Tìm min :
Ta có : \(x^2+y^2-xy=4\)
\(\Leftrightarrow x^2+y^2=4+xy\le4+\frac{x^2+y^2}{2}\) ( vì \(\left(x-y\right)^2\ge0\) )
\(\Leftrightarrow\frac{A}{2}\le4\)
\(\Leftrightarrow A\le8\)
Tìm max
\(x^2+y^2-xy=4\)
\(\Leftrightarrow x^2+y^2=4+xy\)
\(\Leftrightarrow3\left(x^2+y^2\right)=8+\left(x+y\right)^2\ge8\)
\(\Leftrightarrow A\ge\frac{8}{3}\)
Há miệng ra và nói: ''PHỞ SÁNG"
Vì x,y>0x,y>0 nên A4=x2+y2x2+y2−xyA4=x2+y2x2+y2−xy
Đặt xy=a(a>0)xy=a(a>0) thì ta có:
A4=a2+1a2−a+1⇔A(a2−a+1)=4(a2+1)A4=a2+1a2−a+1⇔A(a2−a+1)=4(a2+1)
⇔a2(A−4)−Aa+A−4=0⇔a2(A−4)−Aa+A−4=0
Ta có: Δ=A2−4(A−4)2≥0⇔83≤A≤8
