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13.298. Two athletes are running on the same closed track of a stadium. The speed of each is constant, but the first one takes 10 seconds less to run the entire track than the second one. If they start running from a common starting point in the same direction, they will meet again after 720 seconds. What fraction of t... | Solution.
Let $V_{1}, V_{2}$ be the speeds of the athletes, $l$ be the length of the track. According to the problem, we have the system $\left\{\begin{array}{l}\frac{l}{V_{2}}-\frac{l}{V_{1}}=10, \\ 720 V_{1}-720 V_{2}=l,\end{array}\right.$ from which $\frac{l}{V_{1}}=80, \frac{l}{V_{2}}=90$.
Then the first athlete ... | \frac{1}{80} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,384 |
13.299. Two points rotate uniformly along two concentric circles. One of them completes a full revolution 5 seconds faster than the other, and therefore manages to make two more revolutions per minute. Let the rays directed from the center of the circle to these points coincide at the beginning of the motion. Calculate... | Solution.
Let $V_{1}, V_{2}$ be the rotational speeds of the points, $l$ be the length of the circumference, and $x$ be the length of the arc between the points after 1 second of rotation. Then the angle between the rays will be $\alpha=\frac{x \cdot 360^{\circ}}{l}$ (1). According to the problem, we have the system $... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,385 |
13.300. The smaller arc between points $A$ and $B$, located on a circle, is 150 m. If the points start moving towards each other along the smaller arc, they will meet after $10 \mathrm{c}$, and if they move along the larger arc, the meeting will occur after $14 \mathrm{c}$. Determine the speeds of the points and the le... | ## Solution.
Let $V_{a}$ and $V_{b}$ be the speeds of points $A$ and $B$, and $l$ be the length of the circumference. According to the conditions, we have the system $\left\{\begin{array}{l}10\left(V_{a}+V_{b}\right)=150, \\ 14\left(V_{a}+V_{b}\right)=l-150, \\ \frac{l}{V_{a}}=\frac{90}{V_{b}},\end{array}\right.$
fro... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,386 |
13.301. In a certain mechanism, three gears of different diameters are connected in such a way that the largest one touches both of the smaller ones, and all three gears together have 60 teeth. When the largest gear is 20 teeth short of completing four full revolutions, the second and third gears make 5 and 10 full rev... | ## Solution.
Let $x, y, z$ be the number of teeth on three gears, where $x>y>z$ and $x+y+z=60$ (1). Over the time the gears rotate, the same number of teeth of each gear will come into contact, i.e., $10 z=5 y=4 x-20$ (2). Solving the system of equations (1),(2), we find $x=30, y=20, z=10$.
Answer: 10, 20, and 30 tee... | 10,20,30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,387 |
13.302. Two points move uniformly in the same direction along a circle 60 m in circumference. One of them completes a full revolution 5 seconds faster than the other. At the same time, the points coincide every 1 minute. Determine the speeds of the points. | Solution.
Let $V_{1}, V_{2}$ be the speeds of the points. According to the problem, we have the system
$$
\left\{\begin{array}{l}
\frac{60}{V_{2}}-\frac{60}{V_{1}}=5, \\
60\left(V_{1}-V_{2}\right)=60,
\end{array} \text { from which } V_{1}=4(\mathrm{M} / \mathrm{c}), V_{2}=3(\mathrm{M} / \mathrm{c})\right.
$$
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,388 |
13.303. Two wheels are connected by an infinite belt; the smaller one makes 300 more revolutions per minute than the second. The larger wheel makes 10 revolutions in a time interval that is 1 second longer than the time for the same number of revolutions of the smaller wheel. How many revolutions per minute does each w... | ## Solution.
Let $x$ be the number of revolutions of the larger wheel per minute. Then the smaller wheel makes $x+300$ revolutions. According to the condition, we have $\frac{10}{x}-\frac{1}{60}=\frac{10}{x+300}$, from which $x=300$.
Answer: 300 and 600. | 300600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,389 |
13.304. Two meshing gears $A$ and $B$ are mounted tightly: the first on shaft $O_{1}$, and the second on shaft $O_{2}$. Gear $A$ has 10 more teeth than $B$. At a certain rotational speed of shaft $O_{1}$, shaft $O_{2}$ makes 63 revolutions per minute. If the gears are swapped, then at the same speed of shaft $O_{1}$, s... | Solution.
Let gear $B$ have $x$ teeth, and $A-(x+10), V$ be the speed of shaft $O_{1}$. According to the condition, we have the system $\left\{\begin{array}{l}(x+10) V=63 x, \\ x V=28(x+10)\end{array}\right.$ from which $x=20$.
Answer: 20 and 30. | 2030 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,390 |
13.305. Find two two-digit numbers $A$ and $B$ under the following conditions. If the number $A$ is written before the number $B$, and the resulting four-digit number is divided by the number $B$, the quotient is 121. If the number $B$ is written before the number $A$, and the resulting four-digit number is divided by ... | Solution.
According to the condition, we have the system $\left\{\begin{array}{l}100 A+B=121 B, \\ 100 B+A=84 A+14,\end{array}\right.$ from which $A=42, B=35$.
Answer: 42 and 35. | 4235 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,391 |
13.306. Two hours after departure, the train stopped for 30 minutes. On the remaining part of the route to the station, repair work was being carried out, and the train was allowed a speed that was $1 / 3$ of its initial speed, as a result of which the train arrived at the station 1 hour and 10 minutes late. The next d... | Solution.
Let $x$ be the distance between the stations, $V$ be the speed of the train. According to the problem, we have the system
$$
\left\{\begin{array}{l}
2+\frac{30}{60}+\frac{x-2 V}{\frac{1}{3} V}=\frac{x}{V}+1+\frac{10}{60} \\
\frac{2 V+14}{V}+\frac{30}{60}+\frac{x-(2 V+14)}{\frac{1}{4} V}=\frac{x}{V}+\frac{50... | 196 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,392 |
13.307. Find a three-digit number whose digits form a geometric progression, given that after reducing it by 495, the resulting number is written with the same digits as the desired number but in reverse order; if the digits of the number obtained after subtraction are decreased (from left to right) by 1, 1, and 2, res... | Solution.
Let $100 x+10 y+z$ be the desired number. According to the condition, we have the system
$$
\left\{\begin{array}{l}
x z=y^{2}, \\
100 x+10 y+z-495=100 z+10 y+x, \text { from which } x=9, y=6, z=4 \\
(z-1)+(x-2)=2(y-1)
\end{array}\right.
$$
Answer: 964. | 964 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,393 |
13.308. What two-digit number is less than the sum of the squares of its digits by 11 and greater than their doubled product by 5? | Solution.
Let $10 x+y$ be the desired number. According to the condition, we have the system $\left\{\begin{array}{l}x^{2}+y^{2}=10 x+y+11 \\ 2 x y=10 x+y-5\end{array}\right.$
Subtracting the second equation from the first, we find $(x-y)^{2}=16$ and obtain two systems:
$$
\left\{\begin{array} { l }
{ x - y = 4 } \... | 95or15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,394 |
13.309. There are two gold and silver alloys. In one alloy, the quantities of these metals are in the ratio $1: 2$, in the other - $2: 3$.
How many grams should be taken from each alloy to obtain 19 g of an alloy in which gold and silver are in the ratio 7:12? | Solution.
Let $x$ (g) be the amount of the first alloy taken, which contains $\frac{1}{3} x$ gold and $\frac{2}{3} x$ silver. According to the problem, we have the system
$$
\left\{\begin{array}{l}
x+y=19 \\
\frac{1}{3} x+\frac{2}{5} y=\frac{7}{12}\left(\frac{2}{3} x+\frac{3}{5} y\right), \text{ from which } x=9(\mat... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,395 |
13.310. There is scrap steel of two grades with nickel content of 5 and $40 \%$. How much of each of these grades of metal is needed to obtain 140 tons of steel with a $30 \%$ nickel content? | Solution.
Let $x, y$ be the mass of each type of metal taken. According to the condition
Answer: 40 and 100 tons. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,396 |
13.311. Two trains, a passenger train and an express train, set off towards each other simultaneously from two points that are 2400 km apart. Each train travels at a constant speed, and at some point in time, they meet. If both trains had traveled at the speed of the express train, they would have met 3 hours earlier t... | Solution.
