Đáp án:

\(\begin{array}{l}
a,\\
5xy\left( {x - 3} \right)\\
b,\\
\left( {2x - 7} \right)\left( {2x + 7} \right)\\
c,\\
\left( {x - 3y} \right)\left( {x - 5} \right)\\
d,\\
\left( {x + y - 2} \right)\left( {x + y + 2} \right)\\
e,\\
x\left( {{x^2} - 6xy + 9} \right)\\
f,\\
\left( {x - 2y} \right)\left( {x + 2y + 2} \right)\\
g,\\
\left( {x - 5} \right)\left( {x - 3} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
5{x^2}y - 15xy\\
 = 5xy.x - 5xy.3\\
 = 5xy\left( {x - 3} \right)\\
b,\\
4{x^2} - 49\\
 = {\left( {2x} \right)^2} - {7^2}\\
 = \left( {2x - 7} \right)\left( {2x + 7} \right)\\
c,\\
{x^2} - 3xy - 5x + 15y\\
 = \left( {{x^2} - 3xy} \right) + \left( { - 5x + 15y} \right)\\
 = x.\left( {x - 3y} \right) - 5.\left( {x - 3y} \right)\\
 = \left( {x - 3y} \right)\left( {x - 5} \right)\\
d,\\
{x^2} + 2xy - 4 + {y^2}\\
 = \left( {{x^2} + 2xy + {y^2}} \right) - 4\\
 = {\left( {x + y} \right)^2} - {2^2}\\
 = \left[ {\left( {x + y} \right) - 2} \right].\left[ {\left( {x + y} \right) + 2} \right]\\
 = \left( {x + y - 2} \right)\left( {x + y + 2} \right)\\
e,\\
{x^3} - 6{x^2}y + 9x\\
 = x\left( {{x^2} - 6xy + 9} \right)\\
f,\\
{x^2} - 4{y^2} + 2x - 4y\\
 = \left( {{x^2} - 4{y^2}} \right) + \left( {2x - 4y} \right)\\
 = \left[ {{x^2} - {{\left( {2y} \right)}^2}} \right] + 2.\left( {x - 2y} \right)\\
 = \left( {x - 2y} \right)\left( {x + 2y} \right) + 2\left( {x - 2y} \right)\\
 = \left( {x - 2y} \right)\left[ {\left( {x + 2y} \right) + 2} \right]\\
 = \left( {x - 2y} \right)\left( {x + 2y + 2} \right)\\
g,\\
{x^2} - 8x + 15\\
 = \left( {{x^2} - 8x + 16} \right) - 1\\
 = \left( {{x^2} - 2.x.4 + {4^2}} \right) - 1\\
 = {\left( {x - 4} \right)^2} - {1^2}\\
 = \left[ {\left( {x - 4} \right) - 1} \right].\left[ {\left( {x - 4} \right) + 1} \right]\\
 = \left( {x - 5} \right)\left( {x - 3} \right)
\end{array}\)