Функция симметрична относительно х=-1.5. Замена t=x+1.5 приводит к биквадратному уравнению. Дальше дело техники

asnwer=13x

The root x=0 can be guessed immediately knowing that 2^4=16. After that, another root x=-3 can also be guessed from the symmetry of equation (i.e. looking for x where x+2=-1and x+1=-2 - same values under ^4, but swapped and with opposite sign). Expanding the equation, dividing it by 2x and then by x+3 gives x(x+1)(x^2+3x+6)=0. After that, checking that there are no more real roots or finding the remaining complex roots becomes trivial.

Substitute x=y-3/2 for faster solution: (y+1/2)^4+(y-1/2)^4-17=2[y^4+6(1/2)^2y^2+(1/2)^4]-17=0 Rearrange to 16y^4+24y^2-135=0 yields y^2=(-24±96)/32=-15/4 or 9/4 and y=±i√15/2 or ±3/2 This gives x=y-3/2=±i√15/2-3/2 or 0 or -3

X=0 puis il faut rechercher.

Math Olympiad: (x + 2)⁴ + (x + 1)⁴ = 17; x = ? (x + 2)⁴ + (x + 1)⁴ = 17; 0 ≥ x First method: - 1 > x ≥ 0; 17 > (x + 2)⁴ > (x + 1)⁴ (x + 2)⁴ + (x + 1)⁴ = 17 = 16 + 1 = (0 + 2)⁴ + (0 + 1)⁴; x = 0 - 2 > x ≥ - 4; 17 > (x + 1)⁴ > (x + 2)⁴ (x + 2)⁴ + (x + 1)⁴ = 17 = 1 + 16 = (- 3 + 2)⁴ + (- 3 + 1)⁴; x = - 3 x = 0 or x = - 2; Missing two complex value roots Second method: Let: y = x + 2; (x + 2)⁴ + (x + 1)⁴ = y⁴ + (y - 1)⁴ = 17 2y⁴ - 4y³ + 6y² - 4y + 1 = 17, 2(y⁴ - 2y³ + 3y² - 2y - 8) = 0 y³(y - 2) + 3y(y - 2) + 4(y - 2) = (y - 2)(y³ + 3y + 4) = 0 y³ + 3y + 4 = (y³ + 1) + 3(y + 1) = (y + 1)(y² - y + 1 + 3) = (y + 1)(y² - y + 4) y⁴ - 2y³ + 3y² - 2y - 8 = (y - 2)(y + 1)(y² - y + 4) = 0 y - 2 = 0, y = 2; y + 1 = 0, y = - 1 or y² - y + 4 = 0, y = (1 ± i√15)/2 y = x + 2 = 2, x = 0; x = - 3 or x = (1 ± i√15)/2 - 2 = (- 3 ± i√15)/2 Answer check: x = 0; x = - 3: (x + 2)⁴ + (x + 1)⁴ = 17; Confirmed as shown in First method x = (- 3± i√15)/2; y = (- 3 ± i√15)/2, y² - y + 4 = 0 (x + 2)⁴ + (x + 1)⁴ - 17 = 2(y⁴ - 2y³ + 3y² - 2y - 8) = 2(y - 2)(y + 1)(y² - y + 4) (x + 2)⁴ + (x + 1)⁴ - 17 = 0, (x + 2)⁴ + (x + 1)⁴ = 17; Confirmed Final answer: x = 0, x = - 2, Two complex value roots, if acceptable; x = (- 3 + i√15)/2 or x = (- 3 - i√15)/2

(1) (x+a)^4+(x+b)^4=c ; (2) x=t-(a+b)/2 ; (3) [t+(a-b)/2]^4+[t-(a-b)/2]^4=c ; (4) t=v*(a-b)/2 ; (5) [v+1]^4+[v-1]^4=16*c/(a-b)^4=2*d ; (6) d=8*c/(a-b)^4 {{ [v+1]^4=v^4+4*v^3+6*v^2+4*v+1 }} (6) u=v^2 ; (7) u^2+6*u+1-d=0 ……… With respect, Lidiy