A awesome mathematics problem | Olympiad Question | can you solve this problem | x=?

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[@mrinaldas9614]

Problem indicated is different from what is worked out

[@derwolf7810]

Solution for the task on the preview image:
root_7(x+sqrt(x^2+128)) + root_7(x+sqrt(x^2+128)) = 2
root_7(x+sqrt(x^2+128)) = 1
x+sqrt(x^2+128) = 1
sqrt(x^2+128) + (x-1) = 0
(x^2+128) - (x^2-2x+1) = 0
2x = -127
x = -63.5

[@taniacsibi6879]

Atenție la enunt ,!!!!!

[@derwolf7810]

Alterative approach:
(i)
(x+sqrt(x^2+128)) (x-sqrt(x^2+128)) = x^2-(x^2+128) = -128 != 0
root_7(x+sqrt(x^2+128)) != 0 != root_7(x-sqrt(x^2+128))
(ii)
root_7(x+sqrt(x^2+128)) + root_7(x-sqrt(x^2+128)) = 2
root_7(x+sqrt(x^2+128)) = 2 - root_7(x-sqrt(x^2+128))
root_7(x+sqrt(x^2+128))^2 = 2 root_7(x+sqrt(x^2+128)) + 2
(iii)
root_7(x+sqrt(x^2+128)) + root_7(x-sqrt(x^2+128)) = 2
root_7(x-sqrt(x^2+128)) = 2 - root_7(x+sqrt(x^2+128))
root_7(x-sqrt(x^2+128))^2 = 2 root_7(x-sqrt(x^2+128)) + 2
(iv)
a^2 = 2 a + 2
==> a^4 = (2 a + 2)^2 - 4 a^2 + 4 (2 a + 2) = 16 a + 12
==> a^7 = ((16 a + 12)^2 - 12^2/2 (2 a + 2) + 12^2/2 a^2)/ a
= (256 a^2 + 384 a + 144 - (144 a + 144) + 72 a^2 )/a
= 328 a + 240
(v)
==> x = ((x+sqrt(x^2+128)) + (x-sqrt(x^2+128)))/2
= ((328 root_7(x+sqrt(x^2+128)) + 240) + (328 root_7(x-sqrt(x^2+128)) + 240))/2
= 328 (root_7(x+sqrt(x^2+128)) + root_7(x-sqrt(x^2+128)))/2 + 240
= 328 (2)/2 + 240
= 568
q.e.d.

[@user-bk2bi4rv4h]

問題が間違っています。
前が+、後が-

[@guyhoghton399]

_a = ⁷√(x + √(x² + 128), b = ⁷√(x - √(x² + 128)_
∴ _a > b_
_a + b = 2_
_ab = -2_
_(t - a)(t - b) = 0_ has roots _a, b_
⇒ _t² - (a + b)t + ab = 0_
⇒ _t² - 2t - 2 = 0_
⇒ _t = 1 ± √3_
∴ _a = 1 + √3, b = 1 - √3_
_a⁷ + b⁷ = 2x_
⇒ _2x = (1 + √3)⁷ + (1 - √3)⁷_
⇒ _x = (⁷C₀)(√3)⁰ + (⁷C₂)(√3)² + (⁷C₄)(√3)⁴ + (⁷C₆)(√3)⁶_
= _1 + (21)(3) + (35)(9) + (7)(27) = 568_
*_x = 568_*