Question

Se x e y são números reais, tais que x²+y²/x²-y² + x²-y²/x²+y² = K. O valor de x^8 + y^8/x^8 - y^8 + x^8 - y^8/x^8 + y^8 é igual a:

a) k⁴ + 24k² + 16 / 4k³ + 16k
b) k⁴ - 24k² - 16 / 4k³ - 16k
c) k⁴ + 24k² - 16 / 4k³ - 16k
d) k⁴ + 24k² - 16 / 4k³ + 16k
e) k⁴ - 24k² + 16 / 4k³ + 16k​

Answer 1

Resposta:

(x²+y²)/(x²-y²) +(x²-y²)/(x²+y²)=k

((x²+y²)² +(x²-y²)²)/((x²-y²)(x²+y²))=k

(x⁴+2x²y²+y⁴+x⁴-2x²y²+y⁴)/(x⁴-y⁴)=k

(x⁴+y⁴+x⁴+y⁴)/(x⁴-y⁴)=k

2*(x⁴+y⁴)/(x⁴-y⁴)=k

(x⁴+y⁴)/(x⁴-y⁴)=k/2

x⁴+y⁴ =k(x⁴-y⁴)/2

x⁴+y⁴ =k*x⁴/2 -k*y⁴/2

x⁴-k*x⁴/2 =-y⁴-k*y⁴/2

x⁴*(1-k/2)=-y⁴*(1+k/2)

x⁴*(2/2-k/2)=-y⁴*(2/2+k/2)

x⁴*(2-k)=-y⁴*(2+k)

x⁴*(2-k)=-y⁴*(2+k)

x⁴ = -y⁴*(2+k)/(2-k)

x⁴ = y⁴*(2+k)/(k-2)   (i)

___________________________________________

(x⁸ + y⁸)/(x⁸ - y⁸) + (x⁸ - y⁸)/(x⁸+ y⁸) = w =?

((x⁸ + y^8)*(x⁸ + y⁸)+(x⁸ - y⁸)(x⁸ - y⁸))/((x⁸ - y⁸)*(x⁸+ y⁸)) =w

((x⁸ + y⁸)*(x⁸ + y⁸)+(x⁸ - y⁸)(x⁸ - y⁸))/(x¹⁶ - y¹⁶)=w

((x⁸ + y⁸)²+(x⁸ - y⁸)²)/(x¹⁶ - y¹⁶)=w

(2x¹⁶ + 2y¹⁶)/(x¹⁶ - y¹⁶)=w

w=2 * (x¹⁶ + y¹⁶)/(x¹⁶ - y¹⁶)  

w=2 * ((x⁴)⁴ + y¹⁶)/((x⁴)⁴ - y¹⁶)    (ii)

(i)  em (ii)

w=2 * ((y⁴*(2+k)/(2-k) )⁴ + y¹⁶)/((y⁴*(2+k)/(2-k) )⁴ - y¹⁶)

w=2 * (y¹⁶*((2+k)/(2-k) )⁴ + y¹⁶)/(y¹⁶*((2+k)/(2-k) )⁴ - y¹⁶)

simplificar por y¹⁶

w=2 * (((2+k)/(k-2) )⁴ + 1)/(((2+k)/(k-2) )⁴ - 1)

w=2 * ((2+k)⁴/(k-2)⁴  + 1)/((2+k)⁴/(k-2)⁴  - 1)

w=2 * ((2+k)⁴  + (k-2)⁴)/((2+k)⁴  - (k-2)⁴)

w=2 * ((2+k)⁴  + (k-2)⁴)/((2+k)⁴  - (k-2)⁴)

    2*(16 + 32 k + 24 k^2 + 8 k^3 + k^4 +16 - 32 k + 24 k^2 - 8 k^3 + k^4)

w=    ----------------------------------------------------------------------------------------

       (16 + 32 k + 24 k^2 + 8 k^3 + k^4-16 + 32 k - 24 k^2 + 8 k^3 - k^4)

      2*(32  + 48 k^2  + 2k^4 )

w=    ------------------------

        ( 64 k  + 16 k^3 )

         4*(16 + 24k^2  + k^4 )

w=      -----------------------------------

        ( 64 k  + 16 k^3 )

w= ( k⁴ + 24k² + 16) / (4k³ + 16k)

Letra A

Answer 2

Primeiramente, faz-se necessário analisar as Condições de Existência (C.E.) da fração algébrica situada no primeiro membro da seguinte equação:

$\mathsf{\dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}=k\qquad(i)}$

Baseado nisso, obtém-se as seguintes restrições para as incógnitas x e y:

$\mathsf{\qquad\exists\ \ \dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad x^2-y^2\,\neq\, 0\quad~~ e~~\quad x^2+y^2\,\neq\,0}\\\\\\ \mathsf{\Longleftrightarrow\quad x^2\,\neq\,y^2\quad~~e~~\quad x^2+y^2\,\neq\,0}\\\\\\ \mathsf{\Longleftrightarrow\quad \sqrt{x^2}\,\neq\,\sqrt{y^2}\quad~~e~~\quad x^2+y^2\,\neq\,0}\\\\\\ \mathsf{\Longleftrightarrow\quad |x|\,\neq\,|y|\quad~~e~~\quad x\,\neq\,0~~ou~~y\,\neq\,0}$

