Đáp án:
$\begin{array}{l}
1)A = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x + 5} \right)\left( {x + 7} \right) - 24\\
= \left( {x + 1} \right)\left( {x + 7} \right)\left( {x + 3} \right)\left( {x + 5} \right) - 24\\
= \left( {{x^2} + 8x + 7} \right)\left( {{x^2} + 8x + 15} \right) - 24\\
\text{Đặt}:{x^2} + 8x + 7 = a\\
\Rightarrow A = a\left( {a + 8} \right) - 24\\
= {a^2} + 8a - 24\\
= {a^2} + 8a + 16 - 40\\
= {\left( {a + 4} \right)^2} - {\left( {2\sqrt {10} } \right)^2}\\
= \left( {a + 4 + 2\sqrt {10} } \right)\left( {a + 4 - 2\sqrt {10} } \right)\\
= \left( {{x^2} + 8x + 7 + 4 + 2\sqrt {10} } \right)\\
.\left( {{x^2} + 8x + 7 + 4 - 2\sqrt {10} } \right)\\
= \left( {{x^2} + 8x + 11 + 2\sqrt {10} } \right).\left( {{x^2} + 8x + 11 - 2\sqrt {10} } \right)\\
2)B = {\left( {{x^2} + 4x + 8} \right)^2} + 3x\left( {{x^2} + 4x + 8} \right) + 2{x^2}\\
\text{Đặt}:\left\{ \begin{array}{l}
{x^2} + 4x + 8 = a\\
x = b
\end{array} \right.\\
\Rightarrow B = {a^2} + 3ab + 2{b^2}\\
= \left( {a + b} \right)\left( {a + 2b} \right)\\
= \left( {{x^2} + 4x + 8 + x} \right).\left( {{x^2} + 4x + 8 + 2x} \right)\\
= \left( {{x^2} + 5x + 8} \right).\left( {{x^2} + 6x + 8} \right)\\
3)C = \left( {{x^2} + 8x + 7} \right)\left( {{x^2} + 8x + 15} \right) + 15\\
\text{Đặt}:{x^2} + 8x + 7 = a\\
\Rightarrow C = a\left( {a + 8} \right) + 15\\
= {a^2} + 8a + 15\\
= \left( {a + 3} \right)\left( {a + 5} \right)\\
= \left( {{x^2} + 8x + 7 + 3} \right).\left( {{x^2} + 8x + 7 + 5} \right)\\
= \left( {{x^2} + 8x + 10} \right).\left( {{x^2} + 8x + 12} \right)\\
4)D = \left( {{x^2} + x + 1} \right)\left( {{x^2} + x + 2} \right) - 12\\
{x^2} + x + 1 = a\\
\Rightarrow D = a\left( {a + 1} \right) - 12\\
= {a^2} + a - 12\\
= \left( {a + 4} \right)\left( {a - 3} \right)\\
= \left( {{x^2} + x + 1 + 4} \right).\left( {{x^2} + x + 1 - 3} \right)\\
= \left( {{x^2} + x + 5} \right).\left( {{x^2} + x - 2} \right)
\end{array}$
