Đáp án + giải thích các bước giải:
`B(x)+A(x)=x^4+3x^2-x-1`
`->B(x)=x^4+3x^2-x-1-A(x)`
`->B(x)=x^4+3x^2-x-1-(-4x^3+2x^4-x^2-x^3-x^4+3x^2+1+5x^3)`
`=x^4+3x^2-x-1-(2x^4-x^4-4x^3-x^3+5x^3-x^2+3x^2+1)`
`=x^4+3x^2-x-1-(x^4+2x^2+1)`
`=x^4-x^4+3x^2-2x^2-x-1-1`
`=x^2-x-2`
`->B(x)=x^2-x-2`
Đặt `B(x)=0`
`->x^2-x-2=0`
`->x^2+x-2x-2=0`
`->x(x+1)-2(x+1)=0`
`->(x-2)(x+1)=0`
`->`\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy `x=2;-1`