Question

Tìm GTNN của:

\(-x^2+x-5\)

Answer 1

Đề này là tìm GTLN nhé bạn

Đặt \(A=-x^2+x-5\)

\(=-\left(x^2-x+5\right)=-\left(x^2-\dfrac{1}{2}.x.2+\dfrac{1}{4}+\dfrac{19}{4}\right)\)

\(=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\right]\)

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{19}{4}\)

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

\(\Leftrightarrow A=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}\)

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

Vậy \(MAX_A=\dfrac{-19}{4}\) khi \(x=\dfrac{1}{2}\)

Answer 2

\(-x^2+x-5=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{19}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{19}{4}\le\dfrac{-19}{4}\)Vậy ...