Let $V_{1}, V_{2}$ be the speeds of the passenger and express trains, respectively, and $t$ be the time of travel until the trains meet. According to the problem, we have the system $\left\{\begin{array}{l}t\left(V_{1}+V_{2}\right)=2400, \\ 2 V_{2}(t-3)=2400, \text { from which } V_{1}=60(\text { km } / \tex... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,397 |
13.312. When unloading a barge, four lifting cranes of the same power initially worked for 2 hours. Then, two additional cranes of lesser but equal power were put into operation. After this, it took another 3 hours to complete the unloading. If all these cranes had started working simultaneously, the unloading would ha... | Solution.
Let's assume the entire volume of work is 1. Let the time required for the more powerful crane to complete the work alone be $x$ days, for the less powerful crane be $y$ days, and for one more powerful and one less powerful crane working together be $-z$ days. The work efficiency is then $\frac{1}{x}, \frac{... | 14.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,398 |
13.313. The denominator of the fraction is less than the square of its numerator by 1. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $1 / 4$; if 3 is subtracted from both the numerator and the denominator of the original fraction, the value of the fraction will ... | Solution.
Let $\frac{x}{y}$ be the desired fraction. According to the condition, we have the system $\left\{\begin{array}{l}y=x^{2}-1, \\ \frac{x-3}{y-3}=\frac{1}{12}\end{array}\right.$, from which the desired fraction is $\frac{4}{15}$, and the fraction $\frac{8}{63}$ does not satisfy the inequality $\frac{x+2}{y+2}>... | \frac{4}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,399 |
13.314. Two gears are in mesh. Gear $A$ has 12 teeth, and gear $B$ has 54. How many revolutions will each gear make before both return to their initial position? | ## Solution.
Let wheel $A$ make $x$ revolutions, and wheel $B$ make $y$ revolutions. Since the wheels are meshed, the same number of teeth of each wheel will come into contact over the time of their rotation, i.e., $12 x = 54 y$, from which $x = \frac{9}{2} y$. The number $x$ is a natural number, so the smallest value... | 92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,400 |
13.315. The initial cost price of a unit of product was 50 rubles. During the first year of production, it increased by a certain percentage, and during the second year, it decreased (relative to the increased cost price) by the same percentage, as a result of which it became 48 rubles. Determine the percentages of the... | ## Solution.
Let the cost of a unit of production increase by x percent.
After the increase, the cost became $50+\frac{x}{100} \cdot 50=50+\frac{x}{2}$, and after the decrease - $\left(50+\frac{x}{2}\right)-\frac{x}{100} \cdot\left(50+\frac{x}{2}\right)=\left(50+\frac{x}{2}\right)\left(1-\frac{1}{100}\right) \cdot$ A... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,401 |
13.316. An enterprise increased the volume of its production annually by the same percentage. Find this percentage if it is known that over two years, the volume of production increased by two times. | ## Solution.
Let $x$ be the volume of production, and $y$ be the percentage.
In one year, the volume of production becomes $x+\frac{y}{100}=x\left(1+\frac{y}{100}\right)$, and in the second year, it becomes $-x\left(1+\frac{y}{100}\right)+\frac{y}{100}\left(x\left(1+\frac{y}{100}\right)\right)=x\left(1+\frac{y}{100}\... | 41.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,402 |
13.317. One tourist set out at 6 a.m., and the second - towards him at 7 a.m. They met at 8 a.m. and, without stopping, continued their journey. How much time did each of them spend on the entire journey, if the first arrived at the place where the second started 28 minutes later than the second arrived at the place wh... | ## Solution.
Let $V_{1}, V_{2}$ be the speeds of the first and second tourists, respectively, and $x$ be the time it takes for the second tourist to cover the distance that the first tourist covers in 2 hours. According to the problem, we have the system $\left\{\begin{array}{l}x V_{2}=120 V_{1}, \\ V_{2} x+60 V_{2}=1... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,403 |
13.318. The price of one product was reduced twice, each time by $15 \%$. For another product, which initially had the same price as the first, the price was reduced once by $x \%$. What should the number $x$ be so that after all the specified reductions, both products have the same price again? | ## Solution.
Let $z$ be the original price of the product. After the first reduction, the price became $z-0.15z=0.85z$, and after the second $0.85z-0.85 \cdot 0.15z=0.85^2z$. According to the condition, $z-\frac{x}{100}z=0.85^2z$, from which $x=(1-0.85^2) \cdot 100=27.75\%$.
Answer: $27.75\%$. | 27.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,404 |
13.319. A vessel with a capacity of 8 liters is filled with a mixture of oxygen and nitrogen, with oxygen accounting for $16 \%$ of the vessel's capacity. From this vessel, a certain amount of the mixture is released and an equal amount of nitrogen is added, after which the same amount of the mixture is released again ... | Solution.
Initially, the vessel contained $16 \cdot \frac{8}{100}=\frac{32}{25}$ liters of oxygen. The released $x$ liters of the mixture contain $\frac{16 x}{100}=\frac{4 x}{25}$ liters of oxygen. Now, for every 8 liters of the mixture in the vessel, there are $\frac{32-4 x}{25}$ liters of oxygen, which constitutes $... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,405 |
13.320. Impurities constitute $20 \%$ of the total volume of the solution. What is the smallest number of filters through which the solution must be passed so that the final impurity content does not exceed $0.01 \%$, if each filter absorbs $80 \%$ of the impurities? (It is known that $\lg 2 \approx 0.30$.) | Solution.
Impurities constitute $\frac{1}{5}$ of the solution. After the first filtration, $\left(\frac{1}{5}\right)^{2}$ of the impurities will remain, and after the $k$-th filtration, $-\left(\frac{1}{5}\right)^{k+1}$. According to the condition, $\left(\frac{1}{5}\right)^{k+1} \leq 10^{-4} ;-(k+1) \lg 5 \leq-4$, fr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,406 |
13.321. The sum of two three-digit numbers, written with the same digits but in reverse order, is 1252. Find these numbers if the sum of the digits of each of them is 14, and the sum of the squares of the digits is 84. | Solution.
Let $100 x+10 y+z$ be the desired number. According to the condition, we have the system
$$
\left\{\begin{array}{l}
100 x+10 y+z+100 z+10 y+x=1252 \\
x+y+z=14 \\
x^{2}+y^{2}+z^{2}=84
\end{array}\right.
$$
from which $x=8, y=2, z=4$.
Answer: 824 and 428. | 824428 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,407 |
13.322. Bees, when processing flower nectar into honey, remove a significant portion of the water. Studies have shown that nectar usually contains about $70 \%$ water, while the honey obtained from it contains only $17 \%$ water. How many kilograms of nectar do bees need to process to obtain 1 kg of honey? | Solution.
Let 1 kg of honey be obtained from $x$ kg of nectar. After removing water from the nectar, 300 g of other substances remain for each kilogram, and after removing water from the honey - 830 g per kilogram. We have $300 x=830$, from which $x \approx 2.77$ (kg).
Answer: $x \approx 2.77$ kg. | 2.77 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,408 |
13.323. For the production of wheat bread, as many kilograms of flour were taken as the percentage that the leaven constitutes for this flour. For the production of rye bread, 10 kg more flour was taken, i.e., as many kilograms as the percentage that the leaven constitutes for rye flour. How many kilograms of each type... | Solution.
Let $x$ be the mass of flour for wheat bread, and $(x+10)$ be for rye bread.
According to the condition, $x+x \cdot \frac{x}{100}+(x+10)+\frac{x+10}{100}(x+10)=112.5$ from which $x=35$.
Answer: 35 and 45 kg. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,409 |
13.324. In the first week of his vacation, the engineer spent a few rubles less than $3 / 5$ of the amount of money he had taken with him; in the second week, $1 / 4$ of the remainder plus 30 rubles; in the third week, $2 / 5$ of the new remainder plus 12 rubles; after which $6 / 35$ of the amount of money taken remain... | Solution.
Let $S$ be the total amount of money taken. We fill in the table:
| Period | Spent, rubles | Balance, rubles |
| :---: | :---: | :---: |
| first week | $x$ | $S-x$ |
| second week | $\frac{S-x}{4}+30$ | $\frac{3(S-x)}{4}-30$ |
| third week | $\frac{2}{5}\left(\frac{3(S-x)}{4}-30\right)+12$ | $\frac{6}{35} S... | 1160 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,410 |
13.325. 9000 parts can be manufactured on several new machines of the same design and one machine of the old design, which works twice as slowly as each of the new machines. The old machine can also be replaced by a new machine of the same design as the others. In this second option, each machine would produce 200 fewe... | Solution.
