`P=4x^2 +2y^2 -4xy-4x-8y+2050`

`P=4x^2 +y^2 +y^2 -4xy-4x+2y-10y+1+25+2024`

`P=[4x^2 -(4xy+4x)+(y^2 +2y+1)]+(y^2 -10y+25)+2024`

`P=[(2x)^2 -2.2x.(y+1)+(y+1)^2]+(y-5)^2 +2024`

`P=[2x-(y+1)]^2 +(y-5)^2 +2024`

`P=(2x-y-1)^2 +(y-5)^2 +2024`

Vì `(2x-y-1)^2≥0` , `(y-5)^2≥0` nên 

`P=(2x-y-1)^2 +(y-5)^2 +2024≥0+0+2025=2024`

Dấu `=` xảy ra khi:

`{((2x-y-1)^2=0),((y-5)^2=0):}`

`{(2x-y-1=0),(y-5=0):}`

`{(2x=y+1),(y=5):}`

`{(2x=5+1),(y=5):}`

`{(2x=6),(y=5):}`

`{(x=3),(y=5):}`

Vậy P có GTNN là `2024` khi `x=3;y=5`