\(f\left(x\right)=4x^2-9x+1\)
\(f\left(x\right)=\left(4x^2-9x+\dfrac{81}{16}\right)-\dfrac{81}{16}+1\)
\(f\left(x\right)=\left(2x-\dfrac{9}{4}\right)^2-\dfrac{65}{16}\ge-\dfrac{65}{16}\)
Dấu "=" xảy ra `<=>2x-9/4=0`
`<=>x=9/8`
Vậy \(Min_{f\left(x\right)}=-\dfrac{65}{16}\) khi `x=9/8`