Let $x$ parts be manufactured on one new machine, $\frac{x}{2}$ - on the old one, $n$ - the number of machines. According to the condition, we have the system
$$
\left\{\begin{array}{l}
(n-1) x+\frac{x}{2}=9000, \\
\frac{9000}{n}=x-200, \quad \text { from which } n=5
\end{array}\right.
$$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,411 |
13.326. Three cars are dispatched from $A$ to $B$ at equal time intervals. They arrive in $B$ simultaneously, then proceed to point $C$, which is 120 km away from $B$. The first car arrives there one hour after the second. The third car, upon arriving at $C$, immediately turns back and meets the first car 40 km from $C... | Solution.
Let $V_{1}, V_{2}, V_{3}$ be the speeds of the cars. Compare the time intervals (see Fig. 13.16), expressing them as the ratio of distance to speed:
$A P=\frac{x}{V_{1}}, M P=\frac{x}{V_{2}}, N P=\frac{x}{V_{3}}$.
By the condition:
$$
\frac{x}{V_{1}}-\frac{x}{V_{2}}=\frac{x}{V_{2}}-\frac{x}{V_{3}}, \text ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,412 |
13.327. A total of 24 liters of liquid is distributed among three vessels. First, from the first vessel, as much liquid was poured into the other two as was already in each of them. Then, from the second vessel, as much liquid was poured into the other two as was in each of them after the first pouring. Finally, from t... | Solution.
Let's fill in the table:
| Vessel | Initially | After the first transfer | After the second transfer | After the third transfer |
| :--- | :---: | :---: | :---: | :---: |
| First | $x$ | $x-y-z$ | $2(x-y-z)$ | $4(x-y-z)$ |
| Second | $y$ | $2 y$ | $3 y-x-z$ | $6 y-2 x-2 z$ |
| Third | $z$ | $2 z$ | $4 z$... | 13,7,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,413 |
13.328. A fishing brigade planned to catch 1800 tons of fish within a certain period. For $1 / 3$ of this period, there was a storm, as a result of which the planned daily target was underachieved by 20 tons. However, in the remaining days, the brigade managed to catch 20 tons more than the daily norm, and the planned ... | ## Solution.
Let $x$ be the planned period, and $y$ be the number of centners per day according to the plan. According to the conditions, we have the system $\left\{\begin{array}{l}x y=1800, \\ \frac{x}{3}(y-20)+\left(\frac{2 x}{3}-1\right)(y+20)=1800,\end{array}\right.$ from which $y=100$.
Answer: 100 centners. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,414 |
13.329. Two workers were hired for the same period to perform seasonal work with different pay rates per day for each. The first worked $a$ days less than the term and received $r$ rubles, while the second worked $a$ days more than the term and received $s$ rubles. If the first had worked as many days as the second, an... | Solution.
Let $x$ be the duration of work. Then the first earned $\frac{r}{x-a}$ rubles per day, and the second $-\frac{s}{x+a}$. Suppose $s>r$.
According to the condition, we have $\frac{r}{x-a}(x+a)=\frac{s}{x+a}(x-a)$, from which $x=\frac{a(\sqrt{r}+\sqrt{s})}{\sqrt{s}-\sqrt{r}}$.
Answer: $\frac{a(\sqrt{r}+\sqrt{... | \frac{(\sqrt{r}+\sqrt{})}{\sqrt{}-\sqrt{r}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,415 |
13.330. Two trucks were supposed to transport a certain cargo in 6 hours. The second truck was delayed in the garage, and when it arrived at the loading site, the first truck had already transported $3 / 5$ of the total cargo; the remaining part of the cargo was transported by the second truck, and the entire cargo was... | Solution.
Let's take the entire volume of work as 1. Let $x, y$ be the weight of cargo transported by each machine in one trip. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
\frac{1}{x+y}=6, \\
\frac{3}{5 x}+\frac{2}{5 y}=12,
\end{array} \text { from which } x=\frac{1}{10}, y=\frac{1}{15}... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,416 |
13.331. Several balls of equal mass for bearings and several piston rings, also of equal mass, were made from a certain grade of metal. If the number expressing the mass of each ball in grams were 2 less than the number of rings made, and the number expressing the mass of each ring in grams were 2 more than the number ... | Solution.
Let $x$ be the number of balls, $m_{x}$ be the mass of one ball, $y$ be the number of rings, $m_{y}$ be the mass of one ring. According to the conditions, we have the system
$$
\left\{\begin{array}{l}
m_{x}+2=y \\
m_{y}=x+2 \\
x \cdot m_{x}+y \cdot m_{y}-800=2(y-x), \text{ from which } x=25, y=16 \text{ or ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,417 |
13.332. Three boys $A, E$ and $B$ agreed that during their joint boat trip, each would take turns being the captain, with the duration of each boy's captaincy being proportional to the number of points he earns by participating in a geography quiz. In the end, $A$ scored 3 points more than $B; E$ and $B$ together score... | Solution.
Let $x, 18-x, x-3$ be the points received by boys $A, B$ and $B$ respectively, and $k$ be the proportionality coefficient. According to the condition, we have
the system $\left\{\begin{array}{l}\frac{1}{10} k(x+18-x+x-3)=x+18-x+x-3+25, \\ k \cdot(18-x)=160,\end{array}\right.$
from which $x=10, k=20$. Then ... | 200140 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,418 |
13.333. A ball falls from a height of 2 m 43 cm and, upon hitting the ground, bounces back up, each time reaching $2 / 3$ of the height from which it falls again. After how many bounces will the ball rise to a height of 32 cm? | Solution.
The numbers expressing the height to which the ball rises form a geometric progression with $b_{1}=243 q$ and $q=\frac{2}{3}$. According to the condition $q_{n}=32=b_{1} q^{n-1}$, from which $n=5$.
Answer: after 5 bounces. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,419 |
13.335. If a two-digit number is divided by the product of its digits, the quotient is 3 and the remainder is 8. If the number, formed by the same digits but in reverse order, is divided by the product of the digits, the quotient is 2 and the remainder is 5. Find this number. | ## Solution.
Let $10 x+y$ be the desired number. According to the condition, we have the system
$$
\left\{\begin{array}{l}
10 x+y=3 x y+8 \\
10 y+x=2 x y+5
\end{array}\right.
$$
Answer: 53. | 53 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,420 |
13.336. Coal delivered to the warehouse is intended for two plants. Coal delivery to the first plant started on June 1st at a rate of $m$ tons daily, including Sundays, and to the second plant - on June 8th at a rate of $n$ tons daily, including Sundays. By the end of June 16th, half of the initial amount of coal remai... | Solution.
Let the total mass of coal be 1. Let $z$ be the number of days required to deliver half of all the coal to two plants simultaneously.
According to the condition, we have the system $\left\{\begin{array}{l}(8-1) m+(16-8+1)(n+m)=\frac{1}{2}, \\ z(n+m)=\frac{1}{2}, \\ (16+z) m=(16-8+1+z) n,\end{array}\right.$
... | June28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,421 |
13.337. At a plant that manufactures instant coffee, a batch of coffee beans for processing was delivered in late May. One mechanism, which grinds the beans, was put into operation on Monday, June 1st, and ground $m$ kg daily. Starting from June 6th, a second mechanism was added, which ground $n$ kg daily. By the end o... | Solution.
Let the mass of the entire batch of grains be 1. Let $z$ be the number of days required to grind half of all the grains by two mechanisms simultaneously. According to the condition, we have the system
$$
\left\{\begin{array}{l}
(6-1) m+(10-6)(n+m)=\frac{1}{2}, \\
z(n+m)=\frac{1}{2}, \\
(10-1+z) m=(10-6+z) n... | June17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,422 |
13.338. The recording of a six-digit number starts with the digit 2. If this digit is moved from the first position to the last, keeping the order of the other five digits, the newly obtained number will be three times the original number. Find the original number. | Solution.
The original six-digit number has the form $2 \cdot 10^{5}+x$. After moving the digit 2 to the last place, we get the number $10 x+2$. According to the condition, $10 x+2=3\left(2 \cdot 10^{5}+x\right)$, from which $x=85714$.
Answer: 285714. | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,423 |
13.339. It was necessary to take several liters of liquid at a temperature of $a^{\circ}$ and another amount of the same liquid at a temperature of $b^{\circ}$, to obtain a mixture temperature of $c^{\circ}$. However, the second liquid was taken in the amount that was supposed to be taken of the first, and vice versa. ... | Solution.
Let $m_{\mathrm{a}}, m_{\mathrm{b}}$ be the volume of the liquid taken at temperatures $a^{\circ}$ and $b^{\circ}$ respectively, $x$ be the temperature of the resulting mixture. According to the condition, we have the system $\left\{\begin{array}{l}m_{a}(a-c)+m_{b}(b-c)=0, \\ m_{b}(a-x)+m_{a}(b-x)=0,\end{arr... | +b- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,424 |
13.340. It is known that the difference between the variable quantities $z$ and $y$ is proportional to the quantity $x$, and the difference between the quantities $x$ and $z$ is proportional to the quantity $y$. The coefficient of proportionality is the same and is a positive integer $k$. A certain value of the quantit... | Solution.
