Question

tìm GTLN của A=x - x²

Answer 1

\(A=x-x^2=-x^2+x=-\left(x^2-x\right)\)

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

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]\)

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

Ta có: \(-\left(x-\dfrac{1}{2}\right)^2\le0\)

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

Dấu " = " khi \(-\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Vậy \(MAX_A=\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)

Answer 2

x là số thực