Vì \(x,y>0\) nên \(\dfrac{A}{4}=\dfrac{x^2+y^2}{x^2+y^2-xy}\)
Đặt \(\dfrac{x}{y}=a\left(a>0\right)\) thì ta có:
\(\dfrac{A}{4}=\dfrac{a^2+1}{a^2-a+1}\Leftrightarrow A\left(a^2-a+1\right)=4\left(a^2+1\right)\)
\(\Leftrightarrow a^2\left(A-4\right)-Aa+A-4=0\)
Ta có: \(\Delta=A^2-4\left(A-4\right)^2\ge0\Leftrightarrow\dfrac{8}{3}\le A\le8\)
Tìm min:
Ta có: \(x^2+y^2-xy=4\)
\(\Leftrightarrow x^2+y^2=4+xy\le4+\dfrac{x^2+y^2}{2}\) (Vì \(\left(x-y\right)^2\ge0\))
\(\Leftrightarrow\dfrac{A}{2}\le4\)
\(\Leftrightarrow A\le8\)
Tìm Max
\(x^2+y^2-xy=4\)
\(\Leftrightarrow x^2+y^2=4+xy\)
\(\Leftrightarrow3\left(x^2+y^2\right)=8+\left(x+y\right)^2\ge8\)
\(\Leftrightarrow A\ge\dfrac{8}{3}\)
