {x - 2 - y} \right)\left( {x - 2 + y} \right)\\
g){x^2} - 2x - 3\\
= {x^2} - 3x + x - 3\\
= x\left( {x - 3} \right) + x - 3\\
= \left( {x - 3} \right)\left( {x + 1} \right)\\
i)x\left( {x - y} \right) + 3x - 3y\\
= x\left( {x - y} \right) + 3\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + 3} \right)\\
j){x^2} - 9{y^2}\\
= {x^2} - {\left( {3y} \right)^2}\\
= \left( {x - 3y} \right)\left( {x + 3y} \right)
\end{array}$
