Đáp án:

\(\begin{array}{l}
7,\\
 - \left( {x + 11y} \right)\left( {11x + y} \right)\\
8,\\
\left( {x - 1} \right).\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)\\
9,\\
{\left( {2m + 3n} \right)^3}\\
10,\\
{\left( {x - 1} \right)^3}\\
11,\\
\left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\\
12,\\
\left( {\dfrac{1}{2}a - b} \right)\left( {\dfrac{1}{2}a + b} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
7,\\
25{\left( {x - y} \right)^2} - 36.{\left( {x + y} \right)^2}\\
 = {5^2}{\left( {x - y} \right)^2} - {6^2}.{\left( {x + y} \right)^2}\\
 = {\left[ {5.\left( {x - y} \right)} \right]^2} - {\left[ {6.\left( {x + y} \right)} \right]^2}\\
 = {\left( {5x - 5y} \right)^2} - {\left( {6x + 6y} \right)^2}\\
 = \left[ {\left( {5x - 5y} \right) - \left( {6x + 6y} \right)} \right].\left[ {\left( {5x - 5y} \right) + \left( {6x + 6y} \right)} \right]\\
 = \left( {5x - 5y - 6x - 6y} \right).\left( {5x - 5y + 6x + 6y} \right)\\
 = \left( { - x - 11y} \right).\left( {11x + y} \right)\\
 =  - \left( {x + 11y} \right)\left( {11x + y} \right)\\
8,\\
{x^6} - 1\\
 = {\left( {{x^3}} \right)^2} - {1^2}\\
 = \left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right)\\
 = \left( {{x^3} - {1^3}} \right).\left( {{x^3} + {1^3}} \right)\\
 = \left( {x - 1} \right).\left( {{x^2} + x.1 + {1^2}} \right).\left( {x + 1} \right)\left( {{x^2} - x.1 + {1^2}} \right)\\
 = \left( {x - 1} \right).\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)\\
9,\\
8{m^3} + 36{m^2}n + 54m{n^2} + 27{n^3}\\
 = 8{m^3} + 3.4{m^2}.3n + 3.2m.9{n^2} + 27{n^3}\\
 = {\left( {2m} \right)^3} + 3.{\left( {2m} \right)^2}.3n + 3.2m.{\left( {3n} \right)^2} + {\left( {3n} \right)^3}\\
 = {\left( {2m + 3n} \right)^3}\\
10,\\
{x^3} - 3{x^2} + 3x - 1\\
 = {x^3} - 3.{x^2}.1 + 3.x{.1^2} - {1^3}\\
 = {\left( {x - 1} \right)^3}\\
11,\\
{a^4} - {b^4}\\
 = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2}\\
 = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)\\
 = \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\\
12,\\
\dfrac{1}{4}{a^2} - {b^2} = {\left( {\dfrac{1}{2}a} \right)^2} - {b^2} = \left( {\dfrac{1}{2}a - b} \right)\left( {\dfrac{1}{2}a + b} \right)
\end{array}\)