Question

As soluções da equação x-a/x+a + x+a/x-a = 2(a^4+1)/ a²(x²-a²) são?

Answer 1

$\dfrac{x-a}{x+a}+\dfrac{x+a}{x-a}=\dfrac{2(a^{4}+1)}{a^{2}(x^{2}-a^{2})}$

O m.m.c entre (x + a) e (x - a) é o produto (x + a)(x - a), logo:

$\dfrac{(x-a)(x-a)+(x+a)(x+a)}{(x+a)(x-a)}=\dfrac{2(a^{4}+1)}{a^{2}(x^{2}-a^{2})}\\\\\\\dfrac{x^{2}-2ax+a^{2}+x^{2}+2ax+a^{2}}{x^{2}-a^{2}}=\dfrac{2(a^{4}+1)}{a^{2}(x^{2}-a^{2})}\\\\\\\dfrac{2x^{2}+2a^{2}}{(x^{2}-a^{2})}=\dfrac{2(a^{4}+1)}{a^{2}(x^{2}-a^{2})}$

Cortando (x² - a²):

$2(x^{2}+a^{2})=\dfrac{2(a^{4}+1)}{a^{2}}\\\\\\x^{2}+a^{2}=\dfrac{a^{4}+1}{a^{2}}$

Passando a² pro outro lado:

$a^{2}(x^{2}+a^{2})=a^{4}+1\\a^{2}x^{2}+a^{2}a^{2}=a^{4}+1\\a^{2}x^{2}+a^{4}=a^{4}+1\\(ax)^{2}+0=0+1\\(ax)^{2}=1\\\sqrt{(ax)^{2}}=\pm\sqrt{1}\\ax=\pm1\\\\\\\boxed{\boxed{x=\pm\dfrac{1}{a}~~~ou~~~a=\pm\dfrac{1}{x}}}$

Answer 2

.
$\frac{x-a}{x+a} + \frac{x+a}{x-a} = \frac{2(a^4+1)}{a^2(x^2-a^2)} \\ mmc-a^2(x+a)(x-a) \\ \\ a^2(x-a)~2+a^2(x+a)^2=2a^4+2 \\ a^2(x^2-2ax+a^2)+a^2(x^2+2ax+a^2)=2a^4+2 \\ a^2x^2-2a^3x+a^4+a^2x^2+2a^3x+a^4-2a^4=2 \\ a^2x^2+a^2x^2-2a^3x+2a^3x+2a^4-2a^4=2 \\ 2a^2x^2=2 \\ x^2= \frac{2}{2a^2} = \frac{1}{a^2} \\ \\ x=\pm \sqrt{ \frac{1}{a^2} } \\ \\ x=\pm \frac{1}{a}$