Giải thích các bước giải:

\(\begin{array}{l}
e,\\
5{a^2}\left( {x - y} \right) + 10a.\left( {x - y} \right) = \left( {x - y} \right).\left( {5{a^2} + 10a} \right) = \left( {x - y} \right).5a.\left( {a + 2} \right)\\
f,\\
3a\left( {x - y} \right) + 6\left( {x - y} \right) = \left( {x - y} \right)\left( {3a + 6} \right) = \left( {x - y} \right).3\left( {a + 2} \right) = 3.\left( {x - y} \right)\left( {a + 2} \right)\\
g,\\
a\left( {x - 1} \right) + 10{a^2}\left( {1 - x} \right) = \left( {x - 1} \right).\left( {a - 10{a^2}} \right) = \left( {x - 1} \right).a.\left( {1 - 10a} \right)\\
k,\\
{a^2} - 4{b^2} = {a^2} - {\left( {2b} \right)^2} = \left( {a - 2b} \right)\left( {a + 2b} \right)\\
q,\\
9{x^2} + 12x + 4 = {\left( {3x} \right)^2} + 2.3x.2 + {2^2} = {\left( {3x + 2} \right)^2}\\
r,\\
25{x^2} - 20xy + 4{y^2} = {\left( {5x} \right)^2} - 2.5x.2y + {\left( {2y} \right)^2} = {\left( {5x - 2y} \right)^2}\\
l,\\
25{a^2} - 1 = {\left( {5a} \right)^2} - {1^2} = \left( {5a - 1} \right)\left( {5a + 1} \right)\\
m,\\
25{a^2} - 49{b^4} = {\left( {5a} \right)^2} - {\left( {7{b^2}} \right)^2} = \left( {5a - 7{b^2}} \right)\left( {5a + 7{b^2}} \right)\\
n,\\
\dfrac{1}{4}{a^2} - \dfrac{1}{9}{b^2} = {\left( {\dfrac{1}{2}a} \right)^2} - {\left( {\dfrac{1}{3}b} \right)^2} = \left( {\dfrac{1}{2}a - \dfrac{1}{3}b} \right)\left( {\dfrac{1}{2}a + \dfrac{1}{3}b} \right)\\
o,\\
49{a^2} - {\left( {2a - b} \right)^2} = {\left( {7a} \right)^2} - {\left( {2a - b} \right)^2}\\
 = \left[ {7a - \left( {2a - b} \right)} \right].\left[ {7a + \left( {2a - 3b} \right)} \right]\\
 = \left( {5a + b} \right).\left( {9a - 3b} \right)\\
 = 3.\left( {5a + b} \right).\left( {3a - b} \right)\\
p,\\
{x^2} + 10x + 25 = {x^2} + 2.x.5 + {5^2} = {\left( {x + 5} \right)^2}\\
h,\\
a.\left( {x - 3} \right) - {a^2}\left( {3 - x} \right) = \left( {x - 3} \right)\left( {a + {a^2}} \right) = \left( {x - 3} \right).a.\left( {1 + a} \right)\\
i,\\
5{x^2}y\left( {x - 7} \right) - 5xy\left( {7 - x} \right)\\
 = \left( {x - 7} \right).\left( {5{x^2}y + 5xy} \right)\\
 = \left( {x - 7} \right).5xy\left( {x + 1} \right)
\end{array}\)