Question

Tìm GTLN của biểu thức a = -x^2+x+1

Answer 1

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

\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

Answer 2

Ta có: \(-x^2+x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)