Đáp án:
$\begin{array}{l}
B1)\\
a){\left( {{x^2} - 1} \right)^3} - \left( {{x^4} + {x^2} + 1} \right)\left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right)\left[ {{{\left( {{x^2} - 1} \right)}^2} - {x^4} - {x^2} - 1} \right]\\
= \left( {{x^2} - 1} \right)\left( {{x^4} - 2{x^2} + 1 - {x^4} - {x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right).\left( { - 3{x^2}} \right)\\
= 3{x^2} - 3{x^4}\\
b)\left( {{x^4} - 3{x^2} + 9} \right)\left( {{x^2} + 3} \right) - {\left( {3 - {x^2}} \right)^3}\\
= {\left( {{x^2}} \right)^3} + {3^3} - \left( {27 - 27{x^2} + 9{x^4} - {x^6}} \right)\\
= {x^6} + 27 - 27 + 27{x^2} - 9{x^4} + {x^6}\\
= 2{x^6} - 9{x^4} + 27{x^2}\\
B2)\\
a)\\
M = 1 + 8.\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right).\left( {{3^8} + 1} \right)\\
= 1 + \left( {{3^2} - 1} \right).\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right).\left( {{3^8} + 1} \right)\\
= 1 + \left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right).\left( {{3^8} + 1} \right)\\
= 1 + \left( {{3^8} - 1} \right).\left( {{3^8} + 1} \right)\\
= 1 + {3^{16}} - 1\\
= {3^{16}}\\
= {3.3^{15}}\\
N = {\left( {{3^3}} \right)^5} + {\left( {{3^5}} \right)^3}\\
= {3^{15}} + {3^{15}}\\
= {2.3^{15}} < {3.3^{15}}\\
\Leftrightarrow M > N
\end{array}$