From the condition, we have the system $\left\{\begin{array}{l}z-y=k x, \\ x-z=k y, \\ x-y=\frac{3}{5} z\end{array}\right.$
From the first two equations, we express $\frac{x}{z}$ and $\frac{y}{z}$ in terms of $k$, then substitute the obtained expressions for $\frac{x}{z}$ and $\frac{y}{z}$ into the third eq... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,425 |
13.341. Three workers participated in a competition. The first and third of them produced twice as much product as the second, while the second and third produced three times as much as the first. What place did each worker take in the competition? | ## Solution.
Let $x_{1}, x_{2}, x_{3}$ be the volume of production by the first, second, and third worker, respectively. According to the problem, we have the system
$$
\left\{\begin{array}{l}
x_{1}+x_{3}=2 x_{2}, \\
x_{2}+x_{3}=3 x_{1},
\end{array}\right.
$$
Answer: In first place - the third, in second place - the... | 5:4:3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,426 |
13.342. The distance between stations $A$ and $B$ is 360 km. At the same time, two trains depart from $A$ and $B$ towards each other. The train that departs from $A$ arrives at station $B$ no earlier than 5 hours later. If its speed were 1.5 times greater than it actually is, it would meet the second train earlier than... | ## Solution.
Let $V_{A}, V_{B}$ be the speeds of the trains departing from $A$ and $B$ respectively. According to the condition, we have the system
$$
\left\{\begin{array}{l}
V_{A} \leq \frac{360}{5} \\
\frac{360}{\frac{3}{2} V_{A}+V_{B}}180-\frac{3}{2} V_{A}$. At the maximum speed $V_{A}=72$ km/h, we have
$$
V_{B}-... | V_{B}>V_{A} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,427 |
13.343. It is conjectured that the expression
$$
(x+a)(x+2 a)(x+3 a)(x+4 a)+a^{4}
$$
is the square of a quadratic polynomial of the form $x^{2}+p x+q a^{2}$. How can this statement be verified and the coefficients $p$ and $q$ be found? | Solution.
To verify the given statement, we equate the expressions:
$$
(x+a)(x+2a)(x+3a)(x+4a)+a^{4}=\left(x^{2}+p x+q a^{2}\right)^{2}
$$
expand the brackets
$$
\begin{aligned}
& x^{4}+10 a x^{3}+35 a^{2} x^{2}+50 a^{3} x+25 a^{4}=x^{4}+2 p x^{3}+\left(2 q a^{2}+p^{2}\right) x^{2}+ \\
& +2 p q a^{2} x+q^{2} a^{4}
... | p=5q=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,428 |
13.346. The volume of substance A is half the sum of the volumes of substances B and C, and the volume of substance B is 1/5 of the sum of the volumes of substances A and C. Find the ratio of the volume of substance C to the sum of the volumes of substances A and B. | Solution.
Given that $2 V_{A}=V_{B}+V_{C}$ and $5 V_{B}=V_{A}+V_{C}$. Let $V_{A}=x V_{C}$ and $V_{B}=y V_{C}$. Then we get the system $\left\{\begin{array}{l}2 x-y=1, \\ -x+5 y=1,\end{array}\right.$ from which $x=\frac{2}{3}, y=\frac{1}{3}$. Therefore, $\frac{V_{C}}{V_{A}+V_{B}}=\frac{1}{x+y}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,431 |
13.347. Find two numbers under the following conditions: their sum is 1244; if you append the digit 3 to the end of the notation of the first number, and remove the digit 2 from the end of the notation of the second number, two equal numbers will be obtained. | ## Solution.
Let $x, y$ be the required numbers. According to the condition, we have the system $\left\{\begin{array}{l}x+y=1244, \\ 10 x+3=\frac{y-2}{10},\end{array}\right.$
from which $x=12, y=1232$.
Answer: 12 and 1232. | 121232 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,432 |
13.348. A passenger train departed from station $A$ towards station $B$ (Fig. 13.17). After $a$ hours, a train "Strela" departed from station $B$ towards station $A$. The trains met at station $C$.
![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1215.jpg?height=149&width=667&top_left_y=1619&top_le... | ## Solution.
Let $V_{1}, V_{2}$ be the speeds of the train and "Sarrow" respectively. According to the condition $\frac{c V_{2}-a V_{1}}{V_{1}}=b \frac{V_{1}}{V_{2}}$, from which $\frac{V_{2}}{V_{1}}=\frac{a+\sqrt{a^{2}+4 b c}}{2 c}$. Then the train traveled from station $A$ to $B a+\left(c \frac{V_{2}}{V_{1}}-a\right... | \frac{+2b+\sqrt{^{2}+4}}{2}\text{}\frac{2-+\sqrt{^{2}+4}}{2}\text{0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,433 |
13.349. From post office $A$ to village $B$, one needs to walk 9 km. The postman walks the entire path there and back, without stopping in the village, in 3 hours and 41 minutes. The road from $A$ to $B$ goes uphill first, then on flat ground, and finally downhill. Over what distance does the road run on flat ground, i... | ## Solution.
Let $x, y, z$ be the lengths of the road uphill, on flat ground, and downhill, respectively. According to the problem, we have the system
from which $y=4$ (km).
Answer: 4 k... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,434 |
13.350. Two motorists met halfway between cities $A$ and $B$. At the meeting, it turned out that the first one from $A$ had left earlier than the second one from $B$ by as many hours as half the time (also in hours) that would have passed until their meeting if they had left from the same points at the same time, on th... | Solution.
Let $V_{1}, V_{2}$ be the speeds of the cars from cities $A$ and $B$ respectively, and $x$ be the distance between cities $A$ and $B$. Then the driver from $A$ left
$\frac{x}{2\left(V_{1}+V_{2}\right)}$ (hours) earlier than the driver from $B$ and traveled $\frac{x V_{1}}{2\left(V_{1}+V_{2}\right)}$ (km).
... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,435 |
13.351. The road from post office $A$ to village $B$ goes uphill for 2 km, then on flat ground for 4 km, and finally downhill for 3 km. The postman takes 2 hours and 16 minutes to travel from $A$ to $B$, and 2 hours and 24 minutes to return. If the final destination of his route were located on the same road but twice ... | ## Solution.
Let $V_{1}, V_{2}, V_{3}$ be the speeds of the postman on the uphill, flat, and downhill sections, respectively. According to the problem, we have the system
$$
\left\{\begin{array}{l}
2 V_{1}+4 V_{2}+3 V_{3}=2 \frac{16}{60} \\
3 V_{1}+4 V_{2}+2 V_{3}=2 \frac{24}{60} \\
2 V_{1}+\left(\frac{2+4+3}{2}-2\ri... | )3\mathrm{}/\mathrm{};b)4\mathrm{}/\mathrm{};)5\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,436 |
13.352. A girl, an acquaintance of a young man sitting by the window of a tram, was walking towards the moving tram. 8 seconds after she passed by the window, the young man got off the tram and followed her. How much time passed from that moment until he caught up with the girl? The young man's speed is twice the girl'... | Solution.
Let $x, 2x, 10x$ be the speeds of the girl, the boy, and the tram, respectively. According to the problem, $8 \cdot x + 8 \cdot 10x + t \cdot x = t \cdot 2x$, where $t$ is the desired time. Solving the equation, we find $t=88$ (s).
Answer: 88 s. | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,437 |
13.353. When multiplying two positive numbers, one of which is 75 more than the other, the product was mistakenly obtained as 1000 less than the true product. As a result, when dividing (during verification) the incorrect product by the smaller of the multipliers, a quotient of 227 and a remainder of 113 were obtained.... | Solution.
Let $x, y$ be the required numbers, $z$ be the erroneous product of numbers $x$ and $y$.
According to the condition, we have the system $\left\{\begin{array}{l}x-y=75, \\ x y=z+1000 \\ z=227 y+113,\end{array}\right.$ from which $x=234, y=159$.
Answer: 159 and 234. | 159234 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,438 |
13.354. When multiplying two numbers, one of which is 10 more than the other, a student made a mistake by reducing the tens digit of the product by 4. When dividing the obtained product by the smaller multiplier to check the answer, he got a quotient of 39 and a remainder of 22. Find the multipliers. | ## Solution.
Let $x, y$ be the factors, $z$ be the erroneous product of numbers $x$ and $y$.
