problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
13.182. Water is poured into two vessels of the same mass, with the mass of vessel $A$ with water being $4 / 5$ of the mass of vessel $B$ with water. If the water from vessel $B$ is poured into vessel $A$, the mass of vessel $A$ with water will become 8 times the mass of vessel $B$. Find the mass of the vessels and the... | Solution.
Let $x$ g be the mass of the vessels. The mass of vessel $B$ with water is $y$ g, and the mass of vessel $A$ with water is $\frac{4}{5} y$ g. The mass of water in vessel $A$ is $\left(\frac{4}{5} y - x\right)$ g, and in $B$ is $(y - x)$ g. According to the problem, we have $\left\{\begin{array}{l}\frac{4}{5}... | 50,150,200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,667 |
13.184. If the student had correctly multiplied two numbers written on the board, the product would have been 4500. However, while copying the factors from the board, the student wrote a 3 instead of the last digit 5 in one of them and obtained 4380 after multiplication. What numbers should the student have multiplied? | Solution.
Let $10 x+5$ be the first number, $y$ be the second. Then $\left\{\begin{array}{l}(10 x+5) y=4500, \\ (10 x+3) y=4380,\end{array}\right.$ from which $x=7$, the first number $10 \cdot 7+5=75 ; y=60$.
Answer: 75 and 60. | 7560 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,668 |
13.185. When testing two engines, it was found that the first one consumed 300 g, and the second one 192 g of gasoline, with the second engine working 2 hours less than the first. The first engine consumes 6 g more gasoline per hour than the second. How much gasoline does each engine consume per hour? | ## Solution.
Let $x$ g of gasoline be consumed per hour by the second engine, and $(6+x)$ g per hour - by the first. The first engine worked for $\frac{300}{x+6}$ hours, the second - for $\frac{192}{x}$ hours. According to the condition, $\frac{300}{x+6}-\frac{192}{x}=2$, from which $x=24$ g per hour is consumed by th... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,669 |
13.186. A brigade of masons took on the task of laying $432 \mathrm{~m}^{3}$ of masonry, but in reality, 4 fewer people showed up for work. How many masons are there in the brigade if it is known that each working mason had to lay $9 \mathrm{~m}^{3}$ more than initially planned? | ## Solution.
Let there be $x$ masons in the team. Each mason lays $\frac{432}{x}$ m $^{3}$ of masonry. Since $x-4$ masons were working, each had to lay $\frac{432}{x}+9$ m $^{3}$ of masonry. Therefore, $\left(\frac{432}{x}+9\right)(x-4)=432$, from which $x=16$.
## Answer: 16. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,670 |
13.187. A team of workers was supposed to manufacture 8000 identical parts within a certain period. In fact, the work was completed 8 days ahead of schedule because the team produced 50 more parts daily than planned. What was the original deadline for completing the work, and what was the daily percentage of overachiev... | Solution.
Let the work was supposed to be completed in $x$ days. The team was supposed to produce $\frac{8000}{x}$ pieces daily, but they produced $\left(\frac{8000}{x}+50\right)$ pieces and worked for a total of $(x-8)$ days. Therefore, $\left(\frac{8000}{x}+50\right)(x-8)=8000$, from which $x=40$ days. The team was ... | 40;25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,671 |
13.188. Worker $A$ spends $k$ minutes less on processing one part than worker $B$. How many parts does each of them process in $t$ hours of work, if $A$ processes $n$ more parts than $B$ in this time? | Solution.
Let $B$ process $x$ parts, $A$ processes $x+n$ parts. For processing 1 part, $A$ spends $\frac{t}{x+n}$ hours, $B$ spends $\frac{t}{x}$ hours. According to the condition, $\frac{t}{x}-\frac{t}{x+n}=\frac{k}{60}$, from which we get
$x=\frac{-k n+\sqrt{k^{2} n^{2}+240 t k n}}{2 k}$
$x+n=\frac{-k n+\sqrt{k^{2}... | (-kn+\sqrt{k^{2}n^{2}+240kn})/(2k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,672 |
13.189. The sum of the squares of the roots of the equation $x^{2}-3 a x+a^{2}=0$ is 1.75. Find the value of $a$. | Solution.
Solve the equation with respect to $x: x^{2}-3 a x+a^{2}=0$;
$D=9 a^{2}-4 a^{2}=5 a^{2} . x_{1}=\frac{3 a-a \sqrt{5}}{2} ; x_{2}=\frac{3 a+a \sqrt{5}}{2}$. According to the condition
$\frac{a^{2}}{4}(3-\sqrt{5})^{2}+\frac{a^{2}}{4}(3+\sqrt{5})^{2}=1.75$, from which $a^{2}=\frac{1}{4}, a= \pm 0.5$.
Answer:... | \0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,673 |
13.191. Two forces are applied to a material point, the angle between which is $30^{\circ}$. The magnitude of one of the applied forces is $7 \sqrt{3}$ times the magnitude of the other, and the magnitude of the resultant force is $24 \mathrm{N}$ greater than the magnitude of the smaller force. Determine the magnitude o... | ## Solution.
Let $x \mathrm{H}$ be the modulus of the smaller force (Fig. 13.11). By the cosine rule: $\quad(24+x)^{2}=x^{2}+(7 \sqrt{3} x)^{2}-2 x \cdot 7 \sqrt{3} x \cdot \cos \left(180^{\circ}-30^{\circ}\right)$, from which $x=2 \mathrm{H}$ is the modulus of the smaller force; $24+2=26 \mathrm{H}$ is the modulus of... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,675 |
13.192. There are three vessels containing unequal amounts of liquid. To equalize these amounts, three pourings were made. First, $1 / 3$ of the liquid was poured from the first vessel into the second, then $1 / 4$ of the liquid that ended up in the second vessel was poured into the third, and finally, $1 / 10$ of the ... | Solution.
Let's form a table:
| Cо- container | Initial amount of liquid | After the first transfer | After the second transfer | After the third transfer |
| :---: | :---: | :---: | :---: | :---: |
| 1 | $x$ l | $\frac{2}{3} x$ l | $\frac{2}{3} x \pi$ | $\frac{2}{3} x+\frac{1}{10}\left(z+\frac{1}{4}\left(y+... | 12,8,7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,676 |
13.193. During exercises, a reconnaissance boat approached the lead ship of the squadron and received an order to conduct reconnaissance ahead of the squadron in the direction of its movement at a distance of 70 km. Determine how long it will take for the boat to return to the lead ship of the squadron, which continues... | Solution.
