Nv: [Médio] ÁLGEBRA (FME-01-1985-A.383)

por LinkGyn12 Qua 01 maio 2024, 17:06

Resolver a equação:

[latex] \ \sqrt[3]{x + 1} - \sqrt[3]{x - 1} = \sqrt[6]{x^2 - 1} \[/latex]

Resposta:

S={-V5/2, V5/2}

por **Giovana Martins** Qua 01 maio 2024, 19:55

[latex]\mathrm{Sejam\ a^6=x+1\ e\ b^6=x-1\ \therefore\ a^2-b^2=ab.}[/latex]

[latex]\mathrm{a^6-b^6=2\to (a^3)^2-(b^3)^2=2\to (a^3+b^3)(a^3-b^3)=2}[/latex]

[latex]\mathrm{(a+b)(a^2-ab+b^2)(a-b)(a^2+ab+b^2)=ab(a^2+ab+b^2)(a^2-ab+b^2)=2}[/latex]

[latex]\mathrm{ab(a^2+a^2-b^2+b^2)(a^2-a^2+b^2+b^2)=2\to ab(2a^2)(2b^2)=2\to a^3b^3=\frac{1}{2}}[/latex]

[latex]\mathrm{a^6b^6=\frac{1}{4}\to (x+1)(x-1)=\frac{1}{4}\to x^2-1=\frac{1}{4}\ \therefore\ \boxed {\mathrm{x=\pm \frac{\sqrt{5}}{2}}}}[/latex]

por LinkGyn12 Qua 01 maio 2024, 20:18

Giovana Martins escreveu:
[latex]\mathrm{Sejam\ a^6=x+1\ e\ b^6=x-1\ \therefore\ a^2-b^2=ab.}[/latex]

[latex]\mathrm{a^6-b^6=2\to (a^3)^2-(b^3)^2=2\to (a^3+b^3)(a^3-b^3)=2}[/latex]

[latex]\mathrm{(a+b)(a^2-ab+b^2)(a-b)(a^2+ab+b^2)=ab(a^2+ab+b^2)(a^2-ab+b^2)=2}[/latex]

[latex]\mathrm{ab(a^2+a^2-b^2+b^2)(a^2-a^2+b^2+b^2)=2\to ab(2a^2)(2b^2)=2\to a^3b^3=\frac{1}{2}}[/latex]

[latex]\mathrm{a^6b^6=\frac{1}{4}\to (x+1)(x-1)=\frac{1}{4}\to x^2-1=\frac{1}{4}\ \therefore\ \boxed {\mathrm{x=\pm \frac{\sqrt{5}}{2}}}}[/latex]

Salvou dms, mestra vlwzz