`a)`
Ta có :
`M = x^2+ 2y^2 - 2xy + 2x -10y`
`= [ (x^2 - 2xy + y^2) + 2 (x-y) . 1+ 1 ^2 ] + (y^2 - 8y + 16) - 17`
` = (x-y + 1)^2 + (y-4)^2 - 17`
`\forall x;y` ta có :
`(x-y+1)^2 \ge 0`
`(y-4)^2 \ge 0`
`=> (x-y + 1)^2 + (y-4)^2 \ge 0`
`=> (x-y + 1)^2 + (y-4)^2- 17 \ge -17`
`=> M \ge -17`
Dấu `=` xảy ra` <=> {(x - y + 1 = 0 ),(y-4 = 0):}`
`<=> {(x - y = -1 ),(y = 4):}`
`<=> {(x = 3),(y=4):}`
Vậy `\text{Min}_M = -17 <=> {(x = 3),(y=4):}`
`b)`
Ta có :
`N = (x+1)^2 + (x-3)^2`
`= (x^2 + 2x + 1) + (x^2 - 6x+ 9)`
` = x^2 + 2x + 1 + x^2 - 6x + 9`
` = 2x^2 - 4x + 10`
` = 2 (x^2 - 2x + 1) + 8`
` = 2 (x-1)^2 + 8`
`\forall x` ta có :
`(x-1)^2 \ge 0`
`=> 2 (x-1)^2 \ge 0`
`=> 2 (x-1)^2 + 8 \ge 8`
`=> N \ge 8`
Dấu `=` xảy ra `<=> x - 1 = 0 <=> x=1`
Vậy `\text{Min}_N = 8 <=> x = 1`
`c)`
`D = 2007 - x^2 - 5x`
` = - (x^2 + 5x + 25/4) + 8053/4`
` = - (x + 5/2)^2 + 8053/4`
`\forall x` ta có :
`(x+5/2)^2 \ge 0`
`=> - (x+5/2)^2 \le 0`
`=> - (x+5/2)^2 + 8053/4 \le 8053/4`
`=> D \le 8053/4`
Dấu `=` xảy ra ` <=> x + 5/2 = 0`
`<=> x = -5/2`
Vậy `\text{Max}_D = 8053/4 <=> x = -5/2`