Let the boat return to the lead ship after $x$ hours. In this time, the squadron will travel $14 x$ km. Therefore, the boat must travel $70+70-14 x$ km. According to the condition, $140-14 x=28 x$, from which $x=3 \frac{1}{3}$ hours.
Answer: in 3 hours 20 minutes. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,677 |
13.194. The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $1 / 4$ of its length, and the circumference of the rear wheel is increased by $1 / 5$ of its length, then over the same distance, the front wheel will... | Solution.
The circumference of the wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $\mathrm{Cn}=s$. We will fill in the table of values for these quantities in the order indicated by the numbers (1), (2), ..., (12):
| Wheel | Before change | | | After change | | |
| :---:... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,678 |
13.196. Two hours after leaving the factory, the driver looked at the speedometer and noticed that he had only traveled 112 km. He mentally calculated that if he continued at the same speed, he would be 30 minutes late in delivering the cargo to the station. Therefore, the driver increased his speed and arrived at the ... | Solution.
The initial speed of the car is $\frac{112}{2}=56$ km/h. The driver was on the road for a total of $\frac{280}{56}-1=4$ hours. Let the driver increase the speed by $x$ km/h. With this speed, he drove for 2 hours and covered $280-112=168$ km. Therefore, $2(56+x)=168$, from which the new speed of the car is $8... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,679 |
13.197. A cinema hall has two doors, a wide one and a narrow one. After a screening, the audience exits the hall through both doors in 3 minutes and 45 seconds. If the audience is let out through only the wide door, it takes 4 minutes less than if they are let out through only the narrow door. How much time is required... | ## Solution.
Let $x$ min be necessary to release the audience only through the wide door, $(x+4)$ min - only through the narrow door. In one minute, $\frac{1}{x}$ people exit through the wide door, and $\frac{1}{x+4}$ people exit through the narrow door. According to the condition: $\frac{1}{x}+\frac{1}{x+4}=\frac{1}{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,680 |
13.198. A certain substance absorbs moisture, thereby increasing its mass. To absorb 1400 kg of moisture, 300 kg more of the uncrushed substance is required than of the crushed substance. What percentage of the mass of the substance is the mass of the absorbed moisture in the case of crushed substance and in the case o... | Solution.
Let's take $x$ kg of crushed substance, $(x+300)$ kg of uncrushed.
$$
\begin{aligned}
& x \text { kg is } 100 \% \\
& 1400 \text{ kg } \quad-\quad y_{1} \% \\
& x+300 \text { kg is } 100 \% \\
& 1400 \text { kg } \quad-\quad y_{2} \% . \\
& \left\{\begin{array}{l}
y_{1}=\frac{140000}{x} \\
y_{2}=\frac{14000... | 280175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,681 |
13.199. On the way from the village to the field, the truck's wheel makes 100 fewer revolutions than the bicycle's wheel and 150 more revolutions than the tractor's track. Find the distance between the village and the field, given that the circumference of the truck's wheel is $4 / 3$ of the circumference of the bicycl... | ## Solution.
Let $x$ m be the distance between the village and the field, $y$ m be the circumference of the bicycle wheel; $\frac{4}{3} y$ m be that of the truck, $\left(\frac{4}{3} y+2\right)$ m be that of the tractor.
The truck wheel makes $\frac{x}{\frac{4}{3} y}$ revolutions; the bicycle wheel makes $-\frac{x}{y}... | 600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,682 |
13.200. Two skins with a total cost of 22500 rubles were sold at an auction with a profit of $40 \%$. What is the cost of each skin if a profit of $25 \%$ was made on the first one, and $50 \%$ on the second one? | Solution.
Let $x$ rubles be the cost of the first pelt, and 22500 - $x$ rubles be the cost of the second. The first pelt was sold for $1.25 x$ rubles, and the second for $1.5(22500-x)$ rubles. According to the condition, $1.25 x + 1.5(22500-x) = 1.4 \cdot 22500$, from which $x = 9000$ rubles.
Answer: 9000 and 13500 r... | 9000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,683 |
13.201. A sports field has the shape of a rectangle, the length of which is $b$ m longer than its width. The field is surrounded by a path of uniform width of $a$ m. What are the dimensions of the sports field if its area is equal to the area of the surrounding path? | ## Solution.
Let $x$ m be the width of the plot, $b+x$ m - its length (Fig. 13.12). The area of the plot is $x(x+b)$ m². The area of the path is $2 a x + 2(b + x + 2 a) a$, from which we get $x = \frac{\sqrt{b^{2} + 32 a^{2}} - b + 4 a}{2} \text{ m};$
$$
b + x = \frac{4 a + b + \sqrt{b^{2} + 32 a^{2}}}{2} \text{ m}
$... | \frac{(\sqrt{b^{2}+32^{2}}-4)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,684 |
13.205. If a two-digit number is divided by the sum of its digits, the quotient is 3 and the remainder is 7. If then the sum of the squares of the digits of this number is taken and the product of the same digits is subtracted from it, the original number is obtained. Find this number. | Solution.
Let $10 x+y$ be the desired number. According to the condition $\left\{\begin{array}{l}10 x+y=3(x+y)+7, \\ x^{2}+y^{2}-x y=10 x+y,\end{array}\right.$ from which $x=3, y=7$. The desired number is 37.
Answer: 37. | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,688 |
13.206. A three-digit number ends with the digit 2. If it is moved to the beginning of the number, the resulting number will be 18 more than the original. Find this number. | ## Solution.
Let the desired three-digit number be of the form $100x + 10y + 2$; then after moving the digit 2, it will take the form $200 + 10x + y$ (1). According to the problem, $200 + 10x + y - (100x + 10y + 2) = 18$, from which we get $10x + y = 20$. Substituting this expression into (1), we get $200 + 20 = 220$.... | 202 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,689 |
13.207. The express train travels the distance from Moscow to St. Petersburg 3 hours and 30 minutes faster than the passenger train, as it covers 35 km more in 1 hour. How many kilometers per hour does each of them travel, if the distance between Moscow and St. Petersburg is rounded to 650 km? | ## Solution.
Let $x$ km/h be the speed of the passenger train, and $x+35$ km/h be the speed of the express train. According to the condition, $\frac{650}{x}-\frac{650}{x+35}=3.5$, from which $x=65$ km/h.
