2. Does the equation $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1=0$ have negative roots ( $[4]$, № 564 ) ?

口 5(1081). Find the common fraction with a denominator of 21, lying between the fractions $\frac{5}{14}$ and $\frac{5}{12}$.
Solution. Let's bring these fractions to a common denominator:

$$
\begin{aligned}
& \frac{5}{14}=\frac{5}{2 \cdot 7}=\frac{5 \cdot 6}{2 \cdot 7 \cdot 6}=\frac{30}{84} \\
& \frac{5}{12}=\frac{5}{2 \cdot 6}=\frac{5 \cdot 7}{2 \cdot 6 \cdot 7}=\frac{35}{84} .
\end{aligned}
$$

Between the fractions $\frac{30}{84}$ and $\frac{35}{84}$, there are fractions $\frac{31}{84}, \frac{32}{84}, \frac{33}{84}, \frac{34}{84}$, of which only one fraction $\frac{32}{84}$ satisfies the condition of the problem:

$$
\frac{32}{84}=\frac{8 \cdot 4}{21 \cdot 4}=\frac{8}{21}
$$

Answer: $\frac{8}{21}$.

Remark. It is advisable to ask students: are there other fractions with a denominator of 21 that satisfy the condition of the problem? If not, why? It is also useful to modify the problem: find ordinary fractions with other denominators $(22,23, \ldots)$, enclosed between the given fractions $\frac{5}{14}$ and $\frac{5}{12}$.

м(1082). What digit does the sum $54^{35}+28^{21}$ end with? Solution. Let's determine what digit $54^{35}$ ends with. $54^{1}$ ends with the digit 4,

$54^{2}$ ends with the digit 6,

$54^{3}$ ends with the digit 4,

$54^{4}$ ends with the digit 6.

It is easy to establish the pattern: even powers of 54 end with the digit 6, and odd powers end with the digit 4. Therefore, $54^{35}$ ends with the digit 4.

Let's determine what digit $28^{21}$ ends with.
$28^{1}$ ends with the digit 8,

$28^{2}$ ends with the digit 4,

$28^{3}$ ends with the digit 2,

$28^{4}$ ends with the digit 6,

$28^{5}$ ends with the digit 8.

We notice that the last digits of the powers of 28 repeat every four powers. Thus, $28^{21}=28^{4 \cdot 5+1}$, like $28^{1}$, ends with the digit 8. Therefore, $54^{35}+28^{21}$ ends with the digit $2(4+8=12)$.

Answer: the digit 2.

口 7 (1083). Solve the equation $x^{2}-2 x+y^{2}-4 y+5=0$.

Solution.

$$
\begin{gathered}
x^{2}-2 x+y^{2}-4 y+5=x^{2}-2 x+1+y^{2}-4 y+4= \\
=(x-1)^{2}+(y-2)^{2}=0
\end{gathered}
$$

Since the square of any number is a non-negative number, and the sum of squares is zero by condition, we have:

$$
(x-1)^{2}=0 \text { and }(y-2)^{2}=0
$$

i.e., $x=1, y=2$.

Answer: $x=1, y=2$.