Solve the following equation:

$$
\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}
$$
I. solution. The denominator of the fractions cannot be 0, so $x \neq 50$ and $x \neq 49$. By bringing both sides to a common denominator, we get

$$
\frac{(x-49) 49+(x-50) 50}{50 \cdot 49}=\frac{49(x-49)+50(x-50)}{(x-50)(x-49)}
$$

(Many concluded that the denominators are equal from the equality of the numerators, although this is not necessarily true, i.e., if the numerator is 0. The editor.)

If $(x-49) \cdot 49+(x-50) \cdot 50=0$, then after rearranging, the solution of the resulting linear equation is $x=\frac{4901}{99}$, which is also a solution to (1).

If the numerator in (2) is not 0, then the denominators are equal, i.e.,

$$
50 \cdot 49=(x-50)(x-49)
$$

After rearranging, we get $x^{2}-99 x=x(x-99)=0$, so $x=0$, or $x=99$. It can be seen that both numbers are also solutions to (1).

Thus, the equation has three roots: 0, 99, and $\frac{4901}{99}$.

II. solution. By multiplying with the denominators, we get

$$
\begin{aligned}
& (x-49)^{2}(x-50) \cdot 49+(x-50)^{2}(x-49) \cdot 50= \\
& \quad=49^{2} \cdot 50 \cdot(x-49)+50^{2} \cdot 49 \cdot(x-50)
\end{aligned}
$$

If we do not perform the multiplications, then after grouping,

$$
\begin{gathered}
49(x-49) \cdot[(x-49)(x-50)-49 \cdot 50]+50(x-50)[(x-50)(x-49)-50 \cdot 49]=0 \\
\text{that is, } [49(x-49)+50(x-50)][(x-49)(x-50)-49 \cdot 50]= \\
=(99 x-4901) x(x-99)=0
\end{gathered}
$$

follows.

The roots obtained from the product, $\frac{4901}{99}, 0$ and 99, are clearly also roots of the original equation.