\(y-x=1\Rightarrow x=y-1\)
\(\Rightarrow x^2+y^2=\left(y-1\right)^2+y^2\)
\(=y^2-2y+1+y^2\)
\(=2y^2-2y+1\)
\(=2\left(y^2-y+\frac{1}{2}\right)\)
\(=2\left(y^2-2y\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)
Dấu"=" xảy ra khi \(2\left(y-\frac{1}{2}\right)^2=0\Rightarrow y=\frac{1}{2}\)
Vì \(y-x=1\)nên
\(\Rightarrow\frac{1}{2}-x=1\Rightarrow x=-\frac{1}{2}\)
Vậy \(Min_A=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{1}{2}\)