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

Đặt: `x+1 = t`

`(1) ⇔ t^3 + (t+2)^2 - (t+1)^2 = 0`

`⇔ t^3 + t^2 + 4t + 4 - t^2 - 2t - 1 = 0`

`⇔ t^3 + 2t + 3 = 0`

`⇔ (t^3+1) + (2t+2) = 0`

`⇔ (t+1)(t^2-t+1) + 2(t+1) = 0`

`⇔ (t+1)(t^2 - t + 3) = 0`

`⇔ t = -1.` `(do` `t^2 - t + 3 = (t-1/2 )^2 + 11/4 > 0`

`⇒ x + 1 = -1`

`⇔ x = -2.`

Vậy: `S = {-2}.`

xin 5*+ctlhn ạ, mih đng cày HSG.

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