Question

A=3x²-2xy+2y²;B=x²+2xy

Tính A.B

Answer 1

\(\left\{{}\begin{matrix}A=3x^2-2xy+2y^2\\B=x^2+2xy\end{matrix}\right.\)

\(\Rightarrow A.B=\left(3x^2-2xy+2y^2\right)\left(x^2+2xy\right)\)

\(\Rightarrow A.B=3x^4-2x^3y+2x^2y^2+6x^3y-4x^2y^2+4y^3x\)

\(\Rightarrow A.B=3x^4+\left(6x^3y-2x^3y\right)-\left(4x^2y^2-2x^2y^2\right)+4y^3x\)

\(\Rightarrow A.B=3x^4+4x^3y-2x^2y^2+4y^3x\)

:D

Answer 2

A.B=(3x²-2xy+2y²).(x²+2xy)=3x⁴-2x³y+2x²y²+6x³y-4x²y²+4xy³ = 3x⁴+4x³y-2x²y²+4x³y