Đáp án:
$\begin{array}{l}
a)16{x^2}\left( {x - y} \right) - 10y\left( {y - x} \right)\\
= 2\left( {x - y} \right)\left( {8{x^2} + 5y} \right)\\
b)2x\left( {x + 1} \right) + 2\left( {x + 1} \right)\\
= 2\left( {x + 1} \right)\left( {x + 1} \right)\\
= 2{\left( {x + 1} \right)^2}\\
c){y^2}\left( {{x^2} + y} \right) - z{x^2} - xy\\
= {y^2}\left( {{x^2} + y} \right) - z\left( {{x^2} + y} \right)\\
= \left( {{x^2} + y} \right)\left( {{y^2} - z} \right)\\
d)4x\left( {x - 2y} \right) - 8y\left( {x - 2y} \right)\\
= 4\left( {x - 2y} \right)\left( {x - 2y} \right)\\
= 4{\left( {x - 2y} \right)^2}\\
e){x^2} - 9\\
= \left( {x - 3} \right)\left( {x + 3} \right)\\
f)4{x^2} - 25\\
= \left( {2x - 5} \right)\left( {2x + 5} \right)\\
g){x^2} - 5\\
= \left( {x - \sqrt 5 } \right)\left( {x + \sqrt 5 } \right)\\
h){\left( {3x + 1} \right)^2} - {\left( {x + 1} \right)^2}\\
= \left( {3x + 1 + x + 1} \right)\left( {3x + 1 - x - 1} \right)\\
= \left( {4x + 2} \right)\left( {2x} \right)\\
= 4x\left( {2x + 1} \right)\\
i){\left( {x + y} \right)^2} - {x^2}\\
= \left( {x + y + x} \right)\left( {x + y - x} \right)\\
= \left( {2x + y} \right).y\\
k) - {y^2} + \dfrac{1}{9}\\
= \dfrac{1}{9} - {y^2}\\
= \left( {\dfrac{1}{3} - y} \right)\left( {\dfrac{1}{3} + y} \right)
\end{array}$
