Đáp án:
$\begin{array}{l}
B2)a){\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2} - 2\left( {1 + 6x} \right)\left( {6x - 1} \right)\\
= {\left( {6x + 1 - 6x + 1} \right)^2}\\
= {2^2}\\
= 4\\
b)3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right)\\
= {2^{32}} - 1\\
B3)a){x^3} - 3{x^2} - 4x + 12\\
= {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
b){x^4} - 5{x^2} + 4\\
= {\left( {{x^2}} \right)^2} - 4{x^2} - {x^2} + 4\\
= {x^2}\left( {{x^2} - 4} \right) - \left( {{x^2} - 4} \right)\\
= \left( {{x^2} - 4} \right)\left( {{x^2} - 1} \right)\\
= \left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
c){\left( {x + y + z} \right)^3} - {x^3} - {y^3} - {z^3}\\
= {\left( {x + y} \right)^3} + 3{\left( {x + y} \right)^2}z + 3\left( {x + y} \right).{z^2} + {z^3}\\
- \left( {{x^3} + {y^3}} \right) - {z^3}\\
= {\left( {x + y} \right)^3} + 3{\left( {x + y} \right)^2}z + 3\left( {x + y} \right).{z^2}\\
- \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\
= \left( {x + y} \right)\left[ \begin{array}{l}
{\left( {x + y} \right)^2} + 3\left( {x + y} \right).z + 3{z^2}\\
- {x^2} + xy - {y^2}
\end{array} \right]\\
= \left( {x + y} \right)\left( \begin{array}{l}
{x^2} + 2xy + {y^2} + 3xz + 3yz + 3{z^2}\\
- {x^2} + xy - {y^2}
\end{array} \right)\\
= \left( {x + y} \right)\left( {3xy + 3xz + 3yz + 3{z^2}} \right)\\
= \left( {x + y} \right).3.\left( {x + z} \right)\left( {y + z} \right)\\
= 3\left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right)
\end{array}$
