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Álgebra Básica

Enviado: 16 Fev 2026, 20:38
por cicero444
Dado: [tex3]\frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=3[/tex3]
O valor de [tex3]\frac{x^{4}+y^{4}}{x^{4}-y^{4}}+\frac{x^{4}-y^{4}}{x^{4}+y^{4}}[/tex3] é:

a) [tex3]3[/tex3]
b) [tex3]\frac{3}{2}[/tex3]
c) [tex3]\frac{13}{6}[/tex3]
d) [tex3]\frac{2}{3}[/tex3]
e) [tex3]\frac{11}{6}[/tex3]

Re: Álgebra Básica

Enviado: 18 Fev 2026, 09:54
por petras
cicero444,

[tex3]\mathtt{ \frac{A}{B}+\frac{B}{A} = 3 \implies A^2+B^2 = 3AB\\\\(x^2+y^2)^2+(x^2-y^2)^2 = 3(x^2+y^2)(x^2-y^2)\\\\
x^4+2x^2y^2+y^4+ x^4-2x^2y^2+y^4=3(x^4-y^4)\\\\
2(x^4+y^4) = 3(x^4-y^4)\implies \frac{x^4+y^4}{x^4-y^4} = \frac{3}{2} \\\\
\therefore \frac{x^{4}+y^{4}}{x^{4}-y^{4}}+\frac{x^{4}-y^{4}}{x^{4}+y^{4}} = \frac{3}{2}+ \frac{1}{\frac{3}{2}} = \frac{3}{2}+\frac{2}{3} = \boxed{\frac{13}{6}_{//}}}



[/tex3]