Bài `11:`
`d)`
`-6- 8x - 16x^2`
`= - (16x^2 + 8x + 1) -5`
` = - [ (4x)^2 + 2 . 4x . 1 + 1^2] -5`
` = - (4x +1)^2 - 5`
`\forall x \in RR` ta có :
`(4x+1)^2 \ge 0`
`=> - (4x+1)^2 \le 0`
`=> - (4x+1)^2 - 5 \le -5 < 0`
`=> -6 - 8x - 16x^2 < 0`
Vậy `-6 - 8x - 16x^2 < 0 \forall x \in RR`
Bài `12`
`b)`
`B = 4x^2 + 4x + 2`
` = (4x^2 + 4x + 1) + 1`
` =[ (2x)^2 + 2 . 2x . 1 + 1^2] + 1`
` = (2x+1)^2 + 1`
`\forall x` ta có :
`(2x+1)^2 \ge 0`
`=> (2x+1)^2 + 1 \ge 1`
`=> B \ge 1`
Dấu `=` xảy ra ` <=> 2x + 1 =0`
`<=> 2x = -1 <=> x = -1/2`
Vậy `\text{Min}_B = 1 <=> x = -1/2`
`d)`
`D = (x-1)(x+2)(x+3)(x+6)`
` = [ (x-1)(x+6)] . [ (x+2)(x+3)]`
` = (x^2 + 6x - x - 6)(x^2 + 3x + 2x + 6)`
` = (x^2 + 5x - 6)(x^2 + 5x + 6)`
` = (x^2 + 5x)^2 - 6^2`
` = (x^2 + 5x)^2 - 36`
`\forall x` ta có :
`(x^2+ 5x)^2 \ge 0`
`=> (x^2 +5x)^2 - 36 \ge -36`
`=> D \ge -36`
Dấu `=` xảy ra `<=> x^2 + 5x = 0`
`<=> x (x+5) = 0`
`<=> x \in {0 ; -5}`
Vậy `\text{Min}_D = -36 <=> x \in {0;-5}`
