Tìm giá trị lớn nhất: a) A = 6x – x ² – 5 b) B = 4x – x ² + 3 c) C = x – x ² d) D = 2x – 2x ² – 5

Đáp án:

 

Giải thích các bước giải:

a) `A = 6x – x^2 – 5

`A=-x^2+6x-5`

`A=-(x^2-6x+5)`

`A=-(x^2-6x+9-4)`

`A=-[(x-3)^2-4]`

`A=-(x-3)^2+4`

Ta có: `(x-3)^2 \ge 0 ∀x`

`⇒ -(x-3)^2 \le 0 ∀x`

`⇒ -(x-3)^2+4 \le 4`

Vậy `A_{max}=4` khi `x-3=0⇔x=3`

b) `B = 4x – x^2+ 3`

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

`B=-(x^2-4x+4-7)`

`B=-[(x-2)^2-7]`

`B=-(x-2)^2+7`

Ta có: `(x-2)^2 \ge 0 ∀x`

`⇒ -(x-2)^2 \le 0 ∀x`

`⇒ -(x-2)^2+7 \le 7`

Vậy `B_{max}=7` khi `x-2=0⇔x=2`

 c) `C = x – x^2`

`C=-(x^2-x)`

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

`C=-[(x-1/2)^2-1/4]`

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

Ta có: `(x-1/2)^2 \ge 0 ∀x`

`⇒ -(x-1/2)^2 \le 0 ∀x`

`⇒ -(x-1/2)^2+1/4 \le 1/4`

Vậy `C_{max}=1/4` khi `x-1/2=0⇔x=1/2`

d) `D = 2x – 2x^2- 5`

`D=-2(x^2-x+5/2)`

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

`D=-2[(x-1/2)^2+9/4]`

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

Ta có: `(x-1/2)^2 \ge 0 ∀x`

`⇒ -2(x-1/2)^2 \le 0 ∀x`

`⇒ -2(x-1/2)^2-9/2 \le -9/2`

Vậy `D_{max}=-9/2` khi `x-1/2=0⇔x=1/2`

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