Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
A = {\left( {x + 2} \right)^2} + 4.\left( {x + 2} \right).\left( {x - 2} \right) + {\left( {x - 4} \right)^2}\\
 = {x^2} + 4x + 4 + 4.\left( {{x^2} - 4} \right) + \left( {{x^2} - 8x + 16} \right)\\
 = {x^2} + 4x + 4 + 4{x^2} - 16 + {x^2} - 8x + 16\\
 = 6{x^2} - 4x + 4\\
b.\\
B = \left( {3{x^2} - 2x + 1} \right)\left( {3{x^2} + 2x + 1} \right) - {\left( {3{x^2} + 1} \right)^2}\\
 = \left[ {\left( {3{x^2} + 1} \right) - 2x} \right].\left[ {\left( {3{x^2} + 1} \right) + 2x} \right] - {\left( {3{x^2} + 1} \right)^2}\\
 = {\left( {3{x^2} + 1} \right)^2} - {\left( {2x} \right)^2} - {\left( {3{x^2} + 1} \right)^2}\\
 =  - 4{x^2}\\
c,\\
C = {\left( {{x^2} - 5x + 2} \right)^2} + 2.\left( {5x - 2} \right).\left( {{x^2} - 5x + 2} \right) + {\left( {5x - 2} \right)^2}\\
 = {\left[ {\left( {{x^2} - 5x + 2} \right) + \left( {5x - 2} \right)} \right]^2}\\
 = {\left( {{x^2}} \right)^2} = {x^4}\\
d,\\
D = {\left( {5a + 5} \right)^2} + 10.\left( {a - 3} \right).\left( {1 + a} \right) + {a^2} - 6a + 9\\
 = {\left[ {5.\left( {a + 1} \right)} \right]^2} + 2.5.\left( {a + 1} \right).\left( {a - 3} \right) + {\left( {a - 3} \right)^2}\\
 = {\left[ {5\left( {a + 1} \right) + \left( {a - 3} \right)} \right]^2}\\
 = {\left( {6a + 2} \right)^2}\\
e,\\
E = \frac{{{{\left( {x - 1} \right)}^2}}}{4} + {x^2} - 1 + {\left( {x + 1} \right)^2}\\
 = {\left( {\frac{{x - 1}}{2}} \right)^2} + \left( {x - 1} \right).\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\\
 = {\left( {\frac{{x - 1}}{2}} \right)^2} + 2.\frac{{x - 1}}{2}.\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\\
 = {\left( {\frac{{x - 1}}{2} + x + 1} \right)^2}\\
 = {\left( {\frac{3}{2}x + \frac{1}{2}} \right)^2}\\
2,\\
F = {x^2} + 4{y^2} - 2x + 10 + 4xy - 4y\\
 = \left( {{x^2} + 4xy + 4{y^2}} \right) - \left( {2x + 4y} \right) + 10\\
 = {\left( {x + 2y} \right)^2} - 2.\left( {x + 2y} \right) + 10\\
 = {5^2} - 2.5 + 10\\
 = 25
\end{array}\)