Đáp án:
\(11)16{x^2} - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
5){\left( {4{x^2}} \right)^2} - 1 = \left( {4{x^2} - 1} \right)\left( {4{x^2} + 1} \right)\\
6){\left( {2x} \right)^3} - {\left( {3y} \right)^3} = \left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)\\
7){\left( {7x} \right)^2} - {\left( {\dfrac{1}{2}y} \right)^2} = \left( {7x - \dfrac{1}{2}y} \right)\left( {7x + \dfrac{1}{2}y} \right)\\
8){\left( {\dfrac{1}{2}x} \right)^3} - {\left( {5y} \right)^3}\\
= \left( {\dfrac{1}{2}x - 5y} \right)\left( {\dfrac{1}{4}{x^2} + \dfrac{5}{2}xy + 25{y^2}} \right)\\
9){\left( {2x - 5y} \right)^2} = 4{x^2} - 20xy + 25{y^2}\\
10){\left( {{x^2} + 3y} \right)^2} = {x^4} + 6{x^2}y + 9{y^2}\\
11)\left( {4x - 1} \right)\left( {4x + 1} \right)\\
= 16{x^2} - 1
\end{array}\)
