Đáp án:
`min(x^2+y^2-3x-3y+xy)=-3` khi `(x;y)=(1;1)`
Giải thích các bước giải:
`x^2+y^2-3x-3y+xy`
`=x^2-2x+1+y^2-2y+1+xy-x-y+1-3`
`=(x^2-2x+1)+(y^2-2y+1)+x.(y-1)-(y-1)-3`
`=(x-1)^2+(y-1)^2+(x-1).(y-1)-3`
`=1/4 (x-1)^2+(x-1).(y-1)+(y-1)^2+3/4 (x-1)^2-3`
`=[1/2 .(x-1)]^2+2 . 1/2 (x-1) .(y-1)+(y-1)^2 +3/4 (x-1)^2-3`
`=[1/2 .(x-1)+(y-1)]^2+3/4 (x-1)^2-3`
Vì `[1/2 (x-1)+(y-1)]^2>=0∀x;y`
`(x-1)^2≥0∀x->3/4 (x-1)^2≥0∀x`
`->[1/2 (x-1)+(y-1)]^2+3/4 (x-1)^2≥0∀x;y`
`->[1/2 (x-1)+(y-1)]^2+3/4 (x-1)^2 -3≥-3∀x;y`
`->x^2+y^2-3x-3y+xy>=-3∀x;y`
Dấu `'='` xảy ra
`<=>{(1/2 (x-1)+(y-1)=0),(x-1=0):}<=>{(1/2.(1-1)+(y-1)=0),(x=1):}`
`<=>{(y=1),(x=1):}`
Vậy `x^2+y^2-3x-3y+xy` đạt GTNN bằng `=-3` khi `(x;y)=(1;1)`