Answer: 65 and 100 km/h. | 65 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,690 |
$2.312 A=\left(\frac{\frac{x^{3}-1}{x+1} \cdot \frac{x}{x^{3}+1}}{\frac{(x+1)^{2}-x}{(x-1)^{2}+x} \cdot\left(1-\frac{1}{x}\right)}\right)^{-1 / 2}$ | Solution. $A=\left(\frac{(x-1)\left(x^{2}+x+1\right) x}{(x+1)(x+1)\left(x^{2}-x+1\right)} \cdot \frac{(x-1)^{2}+x}{(x+1)^{2}-x} \cdot \frac{x}{x-1}\right)^{-1 / 2}=$
$=\left(\frac{(x-1)\left(x^{2}+x+1\right) x}{(x+1)^{2}\left(x^{2}-x+1\right)} \cdot \frac{x^{2}-x+1}{x^{2}+x+1} \cdot \frac{x}{x-1}\right)^{-1 / 2}=\left... | |\frac{x+1}{x}| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,692 |
2.313 $A=\frac{\left|x^{3}-1\right|+|x+1|}{x^{3}+x}$. | ## Solution.
1) $\left\{\begin{array}{l}x \leq-1, \\ A=\frac{-x^{3}+1-x-1}{x^{3}+x}\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leq-1, \\ A=-1 .\end{array}\right.\right.$
2) $\left\{\begin{array}{l}-1<x<0,0<x<1, \\ A=\frac{-x^{3}+1+x+1}{x^{3}+x}\end{array} \Leftrightarrow\left\{\begin{array}{l}-1<x<0,0<x<1, \\... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,693 |
$2.317 A=\left(\frac{4 m^{2} n^{2}}{4 m n-m^{2}-4 n^{2}}-\frac{2+\frac{n}{m}+\frac{m}{n}}{\frac{4}{m n}-\frac{1}{n^{2}}-\frac{4}{m^{2}}}\right)^{1 / 2}: \frac{\sqrt{m n}}{m-2 n}$.
$2.317 A=\left(\frac{4 m^{2} n^{2}}{4 m n-m^{2}-4 n^{2}}-\frac{2+\frac{n}{m}+\frac{m}{n}}{\frac{4}{m n}-\frac{1}{n^{2}}-\frac{4}{m^{2}}}\ri... | Solution. Let's denote the terms in parentheses as $B$ and $C$.
$$
\begin{aligned}
& B=\frac{4 m^{2} n^{2}}{4 m n-m^{2}-4 n^{2}}=\frac{4 m^{2} n^{2}}{-(2 n-m)^{2}} \\
& C=\frac{2+\frac{n}{m}+\frac{m}{n}}{\frac{4}{m n}-\frac{1}{n^{2}}-\frac{4}{m^{2}}}=\frac{\frac{2 m n+n^{2}+m^{2}}{m n}}{\frac{4 m n-m^{2}-4 n^{2}}{m^{2... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,695 | |
$2.318 A=\left(\sqrt{x^{4}-a^{4}}-\frac{x \sqrt{x^{2}+a^{2}}}{\sqrt{1-\frac{a^{2}}{x^{2}}}}\right) \cdot \frac{\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}+a^{2}}}$.
$2.318 A=\left(\sqrt{x^{4}-a^{4}}-\frac{x \sqrt{x^{2}+a^{2}}}{\sqrt{1-\frac{a^{2}}{x^{2}}}}\right) \cdot \frac{\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}+a^{2}}}$.
The above te... | Solution. $A=\left(\sqrt{x^{2}-a^{2}} \cdot \sqrt{x^{2}+a^{2}}-\frac{x \sqrt{x^{2}+a^{2}}}{\sqrt{\frac{x^{2}-a^{2}}{x^{2}}}}\right) \cdot \frac{\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}+a^{2}}}=$
$$
=\left(\sqrt{x^{2}-a^{2}}-\frac{x \cdot|x|}{\sqrt{x^{2}-a^{2}}}\right) \cdot \sqrt{x^{2}-a^{2}}=\frac{x^{2}-a^{2}-x \cdot|x|}{\sqr... | if\,x>||,\,then\,A=-^{2};\,if\,x<-||,\,then\,A=2x^{2}-^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,696 |
$2.321 A=\left(\frac{\left(a^{3 / 2}-\sqrt{8}\right)(\sqrt{a}+\sqrt{2})}{a+\sqrt{2 a}+2}\right)^{2}+\sqrt{\left(a^{2}+2\right)^{2}-8 a^{2}}$. | Solution. $A=\left(\frac{\left((\sqrt{a})^{3}-(\sqrt{2})^{3}\right)(\sqrt{a}+\sqrt{2})}{a+\sqrt{2 a}+2}\right)^{2}+\sqrt{a^{2}+4 a^{2}+4-8 a^{2}}=$
$$
\begin{aligned}
& =\left(\frac{(\sqrt{a}-\sqrt{2})(a+\sqrt{2 a}+2)(\sqrt{a}+\sqrt{2})}{a+\sqrt{2 a}+2}\right)^{2}+\sqrt{\left(a^{2}-2\right)^{2}}= \\
& =((\sqrt{a}-\sqr... | if\0\leq\sqrt{2},\then\A=-4a+6;\if\\geq\sqrt{2},\then\A=2(-1)^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,698 |
$2.322 A=\sqrt{y^{2}-6 y+9}-|y-9|+2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.322 A=\sqrt{y^{2}-6 y+9}-|y-9|+2$. | Solution. $A=\sqrt{(y-3)^{2}}-|y-9|+2=|y-3|-|y-9|+2$.
Consider 3 cases.
1) $\left\{\begin{array}{l}y9, \\ A=y-3-(y-9)+2\end{array} \Leftrightarrow\left\{\begin{array}{l}y>9, \\ A=8 .\end{array}\right.\right.$
Answer: if $y>9$, then $A=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,699 |
$2.323 A=\sqrt{\frac{4}{x}+\frac{1}{4 x^{-1}}-2}+\sqrt{\frac{1}{4 x^{-1}}+\frac{2^{-2}}{x}+\frac{1}{2}}$. | Solution.
$$
A=\sqrt{\frac{4}{x}+\frac{x}{4}-2}+\sqrt{\frac{x}{4}+\frac{1}{4 x}+\frac{1}{2}}=\sqrt{\frac{x^{2}-8 x+16}{4 x}}+\sqrt{\frac{x^{2}+2 x+1}{4 x}}=
$$
$$
=\sqrt{\frac{(x-4)^{2}}{4 x}}+\sqrt{\frac{(x+1)^{2}}{4 x}} \Leftrightarrow\left\{\begin{array}{l}
x>0 \\
A=\frac{|x-4|}{2 \sqrt{x}}+\frac{|x+1|}{2 \sqrt{x}... | if\0<x<4,\then\A=\frac{5}{2\\sqrt{x}};\if\x\\geq\4,\then\A=\frac{2\x-3}{2\\sqrt{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,700 |
$2.324 A=\sqrt{\frac{x}{2+x+x^{-1}}}+|x-1|$. | Solution.
$$
\left\{\begin{array} { l }
{ A = \sqrt { \frac { x ^ { 2 } } { x ^ { 2 } + 2 x + 1 } } + | x - 1 | , } \\
{ x \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
A=\frac{|x|}{|x+1|}+|x-1| \\
x \neq 0
\end{array}\right.\right.
$$
Consider 4 cases.
