\(=-4\left(x^2-3x+\dfrac{1}{4}\right)\\ =-4\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}-2\right)\\ =-4\left(x-\dfrac{3}{2}\right)^2+8\le8\forall x\)
Dẫu bằng xảy ra khi
\(x-\dfrac{3}{2}=0\\ x=\dfrac{3}{2}\)
Vậy....
\(12x-4x^2-1\)4
\(=-4x^2+12x-1=-4\left(x^2-3x+\dfrac{9}{4}\right)+8\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+8\le8\forall x\)
Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)
Có: `12x-4x^2-1`
`= -(4x^2-12x+1)`
`= -4(x^2-3x+1/4)`
`= -4(x^2-3x+9/4-2)
`= -4(x-3/2)^2+8`
Vì `-4(x-3/2)^2≤0`
`=> -4(x-3/2)^2+8≤8`
Dấu `''=''` xảy ra khi `-4(x-3/2)^2=0`
`=> x=3/2`
