`a)`
Đặt `A = x^2 + y^2 - x - 6y + 10`
` = (x^2 - x + 1/4) + (y^2- 6y + 9) + 3/4`
` = [x^2 - 2 . x .1/2 + (1/2)^2] + (y^2 - 2 . y . 3 + 3^2) + 3/4`
` = (x-1/2)^2 + (y-3)^2 + 3/4`
`\forall x ;y` ta có :
`(x-1/2)^2 \ge 0`
`(y-3)^2 \ge 0`
`=> (x-1/2)^2 + (y-3)^2 \ge 0`
`=> (x-1/2)^2 + (y-3)^2 + 3/4 \ge 3/4`
`=> A \ge 3/4`
Dấu `=` xảy ra `<=>` `{(x-1/2=0),(y-3=0):}`
`<=> {(x=1/2), (y=3):}`
Vậy `\text{Min}_A = 3/4 <=> {(x=1/2), (y=3):}`
`b)`
Đặt `B = x^2+ 4y^2 + 6x - 4y + 2031`
` = (x^2 + 6x + 9) + (4y^2 - 4y + 1) + 2021`
` = (x^2 + 2 . x . 3 +3^2) + [(2y)^2 - 2 . 2y . 1 + 1^2] + 2021`
` = (x+3)^2 + (2y-1)^2 + 2021`
`\forall x ; y` ta có :
`(x+3)^2 \ge 0`
`(2y-1)^2 \ge 0`
`=> (x+3)^2 + (2y-1)^2 \ge 0`
`=> (x+3)^2 + (2y-1)^2 + 2021 \ge 2021`
`=> B \ge 2021`
Dấu `=` xảy ra `<=> {(x+3=0),(2y-1=0):}`
`<=> {(x = -3),(y = 1/2):}`
Vậy `\text{Min}_B = 2021 <=> {(x = -3),(y = 1/2):}`
`c)`
Đặt `C = x^2 - 2x + 5`
`=(x^2 -2x+1) + 4`
`= (x^2 - 2 . x . 1 + 1^2) + 4`
`= (x-1)^2 +4`
`\forall x` ta có :
`(x-1)^2 \ge 0`
`=> (x-1)^2+ 4 \ge 4`
`=> C \ge 4`
Dấu `=` xảy ra `<=> x - 1 = 0 <=> x =1`
Vậy `\text{Min}_C = 4 <=> x= 1`
