Đáp án:

\(\begin{array}{l}
11,\,\,\,\,{\left( {{x^2} + 4} \right)^2}\\
12,\,\,\,\,\left( {x + y} \right)\left( {x - 5} \right)\\
13,\,\,\,\,\left( {x + y} \right)\left( {1 - x} \right)\\
14,\,\,\,\,\left( {x - y} \right)\left( {x - 7} \right)\\
15,\,\,\,\,\left( {a + c} \right).\left( {{x^2} - y + {y^2}} \right)\\
16,\,\,\,\,\left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
17,\,\,\,\,x.{\left( {x + y} \right)^2}\\
18,\,\,\,\,\left( {x + y} \right).\left( {5 - x - y} \right)\\
19,\,\,\,\,{x^2}\left( {x - 1} \right)\left( {{x^2} + 1} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
11,\\
{\left( {{x^2} + 1} \right)^2} - 6\left( {{x^2} + 1} \right) + 9\\
 = {\left( {{x^2} + 1} \right)^2} - 2.\left( {{x^2} + 1} \right).3 + {3^2}\\
 = {\left[ {\left( {{x^2} + 1} \right) + 3} \right]^2}\\
 = {\left( {{x^2} + 4} \right)^2}\\
12,\\
xy + {x^2} - 5x - 5y\\
 = \left( {xy + {x^2}} \right) - \left( {5x + 5y} \right)\\
 = x\left( {y + x} \right) - 5.\left( {x + y} \right)\\
 = x.\left( {x + y} \right) - 5.\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {x - 5} \right)\\
13,\\
x + y - {x^2} - xy\\
 = \left( {x + y} \right) - \left( {{x^2} + xy} \right)\\
 = \left( {x + y} \right) - x.\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {1 - x} \right)\\
14,\\
{x^2} - xy - 7x + 7y\\
 = \left( {{x^2} - xy} \right) + \left( { - 7x + 7y} \right)\\
 = x.\left( {x - y} \right) - 7.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {x - 7} \right)\\
15,\\
a{x^2} + c{x^2} - ay + a{y^2} - cy + c{y^2}\\
 = \left( {a{x^2} + c{x^2}} \right) - \left( {ay + cy} \right) + \left( {a{y^2} + c{y^2}} \right)\\
 = {x^2}.\left( {a + c} \right) - y.\left( {a + c} \right) + {y^2}.\left( {a + c} \right)\\
 = \left( {a + c} \right).\left( {{x^2} - y + {y^2}} \right)\\
16,\\
{x^2} - 16 - 4xy + 4{y^2}\\
 = \left( {{x^2} - 4xy + 4{y^2}} \right) - 16\\
 = \left[ {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right] - {4^2}\\
 = {\left( {x - 2y} \right)^2} - {4^2}\\
 = \left[ {\left( {x - 2y} \right) - 4} \right].\left[ {\left( {x - 2y} \right) + 4} \right]\\
 = \left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
17,\\
{x^3} + 2{x^2}y + x{y^2}\\
 = x.{x^2} + x.2xy + x.{y^2}\\
 = x.\left( {{x^2} + 2xy + {y^2}} \right)\\
 = x.{\left( {x + y} \right)^2}\\
18,\\
5x + 5y - {x^2} - 2xy - {y^2}\\
 = \left( {5x + 5y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
 = 5.\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
 = \left( {x + y} \right).\left[ {5 - \left( {x + y} \right)} \right]\\
 = \left( {x + y} \right).\left( {5 - x - y} \right)\\
19,\\
{x^5} - {x^4} + {x^3} - {x^2}\\
 = \left( {{x^5} - {x^4}} \right) + \left( {{x^3} - {x^2}} \right)\\
 = {x^4}.\left( {x - 1} \right) + {x^2}\left( {x - 1} \right)\\
 = \left( {x - 1} \right).\left( {{x^4} + {x^2}} \right)\\
 = \left( {x - 1} \right).{x^2}.\left( {{x^2} + 1} \right)\\
 = {x^2}\left( {x - 1} \right)\left( {{x^2} + 1} \right)
\end{array}\)