Đáp án:
$\begin{array}{l}
B4)\\
a){x^4} + 64\\
= {x^4} + 16{x^2} + 64 - 16{x^2}\\
= {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2}\\
= \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right)\\
b){x^8} + {x^4} + 1\\
= {x^8} + 2{x^4} + 1 - {x^4}\\
= {\left( {{x^4} + 1} \right)^2} - {\left( {{x^2}} \right)^2}\\
= \left( {{x^4} + 1 - {x^2}} \right)\left( {{x^4} + 1 + {x^2}} \right)\\
B5)\\
a){\left( {{x^2} + x} \right)^2} + 4{x^2} + 4x - 12\\
= {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) - 12\\
Đặt:\left( {{x^2} + x} \right) = a\\
\Leftrightarrow {a^2} + 4a - 12\\
= {a^2} + 6a - 2a - 12\\
= \left( {a + 6} \right)\left( {a - 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {{x^2} + x - 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {x - 1} \right)\left( {x + 2} \right)\\
b)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right) - 24\\
= \left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)\left( {x + 3} \right) - 24\\
= \left( {{x^2} + 5x + 4} \right)\left( {{x^2} + 5x + 6} \right) - 24\\
Đặt:{x^2} + 5x + 4 = a\\
\Leftrightarrow a\left( {a + 2} \right) - 24\\
= {a^2} + 2a - 24\\
= {a^2} + 6a - 4a - 24\\
= \left( {a + 6} \right)\left( {a - 4} \right)\\
= \left( {{x^2} + 5x + 4 + 6} \right)\left( {{x^2} + 5x + 4 - 4} \right)\\
= \left( {{x^2} + 5x + 10} \right).x\left( {x + 5} \right)\\
c)\left( {{x^2} + x + 1} \right)\left( {{x^2} + x + 2} \right) - 12\\
Đặt:{x^2} + x + 1 = a\\
\Leftrightarrow a\left( {a + 1} \right) - 12\\
= {a^2} + a - 12\\
= {a^2} + 4a - 3a - 12\\
= \left( {a + 4} \right)\left( {a - 3} \right)\\
= \left( {{x^2} + x + 1 + 4} \right)\left( {{x^2} + x + 1 - 3} \right)\\
= \left( {{x^2} + x + 5} \right)\left( {{x^2} + x - 2} \right)\\
= \left( {{x^2} + x + 5} \right)\left( {x - 1} \right)\left( {x + 2} \right)
\end{array}$
