The given equation is equivalent to x = ³√(4+6•³√2³(3x/4 +1/2)) x = ³√(4+6 ³√(6x+4)) (1). Let t =³√(4+6x) (2) => (1) : x =³√(4+6t) => x³ = 4+6t (1)', due to (2). From (2) => t³ =4+6x (2)' (1)' -(2)' => x³ - t³ = 6(t -x) => x³ -t³ + 6(x-t) =0 => (x-t)(x²+xt+t²+6)=0 => x=t (*) or x²+xt+t²+6=0 impossible in real numbers. From (*) and (2) => x = ³√(4+6x) => x³ = 4+6x => x³- 6x-4=0 x³-4x-2x-4=0 x(x²-4)-2(x+2)=0 x(x-2)(x+2)-2(x+2)=0 (x+2)(x²-2x-2)=0 => x = -2 or x = 1 ±√3 .
Η εξισωση γραφεται: χ=[4+6(6χ+4)^(1/3)]^(1/3). Πολλαπλασιαζω με το 6 τα 2 μελη και προσθετω το 4. 6χ+4=6[4+6(6χ+4)^(1/3)]^(1/3)+4. Θετω 6χ+4=α^3 και εχω:α^3=6(4+6α)^(1/3)+4. Θετω 6α+4=β^3. Και εχω: 6χ+4=α^3 και 6α+4=β^3. Αφαιρω κατα μελη α^3-β^3=6(β-α)......(α-β)(α^2+β^2+αβ) +6(α-β)=0 (α-β)(α^2+β^2+αβ+6)=0 α=β ή α^2+β^2+αβ =-6 2α^2+2β^2 +2αβ=-12 (α+β)^2+α^2+β^2=-12 ατοπο α=β τοτε χ=α χ^3-6χ-4=0 χ=-2 ή χ=1+-(3)^(1/2).
X1=-2, X2=(3)^(1/2)-1, X3=-(3)^(1/2)-1.
The given equation is equivalent to
x = ³√(4+6•³√2³(3x/4 +1/2))
x = ³√(4+6 ³√(6x+4)) (1).
Let t =³√(4+6x) (2) =>
(1) : x =³√(4+6t) => x³ = 4+6t (1)', due to (2).
From (2) => t³ =4+6x (2)'
(1)' -(2)' =>
x³ - t³ = 6(t -x) =>
x³ -t³ + 6(x-t) =0 =>
(x-t)(x²+xt+t²+6)=0 => x=t (*) or
x²+xt+t²+6=0 impossible in real numbers.
From (*) and (2) =>
x = ³√(4+6x) => x³ = 4+6x =>
x³- 6x-4=0 x³-4x-2x-4=0
x(x²-4)-2(x+2)=0
x(x-2)(x+2)-2(x+2)=0
(x+2)(x²-2x-2)=0 =>
x = -2 or x = 1 ±√3 .
X= -2; +-√3+(1)
x³ = 4 +6y,y³ = 6x+4 => x³-y³ = 4+6y-6x-4 =>(x-y)(x²+y²+xy)=0=>2(x²+y²+xy) =0=(x+y)²+x²+y²>0
=> x = y => x³-6x-4 = 0 , x = -2 or x²-2x-2 = 0 = (x-1)²-3 => x = 1±V3
x = -2 or x = 1±√3
Η εξισωση γραφεται: χ=[4+6(6χ+4)^(1/3)]^(1/3).
Πολλαπλασιαζω με το 6 τα 2 μελη και προσθετω το 4.
6χ+4=6[4+6(6χ+4)^(1/3)]^(1/3)+4.
Θετω 6χ+4=α^3 και εχω:α^3=6(4+6α)^(1/3)+4.
Θετω 6α+4=β^3. Και εχω:
6χ+4=α^3 και 6α+4=β^3. Αφαιρω κατα μελη
α^3-β^3=6(β-α)......(α-β)(α^2+β^2+αβ) +6(α-β)=0
(α-β)(α^2+β^2+αβ+6)=0 α=β ή α^2+β^2+αβ =-6
2α^2+2β^2 +2αβ=-12
(α+β)^2+α^2+β^2=-12 ατοπο
α=β τοτε χ=α χ^3-6χ-4=0 χ=-2 ή χ=1+-(3)^(1/2).