1) $\left\{\begin{array}{l}x<-1, \\ A=\frac{-x}... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,701 | |
$2.325 A=\frac{n^{4}-2 n^{3}+4 n^{2}+2 n-5}{n^{4}-3 n^{3}+7 n^{2}-5 n}$. | Solution. $A=\frac{\left(n^{4}-3 n^{3}+7 n^{2}-5 n\right)+\left(n^{3}-3 n^{2}+7 n-5\right)}{n\left(n^{3}-3 n^{2}+7 n-5\right)}=1+\frac{1}{n}=\frac{n+1}{n}$
for $n^{3}-3 n^{2}+7 n-5 \neq 0$. But $n^{3}-3 n^{2}+7 n-5=n^{3}-n^{2}-2 n^{2}+2 n+5 n-5=$ $=n^{2}(n-1)-2 n(n-1)+5(n-1)=(n-1)\left(n^{2}-2 n+5\right)$, i.e. $n^{3}... | \frac{n+1}{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,702 |
$2.326 A=\frac{\sqrt{a+2 \sqrt{b}+\frac{b}{a}} \cdot \sqrt{2 a-10 \sqrt[6]{8 a^{3} b^{2}}+25 \sqrt[3]{b^{2}}}}{a \sqrt{2 a}+\sqrt{2 a b}-5 a \sqrt[3]{b}-5 \sqrt[6]{b^{5}}}$.
Translate the above text into English, preserving the original text's line breaks and formatting, and output the translation result directly.
$2... | Solution. Note that $a>0, b \geq 0$. Therefore,
$$
\begin{aligned}
& \sqrt{a+2 \sqrt{b}+\frac{b}{a}}=\sqrt{\frac{a^{2}+2 a \sqrt{b}+b}{a}}=\sqrt{\frac{(a+\sqrt{b})^{2}}{a}}=\frac{|a+\sqrt{b}|}{\sqrt{a}}=\frac{a+\sqrt{b}}{\sqrt{a}} \\
& \sqrt{2 a-10 \sqrt[6]{8 a^{3} b^{2}}+25 \sqrt[3]{b^{2}}}=\sqrt{2 a-10 \cdot \sqrt{2... | \frac{1}{\sqrt{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,703 |
$2.328 A=\sqrt{\left(\frac{x^{2}-4}{2 x}\right)^{2}+4}+\sqrt{1+\frac{4}{x^{2}}+\frac{4}{x}}$. | Solution. $A=\sqrt{\left(\frac{x}{2}-\frac{2}{x}\right)^{2}+4}+\sqrt{\left(1+\frac{2}{x}\right)^{2}}=\sqrt{\frac{x^{2}}{4}+2+\frac{4}{x^{2}}}+\left|1+\frac{2}{x}\right|=$
$$
=\sqrt{\left(\frac{x}{2}+\frac{2}{x}\right)^{2}}+\left|\frac{2}{x}+1\right|=\left|\frac{x}{2}+\frac{2}{x}\right|+\left|\frac{2}{x}+1\right|=\frac... | \frac{x^{2}+2x+8}{2x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,705 |
$2.330 A=\frac{(x+2) \sqrt{(x+2)^{2}-8 x}}{x^{2}-4|x-1|}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$2.330 A=\frac{(x+2) \sqrt{(x+2)^{2}-8 x}}{x^{2}-4|x-1|}$. | Solution. $A=\frac{(x+2) \sqrt{x^{2}+4 x+4-8 x}}{x^{2}-4|x-1|}=\frac{(x+2) \cdot \sqrt{(x-2)^{2}}}{x^{2}-4 \cdot|x-1|}=\frac{(x+2)|x-2|}{x^{2}-4|x-1|}$.
Consider 3 cases.
2) $\left\{\begin... | if\x\leq1,\x\neq-2\2\sqrt{2},\then\A=\frac{4-x^{2}}{x^{2}+4x-4};\if\x>2,\then\A=\frac{x+2}{x-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,707 |
$2.331 A=\frac{\sqrt{3} x^{3 / 2}-5 x^{1 / 3}+5 x^{4 / 3}-\sqrt{3 x}}{\sqrt{3 x+10 \sqrt{3} x^{5 / 6}+25 x^{2 / 3}} \cdot \sqrt{1-2 x^{-1}+x^{-2}}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$2.331 A=\frac{\sqrt{3} x^{... | Solution. Obviously, $x>0$. Consider the numerator of the fraction $A$.
$B=\sqrt{3} x^{3 / 2}-\sqrt{3} x^{1 / 2}+5 x^{4 / 3}-5 x^{1 / 3}=\sqrt{3} x^{1 / 2}(x-1)+5 x^{1 / 3}(x-1)=(x-1)(\sqrt{3 x}+5 \sqrt[3]{x})$.
Consider the factors of the denominator of $A$ one by one.
$C=\sqrt{3 x+2 \cdot 5 \cdot \sqrt{3} x^{1 / 2... | if\0<x<1,\then\A=-x;\if\x>1,\then\A=x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,708 |
$2.333 A=\left(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}-4\left(z+\frac{1}{z}\right)^{2}+12\right)^{1 / 4}:(z-1)$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.333 A=\left(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}-4\left(z+\frac{1}... | Solution. Let $y = z + \frac{1}{z}$. Then $y^2 = z^2 + \frac{1}{z^2} + 2$, i.e., $z^2 + \frac{1}{z^2} = y^2 - 2$.
