Question

ME AJUDEM A RESOLVER ESTA EQUAÇÃO DE SEGUNDO GRAU:
$\frac{X+1}{X-1} +\frac{X-1}{X+1} =\frac{13}{6}$

Answer 1

[ (x + 1)² + (x -1)² ] * 6 = 13 * [(x -1) * (x + 1)]

[x² +2x + 1 +x² -2x +1] * 6 = 13 * [x² -1]

12x² +12 = 13x² -13

12 +13 = 13x² -12x²

√25 = √x²

x = ± 5

Answer 2

Explicação passo-a-passo:

• ${\color{blue}{\frac{x+1}{x-1}+\frac{x-1}{x+1}=\frac{13}{6}}}$

• $\frac{(x+1)(x+1)+(x-1)(x-1)}{(x-1)(x+1)}=\frac{13}{6}$

• $\frac{x^2+x+x+1+x^2-x-x+1}{x^2+\cancel{x-x}+1}=\frac{13}{6}$

• $\frac{2x^2\cancel{+2x-2x}+2}{x^2-2}=\frac{13}{6}$

• $\frac{2x^2+2}{x^2-1}=\frac{13}{6}$

• $(2x^2+2).6=13.(x^2-1)$

• $12x^2+12=13x^2-13$

• $12x^2-13x^2+12+13=0$

• $-x^2+25=0\:\:\:/-1$

• $x^2-25=0$

• $x^2=25$

• $\boxed{\begin{array}{Ir}x=\pm\sqrt{25}=\pm5\end{array}}$

• Espero ter ajudado bastante!)