Đáp án:
Ta có :
`x^5 + 1929x^2(x - 1)^2 + 3859x^3 = (x^2 + 1)(1936x + 11580)`
`<=> x^2[x^3 + 1929(x - 1)^2 + 3859x] - (x^2 + 1)(1936x + 11580) = 0`
`<=> x^2(x^3 + 1929x^2 + x + 1929) - (x^2 + 1)(1936x + 11580) = 0`
`<=> x^2[x^2(x + 1929) + (x + 1929)] - (x^2 + 1)(1936x + 11580) = 0`
`<=> x^2(x + 1929)(x^2 + 1) - (x^2 + 1)(1936x + 11580) = 0`
`<=> (x^2 + 1)[x^2(x + 1929) - 1936x - 11580] = 0`
`<=> (x^2 + 1)(x^3 + 1929x^2 - 1936x - 11580) = 0`
`<=> (x^2 + 1)(x^3 - 3x^2 + 1932x^2 - 5796x + 3860x - 11580) = 0`
`<=> (x^2 + 1)[x^2(x - 3) + 1932x(x - 3) + 3860(x - 3)] = 0`
`<=> (x^2 + 1)(x - 3)(x^2 + 1932x+ 3860) = 0`
`<=> (x^2 + 1)(x - 3)(x^2 + 2x + 1930x + 3860) = 0`
`<=> (x^2 + 1)(x - 3)[x(x + 2) + 1930(x + 2)] = 0`
`<=> (x^2 + 1)(x - 3)(x + 2)(x+ 1930) = 0 (1)`
Do `x^2 + 1 >= 1 > 0`
`-> (1) <=>` \(\left[ \begin{array}{l}x=3\\x=-2\\x = -1930\end{array} \right.\)
Vậy `S = {-1930 ; -2 ; 3}`
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