Question

Answer 1

Ta có: \(A=x^2-xy+y^2-2x-2y\)

\(=\frac14\left(4x^2-4xy+4y^2-8x-8y\right)\)

\(=\frac12\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=\frac12\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=\frac12\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack\ge\frac12\cdot\left(-16\right)=-8\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-2=0\\ 2x-y-2=0\end{cases}\Rightarrow\begin{cases}y=2\\ 2x=y+2=2+2=4\end{cases}\Rightarrow\begin{cases}y=2\\ x=2\end{cases}\)