Đáp án:
\(\begin{array}{l}
VD1:\\
a.{x^3} + {\left( {2y} \right)^3} = \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
VD1:\\
a.{x^3} + {\left( {2y} \right)^3} = \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\
b.{\left( {3xy} \right)^3} + {\left( {\dfrac{2}{3}{x^2}{y^2}} \right)^3}\\
= \left( {3xy + \dfrac{2}{3}{x^2}{y^2}} \right)\left( {9{x^2}{y^2} - 3xy.\dfrac{2}{3}{x^2}{y^2} + \dfrac{4}{9}{x^4}{y^4}} \right)\\
= \left( {3xy + \dfrac{2}{3}{x^2}{y^2}} \right)\left( {9{x^2}{y^2} - 2{x^3}{y^3} + \dfrac{4}{9}{x^4}{y^4}} \right)\\
c.\dfrac{1}{{{4^3}}}.{x^6}.{y^3} + {5^3} = {\left( {\dfrac{1}{4}.{x^2}y} \right)^3} + {5^3}\\
= \left( {\dfrac{1}{4}.{x^2}y + 5} \right)\left( {\dfrac{1}{{16}}.{x^4}{y^2} - \dfrac{1}{4}.{x^2}y.5 + 25} \right)\\
= \left( {\dfrac{1}{4}.{x^2}y + 5} \right)\left( {\dfrac{1}{{16}}.{x^4}{y^2} - \dfrac{5}{4}.{x^2}y + 25} \right)\\
d.{\left( {x{y^2}} \right)^3} + {\left( {2{z^2}} \right)^3} = \left( {x{y^2} + 2{z^2}} \right)\left( {{x^2}{y^4} - 2x{y^2}{z^2} + 4{z^4}} \right)\\
e.\dfrac{1}{{{3^3}}} + \left( {{4^3}.{x^6}} \right) = \dfrac{1}{{{3^3}}} + {\left( {4{x^2}} \right)^3}\\
= \left( {\dfrac{1}{3} + 4{x^2}} \right)\left( {\dfrac{1}{9} - 4{x^2}.\dfrac{1}{3} + 16{x^4}} \right)\\
= \left( {\dfrac{1}{3} + 4{x^2}} \right)\left( {\dfrac{1}{9} - \dfrac{4}{3}{x^2} + 16{x^4}} \right)\\
B5:\\
M = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) + 2{x^2} + 4xy + 2{y^2}\\
Thay:x + y = 7\\
\to M = 7\left( {{x^2} - xy + {y^2}} \right) + 2{x^2} + 4xy + 2{y^2}\\
= 7{x^2} - 7xy + 7{y^2} + 2{x^2} + 4xy + 2{y^2}\\
= 9{x^2} - 3xy + 9{y^2}\\
= 9\left( {{x^2} + {y^2}} \right) - 3xy\\
= 9\left( {{x^2} + 2xy + {y^2} - 2xy} \right) - 3xy\\
= 9{\left( {x + y} \right)^2} - 18xy - 3xy\\
= {9.7^2} - 21xy\\
= 441 - 21xy
\end{array}\)
