Question

Tìm giá trị lớn nhất của biểu thức:

P=2x - 2x^2 - 5

Answer 1

\(P=2x-2x^2-5\)

\(=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}+\dfrac{5}{2}\right]\)

\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\)

\(\)\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)

Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2\le0\)

\(\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

hay P ≤ \(-\dfrac{9}{2}\)

Dấu = xảy ra \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(Max_P=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)