a) `A(x) = 2x^2 - x^3 + x -3`
`B(x) = x^3 - x^2 + 4 - 3x`
`P(x) =A(x) + B(x)`
`=> P(x)= (2x^2 - x^3 + x-3) + (x^3 - x^2 + 4 - 3x)`
`=>P(x)= 2x^2 - x^3 + x - 3 + x^3 - x^2 + 4 - 3x`
`=> P(x) = (2x^2 - x^2) - (x^3 - x^3) + (x- 3x) - (3-4)`
`=> P(x) = x^2 - 2x + 1`
b) `Q(x) = 5x^2 - 5 + a^2 + ax`
Để `Q(x)` có nghiệm `x = -1`
`=> 5.(-1)^2 - 5 + a^2 + a.(-1)=0`
`=> 5.1 - 5 + a^2 -a =0`
`=>a^2 -a =0`
`=> a(a-1)=0`
`=>` \(\left[ \begin{array}{l}a=0\\a-1=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}a=0\\a=1\end{array} \right.\)
Vậy `a=0` hoặc `a=1`