Đáp án:
$\begin{array}{l}
1)a)A = \left( {{x^2} + 3} \right).\left( {{x^4} - 3{x^2} + 9} \right) - {\left( {{x^2} + 3} \right)^3}\\
= {\left( {{x^2}} \right)^3} + {3^3} - {\left( {{x^2}} \right)^3} - 3.{x^4}.3 - 3.{x^2}{.3^2} - {3^3}\\
= - 9{x^4} - 27{x^2}\\
b)B = {\left( {x - 1} \right)^3} - {\left( {x + 1} \right)^3} + 6\left( {x + 1} \right)\left( {x - 1} \right)\\
= \left( {x - 1 - x - 1} \right).\left[ {{{\left( {x - 1} \right)}^2} + \left( {x - 1} \right)\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right]\\
+ 6\left( {{x^2} - 1} \right)\\
= - 2.\left( {{x^2} - 2x + 1 + {x^2} - 1 + {x^2} + 2x + 1} \right)\\
+ 6{x^2} - 6\\
= - 2.\left( {3{x^2} + 1} \right) + 6{x^2} - 6\\
= - 6{x^2} - 2 + 6{x^2} - 6\\
= - 8\\
B2)\\
a)81{a^2} - 6bc - 9{b^2} - {c^2}\\
= 81{a^2} - \left( {9{b^2} + 6bc + {c^2}} \right)\\
= 81{a^2} - {\left( {3b + c} \right)^2}\\
= \left( {9a - 3b - c} \right)\left( {9a + 3b + c} \right)\\
b){a^3} - 6{a^2} + 12a - 8\\
= {\left( {a - 2} \right)^3}\\
B3)\\
{\left( {x - 2} \right)^2} - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
+ \left( {2x - 3} \right)\left( {3x - 2} \right) = 0\\
\Leftrightarrow {x^2} - 4x + 4 - {x^3} - {2^3}\\
+ 6{x^2} - 4x - 9x + 6 = 0\\
\Leftrightarrow - {x^3} + 7{x^2} - 17x + 2 = 0\\
\Leftrightarrow x = 0,123\\
Vậy\,x = 0,123\\
B4)\\
A\left( x \right) = {x^3} - 2{x^2} + x - m + 2\\
= {x^3} + 3{x^2} - 5{x^2} - 15x + 16x + 48 - m - 46\\
= \left( {x + 3} \right)\left( {{x^2} - 5x + 16} \right) - m - 46\\
\Leftrightarrow A\left( x \right):B\left( x \right) = {x^2} - 5x + 16\,du\, - m - 46\\
\Leftrightarrow - m - 46 = 5\\
\Leftrightarrow m = - 51\\
Vậy\,m = - 51\\
B5)\\
{a^3} + {b^3} + 3ab\\
= \left( {a + b} \right).\left( {{a^2} - ab + {b^2}} \right) + 3ab\\
= 1.\left( {{a^2} - ab + {b^2}} \right) + 3ab\\
= {a^2} + 2ab + {b^2}\\
= {\left( {a + b} \right)^2}\\
= 1\left( {do:a + b = 1} \right)
\end{array}$