O que acarreta:

$\mathsf{\dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}=k\,\neq\,0}$

Postas todas as condições acima, vamos à resolução do exercício. Sendo assim, a equação (i) tornar-se-á:

$\mathsf{\qquad\quad \dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}=k}\\\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{x^2+y^2}{x^2-y^2} \cdot \dfrac{x^2+y^2}{x^2+y^2}+\dfrac{x^2-y^2}{x^2+y^2}\cdot \dfrac{x^2-y^2}{x^2-y^2}=k}$

$\mathsf{\Longleftrightarrow\quad \dfrac{\big(x^2+y^2\big)^2}{\big(x^2-y^2\big)(x^2+y^2)}+\dfrac{\big(x^2-y^2\big)^2}{\big(x^2+y^2\big)\big(x^2-y^2\big)}=k}$

$\mathsf{\Longleftrightarrow\quad\dfrac{\big(x^2+y^2\big)^2+\big(x^2-y^2\big)^2}{\big(x^2+y^2\big)\big(x^2-y^2\big)}=k\qquad(ii)}$

Para dar seguimento à resolução, é necessário ter conhecimento das três seguintes identidades algébricas:

$\mathsf{\big(a+b\big)^2+\big(a-b\big)^2=2\big(a^2+b^2\big)\,;\ \forall\ a,\,b\ \in\ \mathbb{C}}\\\\ \mathsf{\big(a+b\big)\big(a-b\big)=a^2-b^2\,;\ \forall\ a,\,b\ \in\ \mathbb{C}}\\\\ \mathsf{\big(a+b\big)^2=a^2+2ab+b^2\,;\ \forall\ a,\,b\ \in\ \mathbb{C}}$

Por fim, a equação (ii) será equivalente a:

$\mathsf{\qquad\ \ \ \dfrac{\big(x^2+y^2\big)^2+\big(x^2-y^2\big)^2}{\big(x^2+y^2\big)\big(x^2-y^2\big)}=k}\\\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{2\big(x^4+y^4\big)}{x^4-y^4}=k}$

$\mathsf{\Longleftrightarrow\quad \dfrac{x^4+y^4}{x^4-y^4}=\dfrac{k}{2}\quad~~e~~\quad \dfrac{x^4-y^4}{x^4+y^4}=\dfrac{2}{k}}$

$\mathsf{\Longrightarrow\quad\dfrac{x^4+y^4}{x^4-y^4}+\dfrac{x^4-y^4}{x^4+y^4}=\dfrac{k}{2}+\dfrac{2}{k}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^4+y^4}{x^4-y^4} \cdot \dfrac{x^4+y^4}{x^4+y^4}+\dfrac{x^4-y^4}{x^4+y^4} \cdot \dfrac{x^4-y^4}{x^4-y^4}=\dfrac{k}{2}\cdot \dfrac{k}{k}+\dfrac{2}{k}\cdot\dfrac{2}{2}}$

$\mathsf{\Longleftrightarrow\quad \dfrac{\big(x^4+y^4\big)^2}{\big(x^4-y^4\big)(x^4+y^4)}+\dfrac{\big(x^4-y^4\big)^2}{\big(x^4+y^4\big)\big(x^4-y^4\big)}=\dfrac{k^2}{2k}+\dfrac{4}{2k}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{\big(x^4+y^4\big)^2+\big(x^4-y^4\big)^2}{\big(x^4+y^4\big)\big(x^4-y^4\big)}=\dfrac{k^2+4}{2k}}$

$\mathsf{\Longleftrightarrow\quad \dfrac{2\big(x^8+y^8\big)}{x^8-y^8}=\dfrac{k^2+4}{2k}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}=\dfrac{k^2+4}{4k}\quad~~e~~\quad\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{4k}{k^2+4}}$

$\mathsf{\Longrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{k^2+4}{4k}+\dfrac{4k}{k^2+4}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{k^2+4}{4k}\cdot \dfrac{k^2+4}{k^2+4}+\dfrac{4k}{k^2+4}\cdot \dfrac{4k}{4k}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{\big(k^2+4\big)^2}{4k\big(k^2+4\big)}+\dfrac{16k^2}{\big(k^2+4\big)4k}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{\big(k^2+4\big)^2+16k^2}{4k\big(k^2+4\big)}}$

$\mathsf{\Longleftrightarrow\quad\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{k^4+8k^2+16+16k^2}{4k^3+16k}}$

${\Longleftrightarrow\quad\boxed{\mathsf{\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}=\dfrac{k^4+24k^2+16}{4k^3+16k}}}$

Alternativa correta: a).

Um grande abraço!