Solve the following equation:

$$
\sqrt{\frac{x-1991}{10}}+\sqrt{\frac{x-1990}{11}}=\sqrt{\frac{x-10}{1991}}+\sqrt{\frac{x-11}{1990}} .
$$
By substitution, we can verify that 2001 is a solution to the equation, indeed in this case, under the square roots, 1 stands in every case, and thus on both sides of the equation $1+1=2$ stands. We will show that our equation has no other solution.

The domain of the equation consists of real numbers greater than or equal to 1991. By introducing the variable $y = x - 2001$, we obtain the somewhat simpler equation

$$
\sqrt{\frac{y}{10}+1}+\sqrt{\frac{y}{11}+1}=\sqrt{\frac{y}{1991}+1}+\sqrt{\frac{y}{1990}+1}, \quad y \geqq -10
$$

If $y > 0$, then clearly

$$
\frac{y}{10} > \frac{y}{1991} \quad \text{and} \quad \frac{y}{11} > \frac{y}{1990}
$$

thus on the left side, the sum of the square roots of larger positive numbers stands, compared to the right side, so in this case, the value on the left side is greater.

If $-10 \leq y < 0$, then the above inequalities reverse, and under the square roots, the numbers that are 1 greater than these, i.e., non-negative numbers, stand. In this case, therefore, the value on the right side of the equation will be greater than the value on the left side. Indeed, there is no solution to the equation other than the already found $y=0$, i.e., $x=2001$.

Matolcsi Máté (Bp., Fazekas M. Gyak. Gimn., II. o. t.)