`a)`
`f(x)=5x^3-3x^4-1/2x^5+3x-8+4x^2`
`=-1/2x^5-3x^4+5x^3+4x^2+3x-8`
`g(x)=3x^4-5x^3+1/2x^5-3x^2-3x+7`
`=1/2x^5+3x^4-5x^3-3x^2-3x+7`
`b)`
`h(x)=f(x)+g(x)`
`=-1/2x^5-3x^4+5x^3+4x^2+3x-8+1/2x^5+3x^4-5x^3-3x^2-3x+7`
`=(-1/2x^5+1/2x^5)+(-3x^4+3x^4)+(5x^3-5x^3)+(4x^2-3x^2)+(3x-3x)+(-8+7)`
`=x^2-1`
`c)`
`\text{Để h(x) có nghiệm}`
`-> h(x)=0->x^2-1=0`
`->x^2=1`
`->` \(\left[ \begin{array}{l}\text{x = 1}\\\text{x = -1}\end{array} \right.\)
Vậy `x\in{1;-1}`