Đáp án:
`P=(x^2-3x+1).(x^2-3x-1)`
Đặt `x^2-3x=a`
`=> P=(a+1)(a-1)`
`=a^2-1>=-1`
Dấu "="xảy ra `<=> x^2-3x=0`
`=> x(x-3)=0`
`=>` \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
Vậy `P_(min)=-1 <=>` \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)