According to the condition, we have the system $\left\{\begin{array}{l}x-y=10, \\ x y=z+40, \\ z=39 y+22,\end{array}\right.$ from which $x=41, y=31$.
Answer: 31 and 41. | 3141 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,439 |
13.355. A car, having traveled a distance from $A$ to $B$, equal to 300 km, turned back and after 1 hour 12 minutes from leaving $B$, increased its speed by 16 km/h. As a result, it spent 48 minutes less on the return trip than on the trip from $A$ to $B$. Find the original speed of the car. | Solution.
Let $V$ be the original speed of the car. According to the condition, we have
$$
\frac{300}{V}-\left(1 \frac{12}{60}+\frac{300-1 \frac{12}{60} V}{V+16}\right)=\frac{48}{60}, \quad \text { from which } V=60 \text { (km/h) }
$$
Answer: 60 km/h. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,440 |
13.356. The distance between points $A$ and $B$ is 308 m. A point moves from point $A$ towards $B$, covering 15 m in the first second and 1 m less in each subsequent second. Another point moves from point $B$ in the opposite direction, covering 20 m in the first second and 3 m more in each subsequent second. At what di... | Solution.
Let's find the laws of motion of the points.
For the point from $A:\left\{\begin{array}{l}15=1 \cdot V_{1}-\frac{a_{1} \cdot 1^{2}}{2}, \\ 15+14=2 V_{1}-\frac{a_{1} \cdot 2^{2}}{2},\end{array}\right.$ from which $V_{1}=15.5(\mathrm{~m} / \mathrm{c})$, $a_{1}=1\left(\mathrm{m} / \mathrm{c}^{2}\right)$, and t... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,441 |
13.357. A cyclist traveled 96 km 2 hours faster than he had planned. During this time, for each hour, he traveled 1 km more than he had planned to travel in 1 hour and 15 minutes. At what speed did he travel? | Solution.
Let $V$ be the speed at which the cyclist was traveling, $\frac{1 \cdot V-1}{1 \frac{15}{60}}$ - the speed at which the cyclist intended to travel. According to the condition $\frac{96}{V}+2=\frac{96}{\frac{V-1}{1 \frac{1}{4}}}$, from which $V=16$ (km/h).
Answer: 16 km/h. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,442 |
13.358. Find a six-digit number starting with the digit 1 such that if this digit is moved to the end, the resulting number is three times the original number. | Solution.
The original six-digit number has the form $1 \cdot 10^{5}+x$. After moving the digit 1 to the last place, we get the number $10 x+1$. According to the condition, $10 x+1=3\left(1 \cdot 10^{5}+x\right)$, from which $x=42857$.
Answer: 142857. | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,443 |
13.359. Find two two-digit numbers with the following property: if to the larger of the sought numbers a zero is appended on the right and then the smaller number, and to the smaller number the larger number is appended on the right and then a zero, then from the two five-digit numbers thus obtained, the first, when di... | Solution.
Let $x, y$ be the required numbers. According to the problem, we have the system
$$
\left\{\begin{array}{l}
1000 x+y=2(1000 y+10 x)+590, \\
2 x+3 y=72,
\end{array} \text { from which } x=21, y=10\right.
$$
Answer: 21 and 10.
13. 360. A cyclist departs from $A$ to $B$. The distance from $A$ to $B$ is 60 km... | 0<V\leq20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,444 |
13.361. A red pencil costs 27 kopecks, a blue one - 23 kopecks. No more than 9 rubles 40 kopecks can be spent on purchasing pencils. It is necessary to purchase the maximum possible total number of red and blue pencils. At the same time, the number of red pencils should be as few as possible, but the number of blue pen... | ## Solution.
Let $x$ red and $y$ blue pencils be bought. According to the condition, $27 x + 23 y \leq 940$ and $y - x \leq 10$. Let's construct the lines $27 x + 23 y = 940$ (1) and $y - x = 10$ (2). From Fig. 13.18, it is clear that these lines intersect at point $A$, the coordinates of which satisfy equations (1) a... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,445 |
13.362. A certain alloy consists of two metals in the ratio $1: 2$, while another contains the same metals in the ratio $2: 3$. How many parts of each alloy should be taken to obtain a third alloy containing the same metals in the ratio $17: 27?$ | Solution.
Let $x$ parts of the first metal and $y$ parts of the second be taken. Then
$$
\frac{1}{3} x+\frac{2}{5} y=\frac{17}{44}(x+y), \text { hence } \frac{y}{x}=\frac{35}{9}
$$
Answer: 9 and 35 parts. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,446 |
13.363. A certain alloy contains metals $A$ and $B$ in the ratio $m:n$, another - the same metals in the ratio $p:q$. What quantities of the first and second alloys need to be taken to obtain 1 kg of a third alloy with equal content of metals $A$ and $B$? | Solution.
Let $x$ be the mass of the first alloy, in which $\frac{m}{n+m} x$ is metal $A$ and $\frac{n}{n+m} x$ is metal $B$.
$\frac{p}{p+q}(1-x)$ is metal $A$ and $\frac{q}{p+q}(1-x)$ is metal $B$.
According to the condition, $\frac{n}{n+m} x+\frac{p}{p+q}(1-x)=\frac{n}{n+m} x+\frac{q}{p+q}(1-x)$, from which $x=\fr... | \frac{1}{2}+\frac{p-nq}{2(np-q)};\frac{1}{2}-\frac{p-nq}{2(np-q)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,447 |
13.364. The base of the power was increased by $k$ times, and the exponent was decreased by the same factor, as a result of which the power itself did not change. Find the base of the power that has such a property. | ## Solution.
Let $x, y$ be the base and the exponent. According to the condition, $(k x)_{k}^{\frac{y}{k}}=x^{y}$ or $(k x)_{k}^{\frac{1}{k}}=x$, from which $x=\sqrt[k-1]{k}$.
Answer: $\sqrt[k-1]{k}$ | \sqrt[k-1]{k} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,448 |
13.366. On a river with a current speed of 5 km/h, there are piers $A, B$, and $C$ in the direction of the current, with $B$ located halfway between $A$ and $C$. From pier $B$, a raft and a boat depart simultaneously in the direction of the current towards pier $C$, and the boat heads towards pier $A$, with the boat's ... | Solution.
Let $x$ be half the distance between $A$ and $C$.
According to the condition $\frac{x}{V-5}+\frac{2 x}{V+5}>\frac{x}{2}$, from which $5<V<15$ (km/h).
Answer: $5<V<15$ km/h. | 5<V<15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,450 |
13.367. Several students decided to buy a tape recorder priced between 170 and 195 dollars. However, at the last moment, two of them refused to participate in the purchase, so each of the remaining had to contribute 1 dollar more. What was the price of the tape recorder? | Solution.
Let $x$ be the amount each student was originally supposed to contribute; $y$ be the number of students. According to the problem,
\[
\left\{
\begin{array}{l}
2 x = y - 2, \\
x y \geq 170, \\
x y \leq 195,
\end{array}
\right.
\left\{
\begin{array}{l}
x = \frac{y}{2} - 1, \\
y\left(\frac{y}{2} - 1\right) \geq... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,451 |
13.368. To transport cargo from one place to another, a certain number of trucks of the same capacity were required. Due to road damage, 0.5 tons less cargo had to be loaded onto each truck than initially planned, which is why 4 additional trucks of the same capacity were required. The total weight of the transported c... | Solution.
Let the mass of the transported cargo be $x$ tons, and the number of trucks be $n$. According to the problem, $x=\left(\frac{x}{n}-\frac{1}{2}\right)(n+4)$ or $n^{2}+4 n-8 x=0$, from which $n=2(-1+\sqrt{1+2 x})$ ( $n$ is a natural number). In the interval $55 \leq x \leq 64$, only $x=60$ fits. Then $n=2(-1+1... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,452 |
13.369. Around the house, lindens and birches are planted, and their total number is more than 14. If the number of lindens is doubled and the number of birches is increased by 18, then the number of birches will be greater than the number of lindens. If, however, the number of birches is doubled without changing the n... | Solution.
Let $x, y$ be the number of lindens and birches, respectively. According to the condition, we have the system of inequalities $\left\{\begin{array}{l}x+y>14, \\ y+18>2 x, \\ x>2 y .\end{array}\right.$
Construct the lines $x+y=14$ (1), $y+18=2 x$ (2), $x=2 y$ (3). From Fig. 13.20, it is clear that the point ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,453 |
13.371. A section of an embankment for a railway is being constructed, 100 m long, with a cross-section in the form of an isosceles trapezoid with a lower base of 5 m, an upper base of no less than 2 m, and a slope angle of $45^{\circ}$. What height $h$ should this embankment have so that the volume of earthworks is no... | ## Solution.