From this, $A = \frac{\sqrt[4]{(y^2 - 2)^2 - 4y^2 + 12}}{z - 1} = \frac{\sqrt[4]{y^4 - 8y + 16}}{z - 1} = \frac{\sqrt[4]{(y^2 - 4)^2}}{z - 1} = \frac{\sqrt{|y^2 - 4|}}{z - 1} = \frac{\sqrt{|z^2 + \frac{1}{... | \frac{z+1}{z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,710 |
$2.335 A=\frac{\sqrt{1+z}-\sqrt{1-z}}{\sqrt{1+z}+\sqrt{1-z}}$, if $z=\frac{2 a}{a^{2}+1}$. | Solution. Since $\sqrt{1+z}=\sqrt{1+\frac{2 a}{a^{2}+1}}=\sqrt{\frac{a^{2}+2 a+1}{a^{2}+1}}=\frac{|a+1|}{\sqrt{a^{2}+1}}$,
$\sqrt{1-z}=\sqrt{1-\frac{2 a}{a^{2}+1}}=\frac{|a-1|}{\sqrt{a^{2}+1}}$, then $\quad A=\frac{\frac{|a+1|}{\sqrt{a^{2}+1}}-\frac{|a-1|}{\sqrt{a^{2}+1}}}{\frac{|a+1|}{\sqrt{a^{2}+1}}+\frac{|a-1|}{\sq... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,712 | |
$2.338 \quad A=\frac{\sqrt[3]{8 x-y-6\left(2 \sqrt[3]{x^{2} y}-\sqrt[3]{x y^{2}}\right)} \cdot\left(4 x^{2 / 3}+2 \sqrt[3]{x y}+y^{2 / 3}\right)}{8 x \sqrt[3]{y}-y^{4 / 3}}$. | Solution. Consider the first factor in the numerator of the fraction $A$, denote it as $B$.
$$
\begin{aligned}
& B=\sqrt[3]{8 x-y-6\left(2 \sqrt[3]{x^{2} y}-\sqrt[3]{x y^{2}}\right)}= \\
& =\sqrt[3]{(2 \sqrt[3]{x})^{3}-(\sqrt[3]{y})^{3}-3 \cdot(2 \sqrt[3]{x})^{2} \cdot \sqrt[3]{y}+3 \cdot 2 \sqrt[3]{x} \cdot \sqrt[3]{... | \frac{1}{\sqrt[3]{y}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,715 |
$2.339 A=\left(\frac{a}{3\left(a^{2}+1\right)^{1 / 2}}-\left(2 a^{2}+1+a \sqrt{4 a^{2}+3}\right)^{1 / 2} \cdot\left(2 a^{2}+3+a \sqrt{4 a^{2}+3}\right)^{-1 / 2}\right)^{2}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.339 A=... | Solution. $A=\left(\frac{a}{3 \sqrt{a^{2}+1}}-\frac{\sqrt{2 a^{2}+1+a \sqrt{4 a^{2}+3}}}{\sqrt{2 a^{2}+3+a \sqrt{4 a^{2}+3}}}\right)^{2}$.
Let: $y=2 a^{2}+3, z=a \sqrt{4 a^{2}+3}$.
It is easy to see that $y>z$. Therefore,
$$
B=\frac{\sqrt{2 a^{2}+1+a \sqrt{4 a^{2}+3}}}{\sqrt{2 a^{2}+3+a \sqrt{4 a^{2}+3}}}=\frac{\sqr... | \frac{4^{2}+3}{9(^{2}+1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,716 |
$2.340 A=\frac{\sqrt{a-\sqrt{4(a-1)}}+\sqrt{a+\sqrt{4(a-1)}}}{\sqrt{a^{2}-4(a-1)}}$. | Solution. $A=\frac{\sqrt{(a-1)-2 \sqrt{a-1}+1}+\sqrt{(a-1)+2 \sqrt{a-1}+1}}{\sqrt{a^{2}-4 a+4}}=$
$$
=\frac{\sqrt{(\sqrt{a-1}-1)^{2}}+\sqrt{(\sqrt{a-1}+1)^{2}}}{\sqrt{(a-2)^{2}}}=\frac{|\sqrt{a-1}-1|+|\sqrt{a-1}+1|}{|a-2|}=\frac{|\sqrt{a-1}-1|+\sqrt{a-1}+1}{|a-2|} .
$$
Consider 2 cases.
^{-1 / 2}+(2 x+5-4 \sqrt{2 x+1})^{-1 / 2}}{(2 x+5+4 \sqrt{2 x+1})^{-1 / 2}-(2 x+5-4 \sqrt{2 x+1})^{-1 / 2}}$. | ## Solution.
Since $2 x+5 \pm 4 \sqrt{2 x+1}=(2 x+1) \pm 4 \sqrt{2 x+1}+4=(\sqrt{2 x+1} \pm 2)^{2}$, then
$$
A=\frac{\frac{1}{\mid \sqrt{2 x+1}+2}+\frac{1}{|\sqrt{2 x+1}-2|}}{\frac{1}{\mid \sqrt{2 x+1}+2}-\frac{1}{|\sqrt{2 x+1}-2|}} \Leftrightarrow\left\{\begin{array}{l}
x \neq \frac{3}{2} \\
A=\frac{|\sqrt{2 x+1}-2|... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,719 | |
$2.343 A=\frac{\sqrt{4(x-\sqrt{y})+y x^{-1}} \cdot \sqrt{9 x^{2}+6 \sqrt[3]{2 y x^{3}}+\sqrt[3]{4 y^{2}}}}{6 x^{2}+2 \sqrt[3]{2 y x^{3}}-3 \sqrt{y x^{2}}-\sqrt[6]{4 y^{5}}}$.
Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly.
$2.34... | Solution. Obviously, $y \geq 0$. Then $x>0$, otherwise $4(x-\sqrt{y})+y x^{-1}\sqrt{y}, \\ A=\frac{2 x-\sqrt{y}}{\sqrt{x}(2 x-\sqrt{y})}\end{array} \Leftrightarrow\left\{\begin{array}{l}4 x^{2}>y \geq 0, \\ A=\frac{1}{\sqrt{x}} .\end{array}\right.\right.$
Answer: if $x>0, y>4 x^{2}$, then $A=-\frac{1}{\sqrt{x}}$; if $... | if\x>0,\y>4x^{2},\then\A=-\frac{1}{\sqrt{x}};\if\x>0,\0\leqy<4x^{2},\then\A=\frac{1}{\sqrt{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,720 |
$2.344 A=\sqrt{\frac{1}{6}\left((3 x+\sqrt{6 x-1})^{-1}+(3 x-\sqrt{6 x-1})^{-1}\right)} \cdot|x-1| \cdot x^{-1 / 2}$. | Solution. Let's denote by B and transform the following expression.
$$
\begin{aligned}
& B=(3 x+\sqrt{6 x-1})^{-1}+(3 x-\sqrt{6 x-1})^{-1}=\frac{1}{3 x+\sqrt{6 x-1}}+\frac{1}{3 x-\sqrt{6 x-1}}= \\
& =\frac{6 x}{(3 x+\sqrt{6 x-1})(3 x-\sqrt{6 x-1})}=\frac{6 x}{9 x^{2}-(\sqrt{6 x-1})^{2}} .
