Giải phương trình sau (x+2)^2+(x+3)^3+(x+4)^4=2

`(x+2)^2 + (x+3)^3 + (x+4)^4 = 2`

`⇔ x^4 + 17x^3 + 106x^2 + 287x + 287 = 2`

`⇔ x^4 + 17x^3 + 106x^2 + 287x + 285 = 0`

`⇔ (x+3)(x^4+17x^3+106x^2+287x+285)/(x+3) = 0`

`⇔ x^3 + 14x^2 + 64x + 95 = 0`

`⇔ (x+3)(x+5)(x^2 + 9x + 19) = 0`

`⇔`\(\left[ \begin{array}{l}x+3=0\\x+5=0\\x^2+9x+19=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=-3\\x=-5\\x=\dfrac{-9\pm\sqrt{5}}{2}\end{array} \right.\) 

Vậy `S = {-3,-5,(-9\pm\sqrt{5})/2}`