From Fig. 13.21, we have $x=\frac{a-b}{2} ; h=x \operatorname{tg} 45^{\circ}=\frac{a-b}{2}$. The area of trapezoid $A B C D S=\frac{a+b}{2} h=\frac{a^{2}-b^{2}}{4}$.
According to the condition $400 \leq \frac{a^{2}-b^{2}}{4} \cdot 100 \leq 500$, from which $\sqrt{5} \leq b \leq 3$ (m). Then $\frac{a-3}{2... | 1\leq\leq\frac{5-\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,455 |
1.1 $\frac{\frac{1}{a}-\frac{1}{b+c}}{\frac{1}{a}+\frac{1}{b+c}}\left(1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right): \frac{a-b-c}{a b c}$;
$$
a=0.02, b=-11.05, c=1.07
$$
$1.2\left(\frac{1}{t^{2}+3 t+2}+\frac{2 t}{t^{2}+4 t+3}+\frac{1}{t^{2}+5 t+6}\right)^{2} \frac{(t-3)^{2}+12 t}{2}$. | 1.1 Let's introduce the notations:
$$
\frac{\frac{1}{a}-\frac{1}{b+c}}{\frac{1}{a}+\frac{1}{b+c}}=A ; \quad 1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}=B ; \quad \frac{a-b-c}{a b c}=C
$$
Then the given expression will take the form $A \cdot B: C$. In the expression $A$, the permissible values are $a \neq 0, b \neq -c$. Bringin... | 0.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,456 |
$1.3 \quad \frac{(a-b)^{2}+a b}{(a+b)^{2}-a b}: \frac{a^{5}+b^{5}+a^{2} b^{3}+a^{3} b^{2}}{\left(a^{3}+b^{3}+a^{2} b+a b^{2}\right)\left(a^{3}-b^{3}\right)}$. | ## 1.3 Let
$$
\begin{aligned}
& A=\frac{(a-b)^{2}+a b}{(a+b)^{2}-a b} ; B=a^{5}+b^{5}+a^{2} b^{3}+a^{3} b^{2} \\
& C=\left(a^{3}+b^{3}+a^{2} b+a b^{2}\right)\left(a^{3}-b^{3}\right)
\end{aligned}
$$
Transform each of the written expressions using factorization formulas (1.9), (1.10), (1.13), (1.14), and grouping of t... | -b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,457 |
$1.4 \frac{3 a^{2}+2 a x-x^{2}}{(3 x+a)(a+x)}-2+10 \frac{a x-3 x^{2}}{a^{2}+9 x^{2}}$.
$1.5\left(\left(\frac{x}{y-x}\right)^{-2}-\frac{(x+y)^{2}-4 x y}{x^{2}-x y}\right)^{2} \frac{x^{4}}{x^{2} y^{2}-y^{4}}$. | 1.4 According to formula (1.8), we have $a^{2}-9 x^{2}=(a-3 x)(a+3 x)$. Simplify the fraction $\frac{a x-3 x^{2}}{(a-3 x)(a+3 x)}$ by canceling $a-3 x$, assuming $a \neq 3 x$; we get $\frac{x}{a+3 x}$. Next, using formula (1.15), factor the numerator of the first fraction:
$$
\begin{aligned}
& 3 a^{2}+2 a x-x^{2}=-\le... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,458 |
$1.6 \frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}} ; a=7,4, b=\frac{5}{37}$. | 1.6 Perform the following transformations in the numerator and denominator:
$$
\begin{aligned}
& \frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}=\frac{b+a-2 c}{a b} \\
& \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}=\frac{b^{2}+a^{2}+2 a b-4 c^{2}}{a^{2} b^{2}}= \\
& =\frac{(b+a)^{2}-(2 c)^{2}}{a^{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,459 |
$1.7 \quad \frac{a^{-1}-b^{-1}}{a^{-3}+b^{-3}}: \frac{a^{2} b^{2}}{(a+b)^{2}-3 a b}\left(\frac{a^{2}-b^{2}}{a b}\right)^{1}$;
$$
a=1-\sqrt{2}, b=1+\sqrt{2}
$$ | 1.7 Using formulas (1.7), (1.8), (1.9) and (1.13), we transform the given expression:
$$
\begin{aligned}
& \frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a^{3}}+\frac{1}{b^{3}}} \cdot \frac{a^{2}-a b+b^{2}}{a^{2} b^{2}} \cdot \frac{a b}{(a-b)(a+b)}= \\
& =\frac{b-a}{b^{3}+a^{3}} \cdot \frac{a^{3} b^{3}}{a b} \cdot \frac{a^{2... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,460 |
$1.9 \quad \frac{1+(a+x)^{-1}}{1-(a+x)^{-1}}\left(1-\frac{1-\left(a^{2}+x^{2}\right)}{2 a x}\right) ; x=\frac{1}{a-1}$. | 1.9 We have:
1) $1-\frac{1-\left(a^{2}+x^{2}\right)}{2 a x}=\frac{2 a x-1+a^{2}+x^{2}}{2 a x}=\frac{(a+x)^{2}-1}{2 a x}=$
$$
=\frac{(a+x-1)(a+x+1)}{2 a x}, a \neq 0, x \neq 0
$$
2) $\left(1+\frac{1}{a+x}\right):\left(1-\frac{1}{a+x}\right)=\frac{a+x+1}{a+x-1}, a+x-1 \neq 0$;
3) $\frac{a+x+1}{a+x-1} \cdot \frac{(a+x-... | \frac{^{3}}{2(-1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,462 |
1.10
$$
\frac{\left(m^{2}-\frac{1}{n^{2}}\right)^{m}\left(n+\frac{1}{m}\right)^{n-m}}{\left(n^{2}-\frac{1}{m^{2}}\right)^{n}\left(m-\frac{1}{n}\right)^{m-n}}
$$
Without resorting to approximate calculations, simplify the numerical expressions (1.11-1.13):
1.11 $(4 \sqrt[3]{1+2 \sqrt{3}}-\sqrt[6]{13+4 \sqrt{3}}) \sqr... | 1.10 Using formulas (1.8), (1.5) and (1.3), we find
$$
\frac{\left(m-\frac{1}{n}\right)^{m}\left(m+\frac{1}{n}\right)^{m}\left(n+\frac{1}{m}\right)^{n}\left(m-\frac{1}{n}\right)^{n}}{\left(n-\frac{1}{m}\right)^{n}\left(n+\frac{1}{m}\right)^{n}\left(m-\frac{1}{n}\right)^{m}\left(n+\frac{1}{m}\right)^{m}}
$$
By canceli... | (\frac{}{n})^{+n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,463 |
$1.14 \frac{x^{4}+2 x^{2}-3 x+1}{x^{2}+\sqrt{3 x}+1}+2\left(\sqrt[6]{27 x^{3}}-\frac{1}{2}\right)$. | 1.14 Let's denote the fraction as $A$, and the expression in parentheses as $B$, then the given expression will take the form $A+2 B$. Note that for $\sqrt{3 x}$ and $\sqrt[6]{27 x^{3}}$, only values $x \geqslant 0$ are permissible.
Using formula (1.9), we extract a complete square in the numerator of the fraction $A$... | x^{2}+\sqrt{3x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,464 |
$1.15 \frac{\sqrt{a^{2}-4 a b+4 b^{2}}}{\sqrt{a^{2}+4 a b+4 b^{2}}}-\frac{8 a b}{a^{2}-4 b^{2}}+\frac{2 b}{a-2 b}, 0<a<2 b$. | ### 1.15 We have
$\sqrt{a^{2}-4 a b+4 b^{2}}=\sqrt{(a-2 b)^{2}}=|a-2 b|=2 b-a$,
$\sqrt{a^{2}+4 a b+4 b^{2}}=\sqrt{(a+2 b)^{2}}=|a+2 b|=2 b+a ;$
Here, formulas (1.19), (1.10), and (1.23) were used. Therefore,
$\frac{\sqrt{a^{2}-4 a b+4 b^{2}}}{\sqrt{a^{2}+4 a b+4 b^{2}}}=\frac{2 b-a}{2 b+a}$.
## Now we find
$$
\be... | \frac{}{2b-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,465 |
$1.16 \frac{x^{2}+4 x-5+(x-5) \sqrt{x^{2}-1}}{x^{2}-4 x-5+(x+5) \sqrt{x^{2}-1}}, x>1$.
