Đáp án:
$\begin{array}{l}
m)2x\left( {y - z} \right) + 5y\left( {z - y} \right)\\
= \left( {y - z} \right)\left( {2x - 5y} \right)\\
n)3\left( {x - y} \right) - 5x\left( {y - x} \right)\\
= \left( {x - y} \right)\left( {3 + 5x} \right)\\
q)\frac{5}{7}x\left( {y - 2009} \right) - 3y\left( {2009 - y} \right)\\
= \left( {y - 2009} \right)\left( {\frac{5}{7}x + 3y} \right)\\
s){a^2}b\left( {x + y} \right) + a{b^2}\left( {x + y} \right)\\
= ab\left( {x + y} \right)\left( {a + b} \right)\\
o)10{x^2}\left( {x + y} \right) - 5\left( {2x + 2y} \right){y^2}\\
= 10\left( {x + y} \right)\left( {{x^2} - {y^2}} \right)\\
= 10{\left( {x + y} \right)^2}\left( {x - y} \right)\\
p)3x\left( {x - 1} \right) + 5\left( {1 - x} \right)\\
= \left( {x - 1} \right)\left( {3x - 5} \right)\\
r)10x\left( {x - y} \right) - 8y\left( {y - x} \right)\\
= 2\left( {x - y} \right)\left( {5x + 4y} \right)\\
t){a^2}b\left( {x + y} \right) + a{b^2}\left( {x + y} \right)\\
= ab\left( {x + y} \right)\left( {a + b} \right)
\end{array}$
