Câu `1:`

`a)`

`A = 2x^2 - 6x - 1`

`=  2 (x^2 - 3x + 9/4) - 11/2`

` = 2 (x - 3/2)^2 - 11/2`

`\forall x` ta có :

`(x-3/2)^2 \ge 0`

`=> 2 (x-3/2)^2 \ge 0`

`=> 2 (x-3/2)^2 - 11/2 \ge -11/2`

`=> A \ge -11/2`

Dấu `=` xảy ra `<=> x - 3/2 = 0`

`<=> x = 3/2`

Vậy `\text{Min}_A = -11/2 <=> x = 3/2`

`b)`

`B = x^2 - x - 2`

` = (x^2 - x+  1/4) - 9/4`

`=  (x - 1/2)^2 - 9/4`

`\forall x` ta có :

`(x-1/2)^2 \ge 0`

`=> (x-1/2)^2 - 9/4 \ge -9/4`

`=> B \ge -9/4`

Dấu `=` xảy ra `<=>x-1/2=0`

`<=>x=1/2`

Vậy `text{Min}_B = -9/4 <=> x = 1/2`

Câu `2:`

`a)`

`A = 3 - x^2 - 2x`

`= - (x^2 + 2x + 1) + 4`

`=  - (x+1)^2 + 4`

`\forall x` ta có :

`(x+1)^2 \ge 0`

`=> - (x+1)^2 \le 0`

`=> - (x+1)^2 + 4 \le 4`

`=> A \le 4`

Dấu `=` xảy ra `<=>x+1=0`

`<=> x =-1`

Vậy `\text{Max}_A = 4 <=> x = -1`

`b)`

`B= 2x - 3x^2 - 4`

`=  -3 (x^2 - 2/3x + 1/9) - 11/3`

`= -3 ( x - 1/3)^2 -11/3`

`\forall x` ta có :

`(x-1/3)^2 \ge 0`

`=> -3 (x-1/3)^2 \le 0`

`=> -3 (x-1/3)^2 - 11/3 \le -11/3`

`=> B \le -11/3`

Dấu `=` xảy ra `<=> x- 1/3=0`

`<=>x=1/3`

Vậy `\text{Max}_B = -11/3 <=> x = 1/3`

`1.`

`A=2x²-6x-1`

`=2(x²-3x-1/2)`

`=2(x²-3x+9/4-11/4)`

`=2(x²-3x+9/4)-11/2`

`=2[x²-2.x. 3/2+(3/2)^2]-11/2`

`=2(x-3/2)^2-11/2`

Ta có:`(x-3/2)^2≥0∀x`

`⇒2(x-3/2)^2≥0∀x`

`⇒2(x-3/2)^2-11/2≥-11/2∀x`

Vậy `A_(min)=-11/2` khi `x-3/2=0⇔x=3/2`

`B=x²-x-2`

`=x²-x+1/4-9/4`

`=(x²-x+1/4)-9/4`

`=[x²-2.x. 1/2+(1/2)^2]-9/4`

`=(x-1/2)^2-9/4`

Ta có:`(x-1/2)^2≥0∀x`

`⇒(x-1/2)^2-9/4≥-9/4∀x`

Vậy `B_(min)=-9/4` khi `x-1/2=0⇔x=1/2`

`2.`

`A=3-x²-2x`

`=-(x²+2x-3)`

`=-(x²+2x+1-4)`

`=-(x²+2x+1)+4`

`=-(x+1)²+4`

Ta có:`(x+1)²≥0∀x`

`⇒-(x+1)²≤0∀x`

`⇒-(x+1)²+4≤4∀x`

Vậy `A_(max)=4` khi `x+1=0⇔x=-1`

`B=2x-3x²-4`

`=-3(x²-2/3x+4/3)`

`=-3(x²-2/3x+1/9+11/9)`

`=-3(x²-2/3x+1/9)-11/3`

`=-3[x²-2.x. 1/3+(1/3)^2]-11/3`

`=-3(x-1/3)^2-11/3`

Ta có:`(x-1/3)^2≥0∀x`

`⇒3(x-1/3)^2≥0∀x`

`⇒-3(x-1/3)^2≤0∀x`

`⇒-3(x-1/3)^2-11/3≤-11/3∀x`

Vậy `B_(max)=-11/3` khi `x-1/3=0⇔x=1/3`