Đáp án:
$\begin{array}{l}
7)a){\left( {x + y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
b){\left( {3x - 5} \right)^2} = 9{x^2} - 30x + 25\\
c){\left( {2x + 3y} \right)^3} = 8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}\\
d){\left( {4x - 1} \right)^3} + {\left( {4x + 1} \right)^3}\\
= \left( {4x - 1 + 4x + 1} \right)\left( {{{\left( {4x - 1} \right)}^2} - \left( {4x - 1} \right)\left( {4x + 1} \right) + {{\left( {4x + 1} \right)}^2}} \right)\\
= 8x\left( {16{x^2} - 8x + 1 - 16{x^2} + 1 + 16{x^2} + 8x + 1} \right)\\
= 8x\left( {16{x^2} + 3} \right)\\
e){\left( {3y + 4} \right)^3} + {\left( {3y - 4} \right)^3}\\
= \left( {3y + 4 + 3y - 4} \right)\left( {{{\left( {3y + 4} \right)}^2} - \left( {3y + 4} \right)\left( {3y - 4} \right) + {{\left( {3y - 4} \right)}^2}} \right)\\
= 6y\left( {9{y^2} + 24y + 16 - 9{y^2} + 16 + 9{y^2} - 24y + 16} \right)\\
= 6y\left( {9{y^2} + 48} \right)\\
8)a){\left( {x + y} \right)^2} - {\left( {x - y} \right)^2}\\
= \left( {x + y + x - y} \right)\left( {x + y - x + y} \right)\\
= 2x.2y\\
= 4xy\\
b){\left( {x + 2y} \right)^2} - {\left( {x - 2y} \right)^2}\\
= \left( {x + 2y + x - 2y} \right)\left( {x + 2y - x + 2y} \right)\\
= 2x.4y\\
= 8xy\\
c){\left( {x + y} \right)^2} + 2\left( {x + y} \right)\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
= {\left( {x + y + x - y} \right)^2}\\
= {\left( {2x} \right)^2}\\
= 4{x^2}
\end{array}$
