По заданию. С решением.

[latex]x\in (-1;1)\\\\ \frac{1}{x}+\underbrace {x+x^2+x^3+...}_{geom.progr.}=3 \; ,\; \; ODZ:\; x\ne 0\\\\S=x+x^2+x^3+...\; \; geometr.\; progressiya\; ,\\\\ b_1=x\; ,\; \; b_2=x^2\; \; \to \; \; q=\frac{b_2}{b_1}=\frac{x^2}{x}=x\; ,\; \; -1\ \textless \ x\ \textless \ 1\; ,\\\\-1\ \textless \ q\ \textless \ 1\; \; \to \; \; S=\frac{b_1}{1-q}=\frac{x}{1-x}\\\\ \frac{1}{x} + \frac{x}{1-x} =3\\\\ \frac{1-x+x^2-3(x-x^2)}{x(1-x)} =0\; ,\; \; \; \frac{4x^2-4x+1}{x(1-x)} =0\; ,\; \; \frac{(2x-1)^2}{x(1-x)} =0\\\\x=\frac{1}{2}[/latex]