Đáp án:
$\begin{array}{l}
a)2{x^2}y - 6{x^3}{y^2}\\
= 2{x^2}y\left( {1 - 2xy} \right)\\
b)9{x^2}y{z^2} - 3xyz + 6{x^2}{y^2}\\
= 3xy\left( {3x{z^2} - z + 2xy} \right)\\
c)5x\left( {2x + 1} \right) + 3\left( {2x + 1} \right)\\
= \left( {2x + 1} \right)\left( {5x + 3} \right)\\
d)10x\left( {x + 1} \right) - 5y\left( {1 + x} \right)\\
= \left( {x + 1} \right)\left( {10x - 5y} \right)\\
= 5\left( {x + 1} \right)\left( {2x - y} \right)\\
e)x\left( {x - 3} \right) + y\left( {3 - x} \right)\\
= x\left( {x - 3} \right) - y\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {x - y} \right)\\
g)2{x^2}\left( {y - x} \right) - xy\left( {x - y} \right)\\
= 2{x^2}\left( {y - x} \right) + xy\left( {y - x} \right)\\
= \left( {y - x} \right).x.\left( {2x + y} \right)\\
h)5{x^2}\left( {x - 1} \right) + 3x\left( {1 - x} \right)\\
= 5{x^2}\left( {x - 1} \right) - 3x\left( {x - 1} \right)\\
= \left( {x - 1} \right).x.\left( {5x - 3} \right)\\
k){x^2} + x\\
= x\left( {x + 1} \right)\\
m)2{a^2}b - 4a{b^2}\\
= 2ab\left( {a - 2b} \right)\\
n)a{\left( {x + y} \right)^2} - {\left( {x + y} \right)^2}\\
= {\left( {x + y} \right)^2}\left( {a - 1} \right)\\
j)5\left( {x - 7} \right) - a\left( {7 - x} \right)\\
= 5\left( {x - 7} \right) + a\left( {x - 7} \right)\\
= \left( {x - 7} \right)\left( {5 + a} \right)
\end{array}$
