Question

1 supra x+1 +1 supra ×-1 =1supra ×patrat -1, rezolvați

Answer 1

Răspuns:

$x \in \{\pm \frac{\sqrt{2}}{2} \}$

Explicație pas cu pas:

$\frac{1}{x+1} + \frac{1}{x-1} = \frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)}\\\frac{(x+1)(x-1) + (x-1)(x+1)}{(x-1)(x+1)} =  \frac{1}{(x-1)(x+1)}\\x^2 - 1 + x^2 - 1 = 1\\2x^2 - 1 = 0\\2x^2 = 1\\x^2 = \frac{1}{2}\\x_{12} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$