a) *Thay x=-1 vào P:
1+2+1=4. Vậy P(-1)=4.
* Thay x=\(\dfrac{1}{2}\) vào P:
\(\left(\dfrac{1}{2}\right)^4+2.\left(\dfrac{1}{2}\right)^2+1=\dfrac{1}{16}+\dfrac{1}{2}+1=\dfrac{25}{16}\)
*Thay x=-2 vào Q:
\(\left(-2\right)^4+4.\left(-2\right)^3+2.\left(-2\right)^2-4\left(-2\right)+1=1\)
Vậy Q(-2)=1.
*Thay x=1 vào Q:
1+4+2-4+1=4.
Vậy Q(1)=4.
b) Đặt t=x2\(\left(t\ge0\right)\)
\(\Rightarrow t^2+2t+1=\left(t+1\right)^2\ge0\)
Vậy Pmin=0\(\Leftrightarrow t=-1\left(KTM\right)\)
c) Q(x)-P(x)=\(4x^3-4x\)
Để Q(x)-P(x)=0 thì \(4x^3-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0;-1;1.