Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
81{a^2}{b^2} - {\left( {x + a} \right)^2}\\
 = {\left( {9ab} \right)^2} - {\left( {x + a} \right)^2}\\
 = \left( {9ab - x - a} \right)\left( {9ab + x + a} \right)\\
b,\\
\left( {{x^2} + 2x + 1} \right) - 9{a^2}{b^4}\\
 = {\left( {x + 1} \right)^2} - {\left( {3a{b^2}} \right)^2}\\
 = \left( {x + 1 - 3a{b^2}} \right)\left( {x + 1 + 3a{b^2}} \right)\\
c,\\
\left( {{x^2} - 4x + 4} \right) - 16{y^2}\\
 = {\left( {x - 2} \right)^2} - {\left( {4y} \right)^2}\\
 = \left( {x - 2 - 4y} \right)\left( {x - 2 + 4y} \right)\\
d,\\
25{x^4} - 10{x^2}y + {y^2}\\
 = {\left( {5{x^2}} \right)^2} - 2.5{x^2}.y + {y^2}\\
 = {\left( {5{x^2} - y} \right)^2}\\
2,\\
a,\\
{x^2} - {y^2} + 2x + 1\\
 = \left( {{x^2} + 2x + 1} \right) - {y^2}\\
 = {\left( {x + 1} \right)^2} - {y^2}\\
 = \left( {x + 1 - y} \right)\left( {x + 1 + y} \right)\\
b,\\
{x^2} - 4{x^2}{y^2} + {y^2} + 2xy\\
 = \left( {{x^2} + 2xy + {y^2}} \right) - 4{x^2}{y^2}\\
 = {\left( {x + y} \right)^2} - {\left( {2xy} \right)^2}\\
 = \left( {x + y - 2xy} \right)\left( {x + y + 2xy} \right)\\
c,\\
4{a^2}{c^2} - {\left( {{b^2} - {c^2} - {a^2}} \right)^2}\\
 = {\left( {2ac} \right)^2} - {\left( {{b^2} - {c^2} - {a^2}} \right)^2}\\
 = \left[ {2ac + \left( {{b^2} - {c^2} - {a^2}} \right)} \right].\left[ {2ac - \left( {{b^2} - {c^2} - {a^2}} \right)} \right]\\
 = \left[ {{b^2} - \left( {{a^2} - 2ac + {c^2}} \right)} \right].\left[ {\left( {{c^2} + 2ac + {a^2}} \right) - {b^2}} \right]\\
 = \left[ {{b^2} - {{\left( {a - c} \right)}^2}} \right].\left[ {{{\left( {a + c} \right)}^2} - {b^2}} \right]\\
 = \left( {b - a + c} \right)\left( {b + a - c} \right).\left( {a + c - b} \right)\left( {a + c + b} \right)\\
4,\\
{\left( {a + b + c} \right)^2} + {\left( {a + b - c} \right)^2} - 4{c^2}\\
 = {\left( {a + b + c} \right)^2} + {\left( {a + b - c} \right)^2} - {\left( {2c} \right)^2}\\
 = {\left( {a + b + c} \right)^2} + \left[ {\left( {a + b - c} \right) - 2c} \right].\left[ {\left( {a + b - c} \right) + 2c} \right]\\
 = {\left( {a + b + c} \right)^2} + \left( {a + b - 3c} \right)\left( {a + b + c} \right)\\
 = \left( {a + b + c} \right).\left[ {\left( {a + b + c} \right) + \left( {a + b - 3c} \right)} \right]\\
 = \left( {a + b + c} \right).\left( {2a + 2b - 2c} \right)\\
 = 2.\left( {a + b + c} \right)\left( {a + b - c} \right)
\end{array}\)