Question

Cho \(x+y=1\).Tính \(A=x^3+y^3+3xy\)

Answer 1

Ta có: \(x+y=1\Rightarrow x=1-y\left(1\right)\)

Thay \(\left(1\right)\rightarrow A:\)

\(A=\left(1-y\right)^3+y^3+3\left(1-y\right)y\)

\(=1-3y+3y^2-y^3+y^3+3y-3y^2\)

\(=1\)

Vậy \(A=1.\)

Answer 2

\(x+y=1\Rightarrow x=1-y\)

\(A=\left(1-y\right)^3+y^3+3\left(1-y\right)y\\ A=1+3y^2-3y-y^3+y^3+3y-3y^2\\ A=1\)

Answer 3

\(A=x^3+y^3+3xy\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)

\(A=\left(x+y\right)^3-3xy\left(x+y-1\right)\Leftrightarrow1^3-3xy\left(1-1\right)\)

\(A=1-0=1\)

vậy với \(x+y=1\) thì \(A=1\)