Solución:
$1+x \ge 0 \Rightarrow x \ge -1$
$1-x \ge 0 \Rightarrow x \le 1$
$\Rightarrow $ Uslov: $x \in [-1, 1]$
$\sqrt{1+x} + \sqrt{1-x}=1$
$\sqrt{1+x} =1 - \sqrt{1-x}$
$(\sqrt{1+x})^2 =(1 - \sqrt{1-x})^2$
$1+x =1 - 2\sqrt{1-x} + (1-x)$
$2x-1 = -2\sqrt{1-x}$
$1-2x = 2\sqrt{1-x}$
$(1-2x)^2 = (2\sqrt{1-x})^2$
$1-4x+4x^2 = 4(1-x)$
$1-4x+4x^2 = 4-4x$
$4x^2 = 3$
$x^2 = \frac34$
$x = \pm\frac{\sqrt{3}}{2} \in [-1, 1]$
Verificación:
$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}=1 \Rightarrow \sqrt{1+0,87}+\sqrt{1-0,87}=1
\Rightarrow 1,37+0,36=1 \Rightarrow 1,73 \ne 1 \Rightarrow x=\frac{\sqrt{3}}{2}$ no es una solucion.
$\sqrt{1+(-\frac{\sqrt{3}}{2})}+\sqrt{1-(-\frac{\sqrt{3}}{2})}=1 \Rightarrow \sqrt{1-0,87}+\sqrt{1+0,87}=1
\Rightarrow 0,36+1,37=1 \Rightarrow 1,73 \ne 1 \Rightarrow x=-\frac{\sqrt{3}}{2}$ no es una solucion.
Responder: $x\in\varnothing$