$\mathbf{1 . 1 7}$
$$
9 b^{\frac{4}{3}}-\frac{a^{\frac{3}{2}}}{b^{2}}
$$ | 1.16 Let us denote the given expression by $f(x)$. Further, using formula (1.15), we factorize the quadratic trinomials in the numerator and the denominator of the fraction:
$$
f(x)=\frac{(x+5)(x-1)+(x-5) \sqrt{x^{2}-1}}{(x-5)(x+1)+(x+5) \sqrt{x^{2}-1}}
$$
Since $x>1$, by relation (1.21) we have
$$
\begin{aligned}
&... | \sqrt{\frac{x-1}{x+1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,466 |
$1.24 \frac{\sqrt{\frac{a b c+4}{a}+4 \sqrt{\frac{b c}{a}}}}{\sqrt{a b c}+2} ; a=0.04$. | 1.24 Assuming $a b c \neq 0$, we have:
$$
\begin{aligned}
& \frac{a b c+4}{a}+4 \sqrt{\frac{b c}{a}}=b c+\frac{4}{a}+4 \sqrt{\frac{b c}{a}}= \\
& =(\sqrt{b c})^{2}+4 \sqrt{\frac{b c}{a}}+\left(\frac{2}{\sqrt{a}}\right)^{2}=\left(\sqrt{b c}+\frac{2}{\sqrt{a}}\right)^{2} \\
& \frac{\sqrt{\left(\sqrt{b c}+\frac{2}{\sqrt{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,469 |
1.25 Check the validity of the equality
$\frac{\sqrt{2}-1}{\sqrt{2}+1}=\sqrt[3]{\frac{10-7 \sqrt{2}}{10+7 \sqrt{2}}}$ | 1.25 We will proceed from the fact that if $a^{3}=b$, then $a=\sqrt[3]{b}$.
Let $a=\frac{\sqrt{2}-1}{\sqrt{2}+1}$. Then, using formulas (1.11) and (1.12), we find
$$
a^{3}=\frac{2 \sqrt{2}-3 \cdot 2+3 \sqrt{2}-1}{2 \sqrt{2}+3 \cdot 2+3 \sqrt{2}+1}=\frac{5 \sqrt{2}-7}{5 \sqrt{2}+7}
$$
Multiplying the numerator and th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,470 |
1.26 Given the expression
$$
\begin{aligned}
& \frac{x^{3}-a^{-\frac{2}{3}} \cdot b^{-1}\left(a^{2}+b^{2}\right) x+b^{\frac{1}{2}}}{b^{\frac{3}{2}} x^{2}} \\
& \text { Substitute } x=a^{\frac{2}{3} b}-\frac{1}{2} \text { and simplify the result }
\end{aligned}
$$
Eliminate the irrationality in the denominator of the ... | 1.26 After performing the specified substitution, we find
$$
\begin{aligned}
& \frac{\left(a^{\frac{2}{3}} b^{-\frac{1}{2}}\right)^{3}-a^{-\frac{2}{3}+\frac{2}{3}} b^{-1-\frac{1}{2}}\left(a^{2}+b^{2}\right)+b^{\frac{1}{2}}}{b^{\frac{3}{2}}\left(a^{\frac{2}{3}} b^{\left.-\frac{1}{2}\right)^{2}}\right.}= \\
& =\frac{a^{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,471 |
$1.27 \frac{2-\sqrt{2}-\sqrt{3}}{2+\sqrt{2}-\sqrt{3}}$. | 1.27 To get rid of the irrationality in the denominator, we multiply the numerator and the denominator by the expression conjugate to the denominator:
$$
\begin{aligned}
& \frac{(2-\sqrt{2})-\sqrt{3}}{(2+\sqrt{2})-\sqrt{3}} \cdot \frac{(2+\sqrt{2})+\sqrt{3}}{(2+\sqrt{2})+\sqrt{3}}= \\
& =\frac{(2-\sqrt{2})(2+\sqrt{2})... | \frac{(1+2\sqrt{6})(3-4\sqrt{2})}{23} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,472 |
1.29 Calculate the sum of the cubes of two numbers if their sum and product are 11 and 21, respectively. | 1.29 Let $a$ and $b$ be the required numbers; then $a+b=11$ and $ab=21$. According to formula (1.11), we have $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$. Therefore, $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$, i.e., $a^{3}+b^{3}=$ $=11^{3}-3 \cdot 21 \cdot 11=638$.
Answer: 638. | 638 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,474 |
$1.31 \frac{a^{3}-2 a^{2}+5 a+26}{a^{3}-5 a^{2}+17 a-13}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$1.31 \frac{a^{3}-2 a^{2}+5 a+26}{a^{3}-5 a^{2}+17 a-13}$. | 1.31 I with assistance. The numerator and denominator are transformed so that a common factor can be factored out:
$$
\begin{aligned}
& a^{3}-2 a^{2}+5 a+26=a^{3}+2 a^{2}-4 a^{2}-8 a+13 a+26= \\
& =a^{2}(a+2)-4 a(a+2)+13 a(a+2)=(a+2)\left(a^{2}-4 a+13\right) \\
& a^{3}-5 a^{2}+17 a-13=a^{3}-a^{2}-4 a^{2}+4 a+13 a-13= ... | \frac{+2}{-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,476 |
$1.36 \frac{x^{4}+x^{2}+x \sqrt{2}+2}{x^{2}-x \sqrt{2}+2}-x \sqrt{2}$. | 1.36 To factor the numerator, we perform the following transformation:
$$
\begin{aligned}
& x^{4}+1=x^{4}+2 x^{2}+1-2 x^{2}=\left(x^{2}+1\right)^{2}-(\sqrt{2} x)^{2}= \\
& \left(x^{2}+1+\sqrt{2} x\right)\left(x^{2}+1-\sqrt{2} x\right)
\end{aligned}
$$
Then the numerator will be
$$
\begin{aligned}
& \left(x^{2}+\sqrt... | x^{2}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,478 |
$1.37 \sqrt[3]{\frac{2 x^{2}}{9+18 x+9 x^{2}}} \cdot \sqrt{\frac{(1+x) \sqrt[3]{1-x}}{x}} \cdot \sqrt[3]{\frac{3 \sqrt{1-x^{2}}}{2 x \sqrt{x}}}$. | 1.37 Let's write all three factors under one common root:
$$
\begin{aligned}
& \sqrt[6]{\frac{4 x^{4}(1+x)^{3} \cdot 9(1-x^{2})(1-x)}{9^{2}(1+2 x+x^{2})^{2} x^{3} \cdot 4 x^{3}}}=\sqrt[6]{\frac{(1+x)^{4}(1-x)^{2}}{9 x^{2}(1+x)^{4}}}= \\
& =\sqrt[6]{\frac{(1-x)^{2}}{9 x^{2}}}=\sqrt[3]{\left|\frac{1-x}{3 x}\right|}
\end... | \sqrt[3]{\frac{1-x}{3x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,479 |
1.38
$$
\frac{\sqrt{1+\left(\frac{x^{2}-1}{2 x}\right)^{2}}}{\left(x^{2}+1\right) \frac{1}{x}}
$$ | ### 1.38 We have
$$
\frac{\sqrt{(2 x)^{2}+\left(x^{2}-1\right)^{2}}}{2|x|\left(x^{2}+1\right) \frac{1}{x}}=\frac{\sqrt{x^{4}+2 x^{2}+1}}{2|x|\left(x^{2}+1\right) \frac{1}{x}}=\frac{x}{2|x|}
$$
Answer: for $x \in(-\infty, 0) \Rightarrow -\frac{1}{2} ;$ for $x \in(0, \infty) \Rightarrow \frac{1}{2}$. | \frac{x}{2|x|} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,480 |
$1.39 \frac{\sqrt{2 b+2 \sqrt{b^{2}-4}}}{\sqrt{b^{2}-4}+b+2}$. | 1.39 It is easy to guess that the expression under the square root in the numerator is a complete square of the sum: $(\sqrt{b+2}+\sqrt{b-2})^{2}$. Then we get
$$
\frac{\sqrt{b+2}+\sqrt{b-2}}{\sqrt{b^{2}-4}+b+2}=\frac{\sqrt{b+2}+\sqrt{b-2}}{\sqrt{b+2}(\sqrt{b+2}+\sqrt{b-2})}=\frac{1}{\sqrt{b+2}}
$$
Answer: $\frac{1}{... | \frac{1}{\sqrt{b+2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,481 |
$1.40 \frac{(x+1)^{-\frac{1}{2}}}{(x-1)^{-\frac{1}{2}}-(x+1)^{-\frac{1}{2}}} ; x=\frac{a^{2}+1}{2 a}$.
