Đáp án:
$\bullet$ `F (x)=3x^4+x^3-x-3x^2+2`
`-> F (x)=3x^4+x^3-3x^2-x+2`
$\bullet$ `G (x) = -x^3+x^2-3x^4+x-1`
`-> G (x) = -3x^4-x^3+x^2+x-1`
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$a,$
$\bullet$ `H (x)= F (x) + G (x)`
`-> H (x) = 3x^4+x^3-3x^2-x+2 -3x^4-x^3+x^2+x-1`
`-> H (x) = (3x^4-3x^4) + (x^3-x^3) + (-3x^2+x^2) + (-x + x) + (2-1)`
`-> H(x)=-2x^2-1`
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$\bullet$ `K (x) = F (x) - G (x)`
`-> K (x) = 3x^4+x^3-3x^2-x+2 + 3x^4 + x^3-x^2-x+1`
`-> K (x) = (3x^4+3x^4)+(x^3+x^3) + (-3x^2 - x^2) + (-x - x) + (2+1)`
`-> K(x)=6x^4+2x^3-4x^2-2x+3`
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$b,$
`H (x)+2x^2+x=0`
`-> -2x^2-1 +2x^2+x=0`
`-> -1 + x=0`
`-> x = 1`
Vậy `x=1` để `H (x)+2x^2+x=0`
