Sửa đề: Tìm GTLN
\(-2x^2+4x+1=-2\left(x^2-2x-\dfrac{1}{2}\right)\)
\(=-2\left(x^2-2x+1-\dfrac{3}{2}\right)\)
\(=-2\left(x-1\right)^2+3< =3\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
có `-2x^2 + 4x +1`
`= -2(x^2 -2x - 1/2)`
`=-2(x^2 - 2x + 1 -3/2)`
`=-2[(x^2 - 2x + 1) - 3/2]`
`=-2(x-1)^2 +3`
Ta thấy :(x-1)^2 >=0`\(\forall x\)
`-2(x-1)^2<=0`\(\forall x\)
`-2(x-1)^2 + 3<=0\(\forall x\)
dấu `"="` xảy ra `<=>x-1=0`
`<=>x = 1`