\end{aligned}
$$
From this, ... | {\begin{pmatrix}\frac{x-1}{3x-1},& | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,721 |
$2.345 A=\sqrt[4]{\left(x^{2}+4 x^{-2}\right)^{2}-8\left(x+2 x^{-1}\right)^{2}+48} \cdot\left(x^{2}-2\right)^{-1}$. | Solution. $A=\sqrt[4]{\left(x^{2}+\frac{4}{x^{2}}\right)^{2}-8\left(x+\frac{2}{x}\right)^{2}+48} \cdot \frac{1}{x^{2}-2}=$
$$
\begin{aligned}
& =\sqrt[4]{\left(x^{2}+\frac{4}{x^{2}}\right)^{2}-8\left(x^{2}+4+\frac{2}{x^{2}}\right)+48} \cdot \frac{1}{x^{2}-2}= \\
& =\sqrt[4]{\left(x^{2}+\frac{4}{x^{2}}\right)^{2}-8\lef... | \frac{1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,722 |
$2.346 A=\left(\frac{x^{2}+x-2 \sqrt{x}+6}{x+2 \sqrt{x}+3}-1\right)^{1 / 2}$. | Solution.
$$
A=\left(\frac{x^{2}+x-2 \sqrt{x}+6-x-2 \sqrt{x}-3}{x+2 \sqrt{x}+3}\right)^{1 / 2}=\left(\frac{x^{2}-4 \sqrt{x}+3}{x+2 \sqrt{x}+3}\right)^{1 / 2}
$$
Let $\sqrt{x}=y$. Then $A=\left(\frac{y^{4}-4 y+3}{y^{2}+2 y+3}\right)^{1 / 2}$.
Divide the polynomial $y^{4}-4 y+3$ by the polynomial $y^{2}+2 y+3$.
$$
\b... | |\sqrt{x}-1| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,723 |
$2.347 A=\sqrt{x\left(x^{-1}+4 x-4\right)^{-1}}-\frac{2 x^{2}}{|2 x-1|}$, where $x>0$. | Solution. Let's denote by $B$ and transform the expression
$$
B=\sqrt{x\left(x^{-1}+4 x-4\right)^{-1}}=\sqrt{\frac{x}{\frac{1}{x}+4 x-4}}=\sqrt{\frac{x^{2}}{4 x^{2}-4 x+1}}=\frac{|x|}{|2 x-1|}=\frac{x}{|2 x-1|} .
$$
Therefore, $A=B-\frac{2 x^{2}}{|2 x-1|}=\frac{x-2 x^{2}}{|2 x-1|}=\frac{x(1-2 x)}{|1-2 x|}$.
1) $\lef... | if\0<x<\frac{1}{2},\then\A=x;\if\x>\frac{1}{2},\then\A=-x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,724 |
$2.348 A=\left|\frac{|x-2|+4}{x-2}\right| \cdot\left(x^{2}-4\right)$. | Solution. $A=\frac{(|x-2|+4)}{|x-2|} \cdot(x-2)(x+2)$.
Consider 2 cases.
.1) $\left\{\begin{array}{l}x>2, \\ A=\frac{(x-2+4)}{x-2} \cdot(x+2)(x-2)\end{array} \Leftrightarrow\left\{\begin{array}{l}x>2, \\ A=(x+2)^{2} .\end{array}\right.\right.$
Answer: if $x>2$, then $A=(x+2)^{2}$. | (x+2)^2ifx>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,725 |
$2.349 A=\left(\frac{x^{8}+x^{4}-x^{2} \sqrt{2}+2}{x^{4}-x^{2} \sqrt{2}+1}+x^{2} \sqrt{2}\right)^{1 / 2}$. | Solution. Let's divide the polynomial $x^{8}+x^{4}-x^{2} \sqrt{2}+2$ by the polynomial $x^{4}-x^{2} \sqrt{2}+1$
$$
\begin{aligned}
& \left.\begin{array}{l|l}
& x^{8}+x^{4}-x^{2} \sqrt{2}+2 \\
- & \frac{x^{8}-x^{6} \sqrt{2}+x^{4}}{x^{6} \sqrt{2}-x^{2} \sqrt{2}+2}
\end{array} \right\rvert\, \frac{x^{4}-x^{2} \sqrt{2}+1... | x^{2}+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,726 |
$2.351 A=\frac{x^{8}+x^{4}-2 x^{2}+6}{x^{4}+2 x^{2}+3}+2 x^{2}-2$. | Solution. Let's divide the polynomial $x^{8}+x^{4}-2 x^{6}+6$ by the polynomial $x^{4}+2 x^{2}+3$.
$$
\begin{aligned}
& \begin{array}{r}
-\frac{-2 x^{6}-4 x^{4}-6 x^{2}}{-2 x^{4}+4 x^{2}+6} ... | x^{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,728 |
$2.353 A=(3 a+\sqrt{6 a-1})^{1 / 2}+(3 a-\sqrt{6 a-1})^{1 / 2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$2.353 A=(3 a+\sqrt{6 a-1})^{1 / 2}+(3 a-\sqrt{6 a-1})^{1 / 2}$. | Solution. Let's denote $B$ and transform the expression
$$
B=3 a \pm \sqrt{6 a-1}=\frac{1}{2}(6 a \pm 2 \sqrt{6 a-1})=\frac{1}{2}((6 a-1) \pm 2 \sqrt{6 a-1}+1)=\frac{1}{2}(\sqrt{6 a-1} \pm 1)^{2}
$$
Then $B^{-1 / 2}=\frac{\sqrt{2}}{|\sqrt{6 a-1} \pm 1|} \cdot$ Therefore $A=\frac{1}{\sqrt{3 a+\sqrt{6 a-1}}}+\frac{1}{\... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,730 |
$2.355 A=\sqrt{\frac{a-8 \sqrt[6]{a^{3} b^{2}}+4 \sqrt[3]{b^{2}}}{\sqrt{a}-2 \sqrt[3]{b}+2 \sqrt[12]{a^{3} b^{2}}}+3 \sqrt[3]{b}}$, where $b>0$. | Solution. Let $x=\sqrt[4]{a}, y=\sqrt[6]{b}$.
