Question

Cho x + y = 1. Tính x³ + y³ + 3xy (x³ + y³ ) + 6x²y² ( x + y )

Answer 1

ta có : \(x^3+y^3+3xy\left(x^3+y^3\right)+6x^2y^2\left(x+y\right)\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)+6x^2y^2\left(x+y\right)\)

\(=\left(1\right)^3-3xy.\left(1\right)+3xy\left(\left(1\right)^3-3xy.\left(1\right)\right)+6x^2y^2.\left(1\right)\)

\(=1-3xy+3xy\left(1-3xy\right)+6x^2y^2\)

\(=1-3xy+3xy-9x^2y^2+6x^2y^2=1-3x^2y^2\)

vậy \(x^3+y^3+3xy\left(x^3+y^3\right)+6x^2y^2\left(x+y\right)=1-3x^2y^2\) khi \(x+y=1\)

Answer 2

lolololololol