`a)`
`A = x^2 - 4x + y^2 + 6y + 15`
`= (x^2 - 4x + 4) + (y^2 + 6y +9) + 2`
`= (x^2 - 2 . x . 2 + 2^2) + (y^2 + 2 . y . 3 + 3^2) + 2`
` = (x-2)^2 + (y+3)^2 + 2`
`\forall x ; y` ta có :
`(x-2)^2 \ge 0`
`(y+3)^2 \ge 0`
`=> (x-2)^2+ (y+3)^2 \ge 0`
`=> (x-2)^2 + (y+3)^2 + 2 \ge 2`
`=> A \ge 2`
Dấu `=` xảy ra `<=> {(x-2=0),(y+3=0):}`
`<=> {(x =2),(y=-3):}`
Vậy `\text{Min}_A = 2 <=> {(x =2),(y=-3):}`
`b)`
`B = (3-x)(x+5)`
`= 3x + 15 - x^2 - 5x`
` = -x^2 - 2x + 15`
` = -(x^2 + 2x + 1) + 16`
` = -(x^2 + 2 . x . 1 + 1^2) + 16`
` = -(x+1)^2 + 16`
`\forall x` ta có :
`(x+1)^2 \ge 0`
`=> -(x+1)^2 \le 0`
`=> -(x+1)^2+ 16 \le 16`
`=> B \le 16`
Dấu `=` xảy ra `<=> x+1=0<=>x=-1`
Vậy `\text{Max}_B = 16 <=> x = -1`