`C=-2x^2+x+1`
`C=-2(x^2-x/2)+1`
`C=-2(x^2-2*x*1/4+1/16)+1+1/8`
`C=-2(x-1/4)^2+9/8<=9/8`
Dấu "=" `<=>x=1/4.`
Ta có: \(C=-2x^2+x+1\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)
\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)