Câu `1:`
`a) A = x^2 - 4xy + 5y^2 + 10x - 22y +28`
` = x^2 - 4xy + 4y^2 + y^2 + 10x - 20y - 2y + 25 + 2 + 1`
` = (x^2 - 4xy + 4y^2) + (10x - 20y) + 25 + (y^2 - 2y + 1) + 2`
` = [ (x-2y)^2 + 2 . 5 . (x-2y) + 5^2 ] + (y-1)^2 + 2`
` = (x- 2y + 5)^2 + (y-1)^2 + 2`
`\forall x ; y` ta có :
`(x-2y+5)^2 \ge 0`
`(y-1)^2 \ge 0`
`=> (x- 2y + 5)^2 + (y-1)^2 \ge 0`
`=> (x- 2y + 5)^2 + (y-1)^2 + 2 \ge 2`
`=> A \ge 2`
Dấu `"="` xảy ra `<=>`$\begin{cases} y - 1 = 0 \\ x - 2y + 5 = 0 \end{cases}$
`<=>` $\begin{cases} y = 1 \\ x = -3 \end{cases}$
Vậy `\text{Min}_A = 2 <=>` $\begin{cases} y = 1 \\ x = -3 \end{cases}$
`b) B = 2x^2 - 6x`
` = 2 . (x^2 - 3x)`
` = 2 . (x^2 - 3x + 9/4) - 9/2`
` = 2 . (x - 3/2)^2 - 9/2`
`\forall x` ta có :
`(x-3/2)^2 \ge 0`
`=> 2 . (x-3/2)^2 \ge 0`
`=> 2 . (x-3/2)^2 - 9/2 \ge -9/2`
`=> B \ge -9/2`
Dấu `"="` xảy ra `<=> x - 3/2 = 0 <=> x = 3/2`
Vậy `\text{Min}_B = -9/2 <=> x = 3/2`
Bài `2:`
`a) A = 4x - 2x^2 + 3`
` = - (2x^2 - 4x) + 3`
` = -2. (x^2 + 2x) + 3`
` = -2 . (x^2 + 2x +1) +5`
` = -2 . (x+1)^2 + 5`
`\forall x` ta có :
`(x+1)^2 \ge 0`
`=> -2 . (x+1)^2 \le 0`
`=> -2 . (x+1)^2 + 5 \le 5`
`=> A \le 5`
Dấu `"="` xảy ra `<=> x + 1 = 0 <=> x = -1`
Vậy `\text{Max}_A = 5 <=> x = -1`
`b) B = x - x^2`
` = - (x^2 - x)`
` = - (x^2 - x + 1/4) + 1/4`
` = -(x-1/2)^2 + 1/4`
`\forall x` ta có :
`(x-1/2)^2 \ge 0`
`=> -(x - 1/2)^2 \le 0`
`=> -(x-1/2)^2 + 1/4 \le 1/4`
`=> B \le 1/4`
Dấu `"="` xảy ra `<=> x-1/2 =0<=> x=1/2`
Vậy `\text{Max}_B = 1/4 <=> x = 1/2`
`c) C = 2x - 2x^2 - 5`
` = -(2x^2 - 2x) - 5`
` = -2.(x^2 - x) - 5`
` = -2 . (x^2 - x + 1/4) - 9/2`
` = -2. (x-1/2)^2 - 9/2`
`\forall x` ta có :
`(x-1/2)^2 \ge 0`
`=> -2 . (x-1/2)^2 \le 0`
`=>-2 . (x-1/2)^2 - 9/2 \le -9/2`
`=> C \le -9/2`
Dấu `"="` xảy ra `<=> x - 1/2 = 0 <=> x = 1/2`
Vậy `\text{Max}_C = -9/2 <=> x = 1/2`
