`a)`
`A(x)=5x^4-5+6x^3+x^4-5x-12`
`=(5x^4+x^4)+6x^3-5x+(-5-12)`
`=6x^4+6x^3-5x-17`
``
`B(x)=8x^4+2x^3-2x^4+4x^3-5x-15-2x^2`
`=(8x^4-2x^4)+(2x^3+4x^3)-2x^2-5x-15`
`=6x^4+6x^3-2x^2-5x-15`
``
`b)`
`C(x)=A(x)-B(x)`
`=(6x^4+6x^3-5x-17)-(6x^4+6x^3-2x^2-5x-15)`
`=6x^4+6x^3-5x-17-6x^4-6x^3+2x^2+5x+15`
`=(6x^4-6x4)+(6x^3-6x^3)+2x^2+(-5x+5x)+(-17+15)`
`=2x^2-2`
Để `C(x)` có nghiệm
`\toC(x)=0`
`\to2x^2-2=0`
`\to2(x^2-1)=0`
`\tox^2-1=0`
`\tox^2=1`
`\to` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `x\in{-1;1}`
