1. Find all $x$ for which $x^{2}-10[x]+9=0$, where $[x]$ is the integer part of $x$.
Answer: $\{1 ; \sqrt{61} ; \sqrt{71} ; 9\}$.

## Solution:

Estimate the left side of the equation from below:

$$
x^{2}-10 x+9 \leq x^{2}-10[x]+9=0
$$

Therefore, $1 \leq x \leq 9$. On the other hand, $x^{2}+9=10[x]$, which means $x^{2}+9$ is an integer and divisible by 10. By checking the values of $x^{2}+9$ from 10 to 90, we find all solutions $\{1 ; \sqrt{61} ; \sqrt{71} ; 9\}$, satisfying the original equation.