Then $A=\sqrt{\frac{x^{4}-8 x^{2} y^{2}+4 y^{4}}{x^{2}+2 x y-2 y^{2}}+3 y^{2}}=\sqrt{\frac{x^{4}-5 x^{2} y^{2}+6 x y^{3}-2 y^{4}}{x^{2}+2 x y-2 y^{2}}}$
$=\sqrt{\frac{\frac{x^{4}-5 x^{2} y^{2}+6 x y^{3}-2 y^{4}}{y^{4}} \cdot y^{4}}{\frac{x^{2}+2 x y-2 y^{2}}{y^{2}} \cdot y... | \begin{cases}\frac{4x}{x-4},&4<x<8\\\frac{2x}{\sqrt{x-4}},&x\geq8\end{cases} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,731 |
2.357 Given: $\frac{x}{m}+\frac{y}{n}+\frac{z}{p}=1 ; \frac{m}{x}+\frac{n}{y}+\frac{p}{z}=0$.
$$
\text { Prove that } \frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}=1 \text {. }
$$ | Solution. Let $\frac{x}{m}=a, \frac{y}{n}=b, \frac{z}{p}=c$.
Then from the condition of the problem we get: $\left\{\begin{array}{l}a+b+c=1, \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 .\end{array}\right.$
Square both sides of the first equation:
$a^{2}+b^{2}+c^{2}+2(a b+b c+a c)=1$.
From the relation $\frac{1}{a}+\fr... | \frac{x^{2}}{^{2}}+\frac{y^{2}}{n^{2}}+\frac{z^{2}}{p^{2}}=1 | Algebra | proof | Yes | Yes | olympiads | false | 51,732 |
### 2.359 Factorize:
$A=x\left(y^{2}-z^{2}\right)+y\left(z^{2}-x^{2}\right)+z\left(x^{2}-y^{2}\right)$ | Solution. $A=x y^{2}-x z^{2}+y z^{2}-y x^{2}+z x^{2}-z y^{2}$.
The factorization of the polynomial $A$ is given in the solution to problem 2.358. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,733 |
2.360 The arithmetic mean of two positive numbers $a$ and $b$ ($a > \boldsymbol{b}$) is $\boldsymbol{m}$ times their geometric mean.
Prove that $\frac{a}{b}=\frac{m+\sqrt{m^{2}+1}}{m-\sqrt{m^{2}-1}}$. | Solution. Transform the expression $\frac{m+\sqrt{m^{2}-1}}{m-\sqrt{m^{2}-1}}=:$
$$
=\frac{\left(m+\sqrt{m^{2}-1}\right)^{2}}{\left(m-\sqrt{m^{2}-1}\right)\left(m+\sqrt{m^{2}-1}\right)}=\frac{\left(m+\sqrt{m^{2}-1}\right)^{2}}{m^{2}-\left(m^{2}-1\right)}=\left(m+\sqrt{m^{2}-1}\right)^{2}
$$
Now let's prove that $\sqr... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,734 |
$6.257 x^{3}-(2 a+1) x^{2}+\left(a^{2}+2 a-b^{2}\right) x+\left(b^{2}-a^{2}\right)=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.257 x^{3}-(2 a+1) x^{2}+\left(a^{2}+2 a-b^{2}\right) x+\left(b^{2}-a^{2}\right)=0$. | Solution. The given equation is a particular case of the model equation (1) if we set $v=1, u=a, w=b$.
Answer: $x_{1}=1, x_{2,3}=a \pm b$. | x_{1}=1,x_{2,3}=\b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,735 |
$6.258 x^{3}-2 x^{2}-\left(a^{2}-a-1\right) x+\left(a^{2}-a\right)=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.258 x^{3}-2 x^{2}-\left(a^{2}-a-1\right) x+\left(a^{2}-a\right)=0$. | Solution. The given equation is a particular case of the model equation (1) if we set $v=1, u=\frac{1}{2}, w^{2}=\left(a-\frac{1}{2}\right)^{2}$.
Answer: $x_{1}=1, x_{2}=a, x_{3}=1-a$. | x_{1}=1,x_{2}=,x_{3}=1- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,736 |
$6.259 x^{3}-(3 a-1) x^{2}+\left(2 a^{2}-3 a\right) x+2 a^{2}=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.259 x^{3}-(3 a-1) x^{2}+\left(2 a^{2}-3 a\right) x+2 a^{2}=0$. | Solution. The given equation is a special case of the model equation (1) if we set $v=-1, u=\frac{3 a}{2}, w^{2}=\frac{a^{2}}{4}$.
Answer: $x_{1}=-1, x_{2}=a, x_{3}=2 a$.
$6.260(x-1)^{5}+(x+3)^{5}=242(x+1)$.
Solution. Let $y=x-1, z=x+3$. Then
$$
\left\{\begin{array} { l }
{ z - y = 4 , } \\
{ y ^ { 5 } + z ^ { 5 }... | x_{1}=-1,x_{2}=0,x_{3}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,737 |
$6.261 x^{3}-(2 a+1) x^{2}+\left(a^{2}+a\right) x-\left(a^{2}-a\right)=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.261 x^{3}-(2 a+1) x^{2}+\left(a^{2}+a\right) x-\left(a^{2}-a\right)=0$. | Solution. This equation is a particular case of the model equation (1) if we set $v=1, u=a, w^{2}=a$.
Answer: $x_{1}=1, x_{2}=a+\sqrt{a}, x_{3}=a-\sqrt{a}$.
$6.262\left(x^{3}+x^{-3}\right)+\left(x^{2}+x^{-2}\right)+\left(x+x^{-1}\right)=6$.
Solution. $\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right)+\l... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,738 |
$6.264 x^{3}-x^{2}-\frac{8}{x^{3}-x^{2}}=2$. | Solution. Let $y=x^{3}-x^{2}$, then $y-\frac{8}{y}=2 \Leftrightarrow y^{2}-2 y-8=0 \Leftrightarrow$
$\Leftrightarrow\left[\begin{array}{l}y=-2, \\ y=4\end{array} \Leftrightarrow\left[\begin{array}{l}x^{3}-x^{2}+2=0, \\ x^{3}-x^{2}-4=0\end{array} \Leftrightarrow\left[\begin{array}{l}(x+1)\left(x^{2}-2 x+2\right)=0, \\ (... | x_{1}=-1,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,739 |
$6.265 x^{3}-(2 a+1) x^{2}+\left(a^{2}+2 a-m\right) x-\left(a^{2}-m\right)=0$. | Solution. The given equation is a particular case of the model equation (1) if we set $v=1, u=a, w^{2}=m(m \geq 0)$.
