Question

Answer 1

`A=1/3-(x+1)^2`

Vì `(x+1)^2>=0`

`=>1/3-(x+1)^2<=1/3-0=1/3`

Hay `A<=1/3`

Dấu "=" xảy ra khi `x+1=0<=>x=-1`

Vậy `max_A=1/3<=>x=-1`

Answer 2

\(A=-\left(x+1\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\forall x\)

Dấu ''='' xảy ra khi x = -1

Answer 3

Ta có:

\(-\left(x+1\right)^2\le0\forall x\\ =>A=\dfrac{1}{3}-\left(x+1\right)^2\le\dfrac{1}{3}\forall x\)

Đấu "=" xảy ra: `x+1=0=>x=-1`

Vậy: ...