Đáp án:
B2:
d. \(5\left( {x - y} \right)\left( {x + y} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2\left( {8{x^3} - 27{y^3}} \right)\\
= 2\left( {2x - 3y} \right)\left( {4{x^2} + 2x.3y + 9{y^2}} \right)\\
= 2\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)\\
b.y\left( {16{x^3} + \dfrac{1}{4}{z^5}} \right)\\
c.5\left( {{x^2} - {y^2}} \right)\\
= 5\left( {x - y} \right)\left( {x + y} \right)\\
d.2\left( {{x^2} - 16} \right)\\
= 2\left( {x - 4} \right)\left( {x + 4} \right)\\
B2:\\
a.{x^2}\left( {{x^2} - 4x + 4} \right)\\
= {x^2}{\left( {x - 2} \right)^2}\\
b.y\left( {2xy - {x^2} - {y^2}} \right)\\
= - y\left( {{x^2} - 2xy + {y^2}} \right)\\
= - y{\left( {x - y} \right)^2}\\
c.2\left( {{x^2} + 2xy + {y^2}} \right)\\
= 2{\left( {x + y} \right)^2}\\
d.5\left( {{x^4} - 2{x^2}{y^2} + {y^4}} \right)\\
= 5\left( {{x^2} - {y^2}} \right)\\
= 5\left( {x - y} \right)\left( {x + y} \right)
\end{array}\)
