Đáp án:

\(\begin{array}{l}
1,\\
a,\\
8ab\left( {3a - 2b + 1} \right)\\
b,\\
\left( {x - 10} \right).\left( {x - 1} \right)\\
c,\\
\left( {4 - 4x + y} \right).\left( {4 + 4x - y} \right)\\
d,\\
\left( {x - 1} \right)\left( {x - 8} \right).\left( {{x^2} - 9x + 26} \right)\\
2,\\
{C_{\max }} = 2021 \Leftrightarrow x = \dfrac{1}{2}\\
3,\\
a,\\
\left[ \begin{array}{l}
x = 0\\
x = 3\\
x =  - 3
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 4
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
24{a^2}b - 16a{b^2} + 8ab\\
 = 8ab.3a - 8ab.2b + 8ab\\
 = 8ab\left( {3a - 2b + 1} \right)\\
b,\\
\left( {x + 2} \right)\left( {x - 10} \right) + 3.\left( {10 - x} \right)\\
 = \left( {x + 2} \right)\left( {x - 10} \right) + 3.\left[ { - \left( {x - 10} \right)} \right]\\
 = \left( {x + 2} \right)\left( {x - 10} \right) - 3\left( {x - 10} \right)\\
 = \left( {x - 10} \right).\left[ {\left( {x + 2} \right) - 3} \right]\\
 = \left( {x - 10} \right).\left( {x - 1} \right)\\
c,\\
16 - {y^2} - 16{x^2} + 8xy\\
 = 16 - \left( {{y^2} + 16{x^2} - 8xy} \right)\\
 = 16 - \left( {16{x^2} - 8xy + {y^2}} \right)\\
 = 16 - \left[ {{{\left( {4x} \right)}^2} - 2.4x.y + {y^2}} \right]\\
 = {4^2} - {\left( {4x - y} \right)^2}\\
 = \left[ {4 - \left( {4x - y} \right)} \right].\left[ {4 + \left( {4x - y} \right)} \right]\\
 = \left( {4 - 4x + y} \right).\left( {4 + 4x - y} \right)\\
d,\\
\left( {x - 7} \right)\left( {x - 5} \right)\left( {x - 4} \right)\left( {x - 2} \right) - 72\\
 = \left[ {\left( {x - 7} \right)\left( {x - 2} \right)} \right].\left[ {\left( {x - 5} \right)\left( {x - 4} \right)} \right] - 72\\
 = \left( {{x^2} - 2x - 7x + 14} \right).\left( {{x^2} - 4x - 5x + 20} \right) - 72\\
 = \left( {{x^2} - 9x + 14} \right).\left( {{x^2} - 9x + 20} \right) - 72\\
 = \left[ {\left( {{x^2} - 9x + 17} \right) - 3} \right].\left[ {\left( {{x^2} - 9x + 17} \right) + 3} \right] - 72\\
 = {\left( {{x^2} - 9x + 17} \right)^2} - {3^2} - 72\\
 = {\left( {{x^2} - 9x + 17} \right)^2} - 81\\
 = {\left( {{x^2} - 9x + 17} \right)^2} - {9^2}\\
 = \left[ {\left( {{x^2} - 9x + 17} \right) - 9} \right].\left[ {\left( {{x^2} - 9x + 17} \right) + 9} \right]\\
 = \left( {{x^2} - 9x + 8} \right).\left( {{x^2} - 9x + 26} \right)\\
 = \left[ {\left( {{x^2} - x} \right) + \left( { - 8x + 8} \right)} \right].\left( {{x^2} - 9x + 26} \right)\\
 = \left[ {x\left( {x - 1} \right) - 8.\left( {x - 1} \right)} \right].\left( {{x^2} - 9x + 26} \right)\\
 = \left( {x - 1} \right)\left( {x - 8} \right).\left( {{x^2} - 9x + 26} \right)\\
2,\\
C = 4x - 4{x^2} + 2020\\
 = 2021 + \left( { - 4{x^2} + 4x - 1} \right)\\
 = 2021 - \left( {4{x^2} - 4x + 1} \right)\\
 = 2021 - \left[ {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right]\\
 = 2021 - {\left( {2x - 1} \right)^2}\\
{\left( {2x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow 2021 - {\left( {2x - 1} \right)^2} \le 2021,\,\,\,\forall x\\
 \Rightarrow C \le 2021,\,\,\,\forall x\\
 \Rightarrow {C_{\max }} = 2021 \Leftrightarrow {\left( {2x - 1} \right)^2} = 0 \Leftrightarrow 2x - 1 = 0 \Leftrightarrow x = \dfrac{1}{2}\\
3,\\
a,\\
3{x^3} - 27x = 0\\
 \Leftrightarrow 3x.{x^2} - 3x.9 = 0\\
 \Leftrightarrow 3x.\left( {{x^2} - 9} \right) = 0\\
 \Leftrightarrow 3x.\left( {{x^2} - {3^2}} \right) = 0\\
 \Leftrightarrow 3x.\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 3 = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 3\\
x =  - 3
\end{array} \right.\\
b,\\
{\left( {2x + 3} \right)^2} - 25 = 0\\
 \Leftrightarrow {\left( {2x + 3} \right)^2} - {5^2} = 0\\
 \Leftrightarrow \left[ {\left( {2x + 3} \right) - 5} \right].\left[ {\left( {2x + 3} \right) + 5} \right] = 0\\
 \Leftrightarrow \left( {2x - 2} \right).\left( {2x + 8} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 2 = 0\\
2x + 8 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 4
\end{array} \right.
\end{array}\)