Ta có: \(4x^2+y^2=8+3xy\Leftrightarrow4x^2-4xy+y^2=8-xy\)

\(\Leftrightarrow\left(2x-y\right)^2=8-xy\ge0\forall x,y\inℝ\Rightarrow xy\le8\)

\(\Rightarrow P=xy+2020\le8+2020=2028\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}2x=y\\xy=8\end{cases}}\Rightarrow\left(x,y\right)\in\left\{\left(2;4\right);\left(-2;-4\right)\right\}\)

Ta co : \(x^2+y^2-4x+3=0\)

\(=>\left(x-2\right)^2+y^2=1\)

\(=>\left(x-2\right)^2\le1=>x\le3\)

Lai co : \(x^2+y^2=4x-3\le4.3-3=9\)

Dau = xay ra \(< =>\hept{\begin{cases}x=4\\y=0\end{cases}}\)

Vay gtln cua P = 9 khi x = 4 ; y = 0

(sai thi bo qua cho minh vi lan dau lam dang nay)

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

\(x^2+y^2-xy=4\)

\(\Leftrightarrow x^2+y^2=4+xy\)

\(\Leftrightarrow3\left(x^2+y^2\right)=8+\left(x+y\right)^2\ge8\)

\(\Leftrightarrow A\ge\frac{8}{3}\)

Ta có: 4x² + 12xy + 10y² + 4x + 4y + 2 = 0

<=> (4x² + 12xy + 9y²) + 2(2x + 3y) + 1 + (y² - 2y + 1) = 0

<=> (2x + 3y)² + 2(2x + 3y) + 1 + (y - 1)² = 0

<=> (2x + 3y + 1)² + (y - 1)² = 0

<=> \(\hept{\begin{cases}2x+3y+1=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-\frac{1+3y}{2}\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)(tm)

Khi đó: P = \(\frac{x^2+y^2+xy}{3xy}=\frac{\left(-2\right)^2+1^2-2.1}{3.\left(-2\right).1}=-\frac{1}{2}\)