Bài 5:
a: Ta có: \(A=-2x^2+x+1\)
\(=-2\left(x^2-\dfrac{1}{2}x-\dfrac{1}{2}\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)
\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)