$1.41\left(\frac{\left(x+\sqrt[3]{2 a x^{2}}\right)\left(2 a+\sqrt[3]{4 a^{2} x}\right)^{-1}-1}{\sqrt[3]{x}-\sqrt[3]{2 a}}-(2 a)^{-\frac{1}{3}}\right)^{-6}$. | 1.40 Here $x>1$, therefore
$$
\frac{a^{2}+1}{2 a}>1 \Rightarrow \frac{a^{2}-2 a+1}{2 a}>0 \Rightarrow \frac{(a-1)^{2}}{2 a}>0 \Rightarrow a>0
$$
Next, we find
$$
x+1=\frac{(a+1)^{2}}{2 a} ; x-1=\frac{(a-1)^{2}}{2 a}
$$
Then the given expression will take the following form:
$\frac{(a+1)^{-1} \sqrt{2 a}}{\frac{\sqr... | \frac{1-}{2}for\in(0,1);\frac{-1}{2}for\in(1,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,482 |
$1.46 \frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2 x+\sqrt{x^{2}-1}\right)}{\sqrt{(x+1)^{3}}-\sqrt{(x-1)^{3}}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$1.46 \frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+... | 1.46 Transform the expression under the square root in the numerator:
$$
\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2=\frac{(x-1)+(x+1)}{\sqrt{x^{2}-1}}-2=
$$
$$
=\frac{(\sqrt{x+1}-\sqrt{x-1})^{2}}{\sqrt{x^{2}-1}}
$$
from which
$$
\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}=\frac{\sqrt{x+1}-\sqrt{x-1}... | \frac{1}{\sqrt[4]{x^{2}-1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,485 |
$1.47 \frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}} ; x=2$. | 1.47 Here in the denominator, the formula should be applied
$$
\sqrt{A \pm B}=\sqrt{\frac{A+\sqrt{A^{2}-B^{2}}}{2}} \pm \sqrt{\frac{A-\sqrt{A^{2}-B^{2}}}{2}} \text {, where } A \geqslant B \text {. }
$$
Substituting $x=2$ and using this formula, we get
$$
\sqrt{\frac{2+\sqrt{4-3}}{2}}+\sqrt{\frac{2-\sqrt{4-3}}{2}}=\... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,486 |
1.50
$\frac{\sqrt{x-4 \sqrt{x-4}}+2}{\sqrt{x+4 \sqrt{x-4}}-2}$
1.51 $\left(\frac{|x-1|}{x-1} \cdot x^{2}-2 x \cdot \frac{|x+1|}{x+1}+2 x-4\right):|x-2|$. | 1.50 We have:
1) $\sqrt{x-4 \sqrt{x-4}}+2=\sqrt{x-4-4 \sqrt{x-4}+4}+2=$
$$
=\sqrt{(\sqrt{x-4}-2)^{2}}+2=|\sqrt{x-4}-2|+2
$$
2) $\sqrt{x+4 \sqrt{x-4}}-2=\sqrt{(\sqrt{x-4}+2)^{2}}-2=$ $=\sqrt{x-4}+2-2=\sqrt{x-4}$
3) $\frac{|\sqrt{x-4}-2|+2}{\sqrt{x-4}}=A$;
If $\sqrt{x-4}-2<0$, then $\sqrt{x-4}<2$, i.e., $4<x<8$. Thus... | forx\in(4,8)\Rightarrow\frac{4}{\sqrt{x-4}}-1;forx\in[8,\infty)\Rightarrow1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,489 |
$1.52 \sqrt{\left(y^{2}+\frac{4}{y^{2}}\right)^{2}-8\left(y+\frac{2}{y}\right)^{2}+48}$.
Check the validity of the equalities (1.53-1.55):
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | 1.52 Instruction. Use the substitution
$$
y+\frac{2}{y}=z ; \text { then } z^{2}-4=y^{2}+\frac{4}{y^{2}}
$$
Answer: $\left(y-\frac{2}{y}\right)^{2}$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,490 | |
$1.53 \frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}-\sqrt{3}=2$. | 1.53 Consider the equality
$$
\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}=2+\sqrt{3}
$$
Obviously, if this equality is true, then the given equality is also true. Let
$$
a=\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}, b=2+\sqrt{3}
$$
It is easy to establish that $a>0$ and ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,491 |
$1.54 \sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=4$. | ### 1.54 Let
$\sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=x$.
Cubing both sides of this equation, and using formula (1.11), we get
$38+\sqrt{1445}+38-\sqrt{1445}+3 \sqrt[3]{(38+\sqrt{1445})(38-\sqrt{1445})} x=x^{3}$,
or $x^{3}+3 x-76=0$. By substitution, we verify that $x=4$ is one of the roots of the resulti... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,492 |
$1.55 \sqrt{6 m+2 \sqrt{9 m^{2}-n^{2}}}-\sqrt{6 m-2 \sqrt{9 m^{2}-n^{2}}}=2 \sqrt{3 m-n}$.
The above text is translated into English, please retain the line breaks and format of the source text, and output the translation result directly.
However, since the provided text is already a mathematical equation, it remain... | ### 1.55 We have
$$
\begin{aligned}
& 6 m + 2 \sqrt{9 m^{2} - n^{2}} = 3 m + n + 2 \sqrt{9 m^{2} - n^{2}} + 3 m - n = \\
& = (\sqrt{3 m + n} + \sqrt{3 m - n})^{2}
\end{aligned}
$$
Similarly,
$$
6 m - 2 \sqrt{9 m^{2} - n^{2}} = (\sqrt{3 m + n} - \sqrt{3 m - n})^{2}
$$
Therefore,
$$
\begin{aligned}
& \sqrt{(\sqrt{3 ... | 2\sqrt{3-n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,493 |
1.56 Simplify the expression
$y=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}$
and then plot the graph of the function $y$ for $1 \leqslant x<\infty$. | 1.56 Since
\[
\begin{aligned}
& \sqrt{x+2 \sqrt{x-1}}=\sqrt{x-1+2 \sqrt{x-1}+1}= \\
& =\sqrt{(\sqrt{x-1}+1)^{2}}=\sqrt{x-1}+1 \\
& \sqrt{x-2 \sqrt{x-1}}=|\sqrt{x-1}-1|
\end{aligned}
\]
the given expression will take the form
\( y=\sqrt{x-1}+1+|\sqrt{x-1}-1| \).
Thus, we obtain:
1) when \(\sqrt{x-1}-1<0 \Rightarrow... | \begin{cases}2& | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,494 |
1.57 Excluding $u$ and $v$ from the equations $u-v=a$, $u^{2}-v^{2}=b$, $u^{3}-v^{3}=c$, find the relation between $a, b$ and $c$. | 1.57 Since $u^{2}-v^{2}=(u-v)(u+v)=b$, then $u+v=\frac{b}{a}$. Thus, we obtain the system
$$
\left\{\begin{array}{l}
u-v=a \\
u+v=\frac{b}{a}
\end{array}\right.
$$
from which
$$
u=\frac{b}{2 a}+\frac{a}{2} \text { and } v=\frac{-b}{2 a}-\frac{a}{2}
$$
Substituting the found values of $u$ and $v$ into the equality $... | 3b^{2}+^{4}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,495 |
1.58 Show that
$$
\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots+\frac{1}{n^{2}+3 n+2}=\frac{n}{2 n+4}
$$ | 1.58 Let's factor the quadratic trinomial:
$n^{2}+3 n+2=(n+1)(n+2)$.
Then we get
$$
\begin{aligned}
& \frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots+\frac{1}{(n+1)(n+2)}= \\
& =\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\ldots+\frac{1}{(n+1)(n+2)}
\end{aligned}
$$
But
$\frac{1}{2 \cdot 3}=\frac{1}{2... | \frac{n}{2n+4} | Algebra | proof | Yes | Yes | olympiads | false | 49,496 |
1.59 Determine $A, B$ and $C$ so that for all permissible values of $x$ the equality holds
$\frac{x^{2}+5}{x^{3}-3 x+2}=\frac{A}{x+2}+\frac{B}{(x-1)^{2}}+\frac{C}{x-1}$. | 1.59 Let's bring the right-hand side of the equation to a common denominator:
\(\frac{x^{2}+5}{x^{3}-3 x+2} \equiv \frac{A(x-1)^{2}+B(x+2)+C(x+2)(x-1)}{x^{3}-3 x+2}\).
If two fractions with the same denominator are equal, then their numerators are also identically equal, i.e.,
\[
x^{2}+5 \equiv A(x-1)^{2}+B(x+2)+C(x... | A=1,B=2,C=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,497 |
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