Đáp án:
$\begin{array}{l}
B2)\\
a){\left( {x + 2{y^2}} \right)^2} = {x^2} + 4x{y^2} + {y^4}\\
b){\left( {a - \dfrac{5}{2}b} \right)^2} = {a^2} - 5ab + \dfrac{{25}}{4}{b^2}\\
c){\left( {m + \dfrac{1}{2}} \right)^2} = {m^2} + m + \dfrac{1}{4}\\
d){x^2} - 16{y^4} = \left( {x + 4{y^2}} \right)\left( {x - 4{y^2}} \right)\\
e)25{a^2} - \dfrac{1}{4}{b^2} = \left( {5a + \dfrac{1}{2}b} \right)\left( {5a - \dfrac{1}{2}b} \right)\\
B3)\\
a){\left( {x + 12} \right)^2} - 9{x^2} = 0\\
\Leftrightarrow \left( {x + 12 - 3x} \right)\left( {x + 12 + 3x} \right) = 0\\
\Leftrightarrow \left( {12 - 2x} \right).\left( {4x + 12} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
12 - 2x = 0\\
4x + 12 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 3
\end{array} \right.\\
Vậy\,x = 6;x = - 3\\
b){\left( {5x + 1} \right)^2} - \left( {5x + 3} \right)\left( {5x - 3} \right) = 30\\
\Leftrightarrow 25{x^2} + 10x + 1 - 25{x^2} + 9 = 30\\
\Leftrightarrow 10x = 20\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
c){\left( {3x + 2} \right)^2} - 2\left( {3x + 2} \right)\left( {2x - 1} \right) + {\left( {2x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {3x + 2 - 2x + 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x + 3} \right)^2} = 0\\
\Leftrightarrow x = - 3\\
Vậy\,x = - 3\\
d)\left( {x + 1} \right)\left( {x + 5} \right) - \left( {x + 3} \right)\left( {x + 4} \right) = 9\\
\Leftrightarrow {x^2} + 6x + 5 - {x^2} - 7x - 12 = 9\\
\Leftrightarrow - x = 16\\
\Leftrightarrow x = - 16\\
Vậy\,x = - 16\\
B4)\\
B = \left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
= \left( {\dfrac{1}{2}} \right).\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
= \dfrac{1}{2}.\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
= \dfrac{1}{2}.\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
= \dfrac{1}{2}.\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
= \dfrac{1}{2}.\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\\
= \dfrac{1}{2}.\left( {{3^{32}} - 1} \right) < {3^{32}} - 1\\
\Leftrightarrow A > B
\end{array}$
