\(P=y^2+4x^2-2y+20x+2048\)
\(=4x^2+20x+25+y^2-2y+1+2022\)
\(=\left(2x+5\right)^2+\left(y-1\right)^2+2022>=2022\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}2x+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=1\end{matrix}\right.\)
`P = y^2 + 4x^2 - 2y + 20x + 2048`
`= (y^2 - 2y + 1) + (4x^2 + 20x + 25) + 2022`
`= (y^2 - 2y + 1) + ((2x)^2 + 2 . 2x . 5 + 5^2) + 2022`
`= (y-1)^2 + (2x + 5)^2 + 2022`
Ta có: `{((y-1)^2 >=0),((2x+5)^2 >=0):}`
`=> (y-1)^2 + (2x + 5)^2 >= 0`
`=> (y-1)^2 + (2x + 5)^2 + 2022 >= 2022`
Hay `P >= 2022`
Dấu = có khi:
`{(y-1=0),(2x+5=0):}`
`<=> {(y=1),(x=-5/2):}`
Vậy ...
P = y² + 4x² + 2y + 20x + 2048
= (y² + 2y + 1) + (4x² + 20x + 25) + 2048 - 1 - 25
= (y + 1)² + (2x + 5)² + 2022
Do (y + 1)² >= 0
(2x + 5)² >= 0
Suy ra (y + 1)² + (2x + 5)² >= 0
Suy ra (y + 1)² + (2x + 5)² + 2022 >= 2022
Vậy GTNN của P là 2022 khi y = -1 và x = -5/2
