Đáp án:
$\begin{array}{l}
1)a){\left( {5{x^2} - 1} \right)^2}\\
= {\left( {5{x^2}} \right)^2} - 2.5{x^2} + 1\\
= 25{x^4} - 10{x^2} + 1\\
b){\left( {3 - 2{y^2}} \right)^3}\\
= {3^3} - {3.3^2}.2{y^2} + 3.3.4{y^4} - 8{y^6}\\
= 27 - 54{y^2} + 36{y^4} - 8{y^6}\\
c){\left( {\dfrac{1}{4}x + 3} \right)^3}\\
= {\left( {\dfrac{1}{4}x} \right)^3} + 3.{\left( {\dfrac{1}{4}x} \right)^2}.3 + 3.\dfrac{1}{4}x + 9 + 27\\
= \dfrac{1}{{64}}{x^3} + \dfrac{9}{{16}}{x^2} + \dfrac{{27}}{4}x + 27\\
B2)\\
A = 2008.2010\\
= \left( {2009 - 1} \right)\left( {2009 + 1} \right)\\
= {2009^2} - 1 < {2009^2}\\
\Leftrightarrow A < B\\
B3)\\
a)A = {x^2} + 4x + 20\\
= {x^2} + 4x + 4 + 16\\
= {\left( {x + 2} \right)^2} + 16 \ge 16\\
\Leftrightarrow GTNN:A = 16\,khi:x = - 2\\
b)B = 4{x^2} - 20x - 13\\
= {\left( {2x} \right)^2} - 2.2x.5 + 25 - 38\\
= {\left( {2x - 5} \right)^2} - 38 \ge - 38\\
\Leftrightarrow GTNN:B = - 38\,khi:x = \dfrac{5}{2}
\end{array}$
