Question

Tìm Max :

B= \(-2x+\sqrt{x+2021}\)

Answer 1

Đặt \(\sqrt{x+2021}=t\ge0\Rightarrow x=t^2-2021\)

\(B=-2\left(t^2-2021\right)+t=-2t^2+t+4042=-2\left(t-\dfrac{1}{4}\right)^2+\dfrac{32337}{8}\le\dfrac{32337}{8}\)