Đáp án ` + ` Giải thích các bước giải:

`#\text{Day 12}`

` b) 2x^3 + 3x^2 - 1 = 0 `

` -> 2x^3 + 2x^2 + x^2 - 1 = 0 `

` -> 2x^2(x + 1) + (x - 1)(x + 1) = 0 `

` -> (x + 1)(2x^2 + x - 1) = 0 `

` -> (x + 1)[2x(x + 1) - (x + 1)] = 0 `

` -> (x + 1)(x + 1)(2x - 1) = 0 `

` -> (x + 1)^2(2x - 1) = 0 `

` -> [(x + 1 = 0),(2x - 1 = 0):} `

` -> [(x = -1),(x = 1/2):} `

Vậy ` S = (-1 ; 1/2) `

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` c) x^3 - 2x^2 + 3x - 6 = 0 `

` -> (x^3 - 2x^2) + (3x - 6) = 0 `

` -> x^2(x - 2) + 3(x - 2) = 0 `

` -> (x - 2)(x^2 + 3) = 0 `

Vì ` x^2 ≥ 0 ∀ x ` nên:

` -> x^2 + 3 ≥ 3 ≥ 0 `

` -> x - 2 = 0 `

` -> x = 2 ` 

Vậy ` S = {2}`

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` b) 3x^3 - 3x^2 + 9x - 1 = 0 `

` -> 3(x^3 - x^2 + 3x - 3) = 0 `

` -> 3[x^2(x - 1) + 3(x - 1)] = 0 `

` -> 3(x - 1)(x^2 + 3) = 0 `

Vì ` x^2 + 3 > 0 ` nên:

` -> x - 1 = 0 `

` -> x = 1 `

Vậy ` S = {1} `

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` b) 2x^4 - 3x^2 - 2 = 0 `

` -> 2x^4 - 4x^2 + x^2 - 2 = 0 `

` -> 2x^2(x^2 - 2) + (x^2 - 2) = 0 `

` -> (x^2 - 2)(2x^2 + 1) = 0 `

Vì ` 2x^2 + 1 ≥ 1 > 0 ` nên:

` -> x^2 - 2 = 0 `

` -> x^2 = 2 `

` -> x = ±sqrt{2} `

Vậy ` S = {±sqrt{2}} `