Question

-x^2 +x+6

Answer 1

Tìm Max:

\(A=-x^2+x+6=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{25}{4}\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\)

Với mọi giá trị của x ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\)

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

Vậy Max A = \(\dfrac{25}{4}\)

Để A = \(\dfrac{25}{4}\) thì \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Answer 2

?? ? đề là j