Đáp án:

\(\begin{array}{l}
1,\\
a,\\
{x^2} - 4xy + 4{y^2}\\
b,\\
{x^2} - 16\\
c,\\
9 + 6x + {x^2}\\
d,\\
{x^3} + 6{x^2} + 12x + 8\\
2,\\
a,\\
 - {x^3} - 7\\
b,\\
8{x^3} - 2x\\
c,\\
2{x^2} - 6x - 11
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {x - 2y} \right)^2} = {x^2} - 2.x.2y + {\left( {2y} \right)^2} = {x^2} - 4xy + 4{y^2}\\
b,\\
\left( {x - 4} \right)\left( {x + 4} \right) = {x^2} - {4^2} = {x^2} - 16\\
c,\\
{\left( {3 + x} \right)^2} = {3^2} + 2.3.x + {x^2} = 9 + 6x + {x^2}\\
d,\\
{\left( {x + 2} \right)^3} = {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = {x^3} + 6{x^2} + 12x + 8\\
2,\\
a,\\
\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 2.\left( {{x^3} + 3} \right)\\
 = \left( {x - 1} \right).\left( {{x^2} + x.1 + {1^2}} \right) - \left( {2{x^3} + 6} \right)\\
 = \left( {{x^3} - {1^3}} \right) - 2{x^3} - 6\\
 = {x^3} - 1 - 2{x^3} - 6\\
 =  - {x^3} - 7\\
b,\\
\left( {2x + 1} \right).\left( {4{x^2} - 2x + 1} \right) - {\left( {x + 1} \right)^2} + {x^2}\\
 = \left( {2x + 1} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.1 + {1^2}} \right] - \left( {{x^2} + 2.x.1 + {1^2}} \right) + {x^2}\\
 = \left[ {{{\left( {2x} \right)}^3} + {1^3}} \right] - \left( {{x^2} + 2x + 1} \right) + {x^2}\\
 = 8{x^3} + 1 - {x^2} - 2x - 1 + {x^2}\\
 = 8{x^3} - 2x\\
c,\\
\left( {x - 5} \right)\left( {x + 3} \right) + {\left( {x - 2} \right)^2}\\
 = \left( {{x^2} + 3x - 5x - 15} \right) + \left( {{x^2} - 2.x.2 + {2^2}} \right)\\
 = {x^2} - 2x - 15 + {x^2} - 4x + 4\\
 = 2{x^2} - 6x - 11
\end{array}\)