1B. In the set of real numbers, solve the equation

$$
\sqrt{\frac{x^{2}-2 x+3}{x^{2}+2 x+4}}+\sqrt{\frac{x^{2}+2 x+4}{x^{2}-2 x+3}}=\frac{5}{2} \text {. }
$$
Solution. Since $x^{2}-2 x+3>0$ and $x^{2}+2 x+4>0, \forall x \in \mathbb{R}$, we have that $D_{x}=\mathbb{R}$. We make the substitution $t=\frac{x^{2}-2 x+3}{x^{2}+2 x+4}$ and we have

$$
\sqrt{t}+\frac{1}{\sqrt{t}}=\frac{5}{2} \Leftrightarrow t+2+\frac{1}{t}=\frac{25}{4} \Leftrightarrow 4 t^{2}-17 t+4=0 \Leftrightarrow t_{1}=4, t_{2}=\frac{1}{4}
$$

Returning to the substitution and checking, we notice that for $t_{1}=4$ the equation has no real solutions, while for $t_{2}=\frac{1}{4}$ we have $x_{1}=2$ and $x_{2}=\frac{4}{3}$.

3B Prove the equality

$$
\frac{1}{\log _{x} 2 \cdot \log _{x} 4}+\frac{1}{\log _{x} 4 \cdot \log _{x} 8}+\cdots+\frac{1}{\log _{x} 2^{n-1} \cdot \log _{x} 2^{n}}=\left(1-\frac{1}{n}\right)\left(\frac{1}{\log _{x} 2}\right)^{2}
$$