33(1109). Find the members of the proportion $x_{1}: x_{2}=x_{3}: x_{4}$, where the first term is 6 more than the second, and the third is 5 more than the fourth. The sum of the squares of all terms is 793.
Solution. By the condition $x_{1}=x_{2}+6, x_{3}=x_{4}+5,\left(x_{2}+6\right): x_{2}=$ $=\left(x_{4}+5\right): x_{4}$.

We have a system of two equations with two unknowns:

$$
\left\{\begin{array}{l}
\left(x_{2}+6\right)^{2}+x_{2}^{2}+\left(x_{4}+5\right)^{2}+x_{4}^{2}=793 \\
\left(x_{2}+6\right) \cdot x_{4}=x_{2} \cdot\left(x_{4}+5\right)
\end{array}\right.
$$

from which

$$
\left\{\begin{array}{l}
x_{2}^{2}+6 x_{2}+x_{4}^{2}+5 x_{4}-366=0 \\
6 x_{4}=5 x_{2}
\end{array}\right.
$$

Substituting the value $x_{4}=\frac{5 x_{2}}{6}$ from the second equation of the system into the first, we get:

$$
\begin{aligned}
& x_{2}^{2}+6 x_{2}-216=0 \\
& x_{2}^{\prime}=-18, x_{2}^{\prime \prime}=12
\end{aligned}
$$

Since $x_{1}=x_{2}+6$, then $x_{1}^{\prime}=-12, x_{1}^{\prime \prime}=18$.

$$
x_{4}^{\prime}=\frac{5 x_{2}^{\prime}}{6}=-15, x_{4}^{\prime \prime}=\frac{5 x_{2}^{\prime \prime}}{6}=10, x_{3}^{\prime}=x_{4}^{\prime}+5=-10, x_{3}^{\prime \prime}=15
$$

We obtained two proportions:

$$
(-12):(-18)=(-10):(-15) \text { and } 18: 12=15: 10 .
$$

Answer: $18: 12=15: 10 ;(-12):(-18)=(-10):(-15)$.

$\triangle 34$ (1110). From a city, two pedestrians set out at different times along two mutually perpendicular roads. The speed of one of them is 4 km $/ h$, and the other's is 5 km $/ h$. Currently, the pedestrian walking at 4 km $/ h$ is 7 km from the city, and the second is 10 km away. In how many hours will the distance between the pedestrians be 25 km?

Solution. Let the distance of 25 km between the pedestrians be after $t$ hours. Then, based on the Pythagorean theorem, we can form the equation $(7+4 t)^{2}+(10+5 t)^{2}=25^{2}$, from which

$$
41 t^{2}+156 t-476=0, t=2
$$

By checking, we easily confirm that the problem is solved correctly. Indeed, after 2 hours, the first pedestrian was 15 km from the city $(7+2 \cdot 4=15)$, and the second - 20 km from the city $(10+2 \cdot 5=20)$. The distance between the pedestrians $\sqrt{15^{2}+20^{2}}=$ $=\sqrt{625}=25$, which meets the condition of the problem.

Answer: in 2 hours.

3 remark. After this problem, it is advisable to solve problem $69 \$ 8$ (No. 1003 from [2]).

$\triangle 1111$. Prove that if $z=\frac{2}{\frac{1}{a}+\frac{1}{b}}(a \neq 0, \quad b \neq 0$, $a+b \neq 0, a-b \neq 0)$, then $\frac{1}{z-a}+\frac{1}{z-b}=\frac{1}{a}+\frac{1}{b}$.

Proof. Transform the given expression $z$ :

$$
z=\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2 a b}{a+b}
$$

Substitute the value $z=\frac{2 a b}{a+b}$ into the left side of the identity to be proven. We will have:

$$
\begin{gathered}
\frac{1}{z-a}+\frac{1}{z-b}=\frac{1}{\frac{2 a b}{a+b}-a}+\frac{1}{\frac{2 a b}{a+b}-b}= \\
=\frac{a+b}{2 a b-a^{2}-a b}+\frac{a+b}{2 a b-a b-b^{2}}=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}= \\
=\frac{a+b}{a(b-a)}+\frac{a+b}{b(a-b)}=\frac{a+b}{a(b-a)}-\frac{a+b}{b(b-a)}=\frac{a b+b^{2}-a^{2}-a b}{a b(b-a)}= \\
=\frac{b^{2}-a^{2}}{a b(b-a)}=\frac{b+a}{a b}=\frac{1}{a}+\frac{1}{b} .
\end{gathered}
$$

We obtained the expression on the right side of the identity. The statement is proven.