Đáp án:

\(\begin{array}{l}
1,\\
A =  - 34\\
2,\\
B =  - 2\\
3,\\
C = 2\\
4,\\
D = 7
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
A = \left( {2x - 5} \right)\left( {2x + 5} \right) - {\left( {2x - 3} \right)^2} - 12x\\
 = \left[ {{{\left( {2x} \right)}^2} - {5^2}} \right] - \left[ {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}} \right] - 12x\\
 = \left( {4{x^2} - 25} \right) - \left( {4{x^2} - 12x + 9} \right) - 12x\\
 = 4{x^2} - 25 - 4{x^2} + 12x - 9 - 12x\\
 = \left( {4{x^2} - 4{x^2}} \right) + \left( {12x - 12x} \right) + \left( { - 25 - 9} \right)\\
 =  - 34\\
2,\\
B = {\left( {x - 1} \right)^3} - {x^3} + 3{x^2} - 3x - 1\\
 = \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - {x^3} + 3{x^2} - 3x - 1\\
 = {x^3} - 3{x^2} + 3x - 1 - {x^3} + 3{x^2} - 3x - 1\\
 = \left( {{x^3} - {x^3}} \right) + \left( { - 3{x^2} + 3{x^3}} \right) + \left( {3x - 3x} \right) + \left( { - 1 - 1} \right)\\
 =  - 2\\
3,\\
C = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {x + 1} \right).\left( {{x^2} - x.1 + {1^2}} \right) - \left( {x - 1} \right).\left( {{x^2} + x.1 + {1^2}} \right)\\
 = \left( {{x^3} + {1^3}} \right) - \left( {{x^3} - {1^3}} \right)\\
 = {x^3} + 1 - {x^3} + 1\\
 = 2\\
4,\\
D = \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - \left( {20 + {x^3}} \right)\\
 = \left( {x + 3} \right).\left( {{x^2} - x.3 + {3^2}} \right) - 20 - {x^3}\\
 = \left( {{x^3} + {3^3}} \right) - 20 - {x^3}\\
 = {x^3} + 27 - 20 - {x^3}\\
 = \left( {{x^3} - {x^3}} \right) + \left( {27 - 20} \right)\\
 = 7
\end{array}\)