`a)`
`f(x)+g(x)=(x^3+3x-x^2-1)+(-x^3+x^2-x+2)`
`=x^3-x^2+3x-1-x^3+x^2-x+2`
`=(x^3-x^3)+(x^2-x^2)+(3x-x)+(2-1)`
`=2x+1`
``
`b)`
`f(x)-g(x)=(x^3+3x-x^2-1)-(-x^3+x^2-x+2)`
`=x^3-x^2+3x-1+x^3-x^2+x-2`
`=(x^3+x^3)-(x^2+x^2)+(3x+x)-(1+2)`
`=2x^3-2x^2+4x-3`
``
`c)`
`r(x)=f(x)+g(x)`
`\to r(x)=2x+1`
Để `r(x)` có nghiệm
`\to r(x)=0`
`\to 2x+1=0`
`\to 2x=-1`
`\to x=-1/2`
Vậy `x=-1/2`
