Đáp án:
$\begin{array}{l}
a){\left( {3x + 5} \right)^2} = 9{x^2} + 30x + 25\\
b){\left( {6{x^2} + \frac{1}{3}} \right)^2} = 36{x^4} + 4{x^2} + \frac{1}{9}\\
c){\left( {5x - 4y} \right)^2} = 25{x^2} - 40xy + 16{y^2}\\
d){\left( {2{x^2}y - 3{y^3}x} \right)^2} = 4{x^4}{y^2} - 12{x^3}{y^4} + 9{y^6}{x^2}\\
e)\left( {5x - 3} \right)\left( {5x + 3} \right) = 25{x^2} - 9\\
f)\left( {6x + 5y} \right)\left( {6x - 5y} \right) = 36{x^2} - 25{y^2}\\
g)\left( { - 4xy - 5} \right)\left( {5 - 4xy} \right)\\
= \left( {4xy + 5} \right)\left( {4xy - 5} \right)\\
= 16{x^2}{y^2} - 25\\
h)\left( {{a^2}b + a{b^2}} \right)\left( {a{b^2} - {a^2}b} \right)\\
= ab\left( {a + b} \right).ab\left( {b - a} \right)\\
= {a^2}{b^2}\left( {{b^2} - {a^2}} \right)\\
= {a^2}{b^4} - {a^4}{b^2}\\
i){\left( {3x - 4} \right)^2} + 2\left( {3x - 4} \right)\left( {4 - x} \right) + {\left( {4 - x} \right)^2}\\
= {\left( {3x - 4 + 4 - x} \right)^2}\\
= {\left( {2x} \right)^2}\\
= 4{x^2}\\
j){\left( {3a - 1} \right)^2} + 2.\left( {9{a^2} - 1} \right) + {\left( {3a + 1} \right)^2}\\
= {\left( {3a - 1} \right)^2} + 2.\left( {3a + 1} \right)\left( {3a - 1} \right) + {\left( {3a + 1} \right)^2}\\
= {\left( {3a - 1 + 3a + 1} \right)^2}\\
= {\left( {6a} \right)^2}\\
= 36{a^2}\\
k)\left( {{a^2} + ab + {b^2}} \right)\left( {{a^2} - ab + {b^2}} \right) - \left( {{a^4} + {b^4}} \right)\\
= {\left( {{a^2} + {b^2}} \right)^2} - {\left( {ab} \right)^2} - {a^4} - {b^4}\\
= {a^4} + 2{a^2}{b^2} + {b^4} - {a^2}{b^2} - {a^4} - {b^4}\\
= {a^2}{b^2}
\end{array}$
