`a)`
`A = x^2 - 7x + 1`
` = (x^2 - 7x + 49/4) - 45/4`
`= [ x^2 - 2 . x . 7/2 + (7/2)^2] - 45/4`
` = (x - 7/2)^2 - 45/4`
`\forall x` ta có :
`(x-7/2)^2 \ge 0`
`=> (x-7/2)^2 - 45/4 \ge -45/4`
`=> A \ge -45/4`
Dấu `=` xảy ra `<=> x - 7/2 = 0 <=> x = 7/2`
Vậy `\text{Min}_A = -45/4 <=> x = 7/2`
`b)`
`B = 4x^2 + 6x + 3`
`= 4 (x^2 + 3/2x + 9/16) + 3/4`
` = 4 [ x^2 + 2 . x . 3/4 + (3/4)^2] + 3/4`
`= 4 (x + 3/4)^2 + 3/4`
`\forall x` ta có :
`(x+3/4)^2 \ge 0`
`=> 4(x+3/4)^2 \ge 0`
`=> 4(x+3/4)^2 + 3/4 \ge 3/4`
`=>B \ge 3/4`
Dấu `=` xảy ra `<=>x + 3/4 =0`
`<=> x = -3/4`
Vậy `\text{Min}_B = 3/4 <=> x = -3/4`
`c)`
`C = 2x^2 + 8x `
` = 2 (x^2 + 4x + 4) - 8`
`= 2(x^2 + 2 . x . 2 + 2^2) - 8`
` = 2 (x+2)^2 - 8`
`\forall x` ta có :
`(x+2)^2 \ge 0`
`=> 2(x+2)^2 \ge 0`
`=> 2(x+2)^2 - 8 \ge -8`
`=> C \ge -8`
Dấu `=` xảy ra `<=> x + 2 = 0`
`<=> x = -2`
Vậy `\text{Min}_C = -8 <=> x = -2`
`d)`
`D = 3x^2 - 9x`
` = 3 (x^2 - 3x + 9/4) - 27/4`
`= 3 [ x^2 - 2 . x . 3/2 + (3/2)^2 ] - 27/4`
`= 3 (x-3/2)^2 - 27/4`
`\forall x` ta có :
`(x-3/2)^2 \ge 0`
`=> 3 (x-3/2)^2 \ge 0`
`=> 3 (x-3/2)^2 - 27/4 \ge -27/4`
`=> D \ge -27/4`
Dấu `=` xảy ra `<=>x -3/2 =0 <=>x=3/2`
Vậy `\text{Min}_D = -27/4 <=> x = 3/2`
