1B. Solve the inequality $2^{\frac{1-x}{x}}<2^{\frac{1-L x}{2 x}}+1$, in the set of real numbers.
The solution. The inequality is written in the equivalent form $2^{\frac{1}{x}-1}<2^{\frac{1}{2 x}}+1$ and by multiplying the last inequality by 2, we obtain the equivalent inequality $2 \frac{1}{x}<2^{\frac{1}{2 x}}+2$. In the last expression, we substitute $2^{\frac{1}{2 x}}=y$ and obtain the inequality $y^{2}<y+2$, that is, $y^{2}-y-2<0 \Leftrightarrow(y+1)(y-2)<0$. Its solution is the interval $-1<y'<2$, that is, $-1<2 \frac{1}{2 x}<2$ which is equivalent to $0<2 \frac{1}{2 x}<2$. From the last, we obtain that the solution of the inequality coincides with the solution of the inequality

$$
\frac{1}{2 x}<1 \Leftrightarrow \frac{1-2 x}{2 x}<0 \Leftrightarrow x \in(-\infty, 0) \cup\left(\frac{1}{2},+\infty\right)
$$