Question

Tìm min của: C=x³+y³+xy biết: x+y=1

Answer 1

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

\(C=x^2-xy+y^2+xy\left(vì.x+y=1\right)\)

\(C=x^2+y^2\)

\(C=x^2+\left(1-x\right)^2\left(vì.y=1-x\right)\)

\(C=x^2+1-2x+x^2\)

\(C=2x^2-2x+1=2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{2}+1\)

\(C=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\left(vì.2\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\in R\right)\)

\(\Rightarrow Min\left(C\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)