`E = x^2 + xy + y^2 - 3x - 3y + 3`

` = (x^2 + xy - 3x + 1/4 y^2 - 3/2y + 9/4) + (3/4y^2 - 3/2y + 3/4)`

` = [ x^2 + 2 . x . 1/2y + (1/2y)^2 - 2 . x . 3/2 - 2 . 1/2y . 3/2 + (3/2)^2] + 3/4 (y^2 - 2y+1)`

` = (x + 1/2y - 3/2)^2 + 3/4 (y-1)^2`

`\forall x;y` ta có :

`(x+1/2y-3/2)^2\ge0`

`3/4(y-1)^2\ge0`

`=>  (x + 1/2y - 3/2)^2 + 3/4 (y-1)^2 \ge0`

`=> E \ge0`

Dấu `=` xảy ra `<=> {(x+1/2y - 3/2 = 0),(y-1=0):} <=> {(x=1),(y=1):}`

Vậy `\text{Min}_E = 0 <=> (x;y)=(1;1)`

Đáp án `+` Giải thích các bước giải:

`E=x^2 + xy +y^2 -3x-3y +3`

`-> E=(x^2 +xy-3x+1/4 y^2 -3/2 y+9/4)+(3/4 y^2 -3/2 y + 3/4)`

`-> E=[x^2 +2x. 1/2 y + (1/2 y)^2 -2x. 3/2 - 2. 1/2 y . 3/2 +(3/2)^2 ]+3/4 (y^2 -2y+1)`

`-> E=(x+1/2 y-3/2)^2  + 3/4 (y-1)^2`

Nhận xét: `(x+1/2 y-3/2)^2 >= 0 AA x`

`<=> (x+1/2 y-3/2)^2  + 3/4 (y-1)^2 >= 0 AA x`

`<=> E >= 0`

Dấu `''=''` xảy ra khi:

`{(x+1/2 y-3/2=0),(y-1=0):}`

`<=> {(x=1),(y=1):}`

Vậy `E_(min) =0 <=> x=1; y=1`