Answer: $x_{1}=1, x_{2,3}=a \pm \sqrt{m}$. | x_{1}=1,x_{2,3}=\\sqrt{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,740 |
$6.266 x^{3}-3 a x^{2}+\left(3 a^{2}-b\right) x-\left(a^{3}-a b\right)=0, b \geq 0$. | Solution. The given equation is a particular case of the model equation (1) if we set $v=a, u=a, w^{2}=b$.
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,741 | |
6.267 x^{3}-\left(a^{2}-a+7\right) x-3\left(a^{2}-a-2\right)=0
$$ | Solution. The given equation is a particular case of the model equation (1) if
we set $v=-3, u=\frac{3}{2}, w^{2}=\left(a-\frac{1}{2}\right)^{2}$.
Answer: $x_{1}=3, x_{2}=a+1, x_{3}=2-a$. | x_{1}=3,x_{2}=+1,x_{3}=2- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,742 |
$6.269 x^{3}-2 a x^{2}+\left(a^{2}+2 \sqrt{3} a-9\right) x-\left(2 a^{2} \sqrt{3}-12 a+6 \sqrt{3}\right)=0$. | Solution. $x^{3}-2 a x^{2}+\left(a^{2}+2 \sqrt{3} a-9\right) x-2 \sqrt{3}(a-\sqrt{3})^{2}=0$.
This equation is a special case of the model equation (1) if we set $v=2 \sqrt{3}, u=a-\sqrt{3}, w=0$.
Answer: $x_{1}=2 \sqrt{3}, x_{2,3}=a-\sqrt{3}$. | x_{1}=2\sqrt{3},x_{2,3}=-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,743 |
$6.27010 x^{3}-3 x^{2}-2 x+1=0$. | ## Solution.
$$
10 x^{3}+5 x^{2}-8 x^{2}-4 x+2 x+1=0 \Leftrightarrow 5 x^{2}(2 x+1)-4 x(2 x+1)+(1+2 x)=0 \Leftrightarrow
$$
$\Leftrightarrow(2 x+1)\left(5 x^{2}-4 x+1\right)=0 \Leftrightarrow 2 x+1=0$.
Answer: $x=-\frac{1}{2}$.
$6.2712\left(x^{2}+x+1\right)^{2}-7(x-1)^{2}=13\left(x^{3}-1\right)$.
Solution. Let $v=... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,744 |
$6.274 x^{2}+\frac{81 x^{2}}{(9+x)^{2}}=40$. | Solution. $x^{2}(9+x)^{2}+81 x^{2}=40(9+x)^{2} \Leftrightarrow x^{2}\left((9+x)^{2}+81\right)=40(9+x)^{2} \Leftrightarrow$ $x^{2}\left(x^{2}+18 x+162\right)=40(9+x)^{2} \Leftrightarrow x^{2}\left(x^{2}+18(x+9)\right)=40(9+x)^{2}$.
Let $u=x^{2}, v=x+9$, then $u(u+18 v)=40 v^{2} \Leftrightarrow$ $u^{2}+18 u v-40 v^{2}=0... | x_{1}=1+\sqrt{19},x_{2}=1-\sqrt{19} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,745 |
6.275 $\frac{2+x}{2-x}+\sqrt{x}=1+x$ | Solution. $\frac{2+x}{2-x}-1=x-\sqrt{x} \Leftrightarrow \frac{2 x}{2-x}=\sqrt{x}(\sqrt{x}-1)$. Let $y=\sqrt{x}$, then $\frac{2 y^{2}}{2-y^{2}}=y(y-1) \Leftrightarrow\left[\begin{array}{l}y=0, \\ y \neq \sqrt{2}, \\ y^{3}-y^{2}+2=0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left[\begin{array}{l}y=0 . \\ (y+1)\l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,746 |
$6.276 \frac{20}{\sqrt{x}}+x \sqrt{x}+x=22$. | Solution. Let $y=\sqrt{x}$, then $\frac{20}{y}+y^{3}+y^{2}=22 \Leftrightarrow$
$\Leftrightarrow\left(\frac{20}{y}-20\right)+\left(y^{3}+y^{2}-2\right)=0 \Leftrightarrow-\frac{20(y-1)}{y}+(y-1)\left(y^{2}+2 y+2\right)=0 \Leftrightarrow$ $\left[\begin{array}{l}y=1, \\ y^{3}+2 y^{2}+2 y-20=0\end{array} \Leftrightarrow\le... | x_{1}=1,x_{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,747 |
6.277 \sqrt{x-1}+\sqrt{x+3}+2 \sqrt{(x-1)(x+3)}=4-2 x
$$
6.277 \sqrt{x-1}+\sqrt{x+3}+2 \sqrt{(x-1)(x+3)}=4-2 x
$$ | Solution. Let $u=\sqrt{x-1}, v=\sqrt{x+3}$, then $u+v+2uv=-(u^2+v^2)+6 \Leftrightarrow$ $(u+v)^2+(u+v)-6=0 \Leftrightarrow\left[\begin{array}{l}u+v=-3, \\ u+v=2\end{array} \Leftrightarrow\left[\begin{array}{l}\varnothing, \\ \sqrt{x-1}+\sqrt{x+3}=2 .\end{array}\right.\right.$
Since $x \geq 1$, then $\sqrt{x+3} \geq 2 ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,748 |
$6.278 \sqrt{2 x+3}+\sqrt{x+1}=3 x+2 \sqrt{2 x^{2}+5 x+3}-16$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.278 \sqrt{2 x+3}+\sqrt{x+1}=3 x+2 \sqrt{2 x^{2}+5 x+3}-16$. | Solution. $\sqrt{2 x+3}+\sqrt{x+1}=3 x-16+2 \sqrt{(2 x+3)(x+1)}$.
Let $u=\sqrt{2 x+3}, v=\sqrt{x+1}$, then $u+\dot{v}=u^{2}+v^{2}-20+2 u v \Leftrightarrow$
$$
\Leftrightarrow(u+v)^{2}-(u+v)-20=0 . \Leftrightarrow\left[\begin{array}{l}
u+v=5, \\
u+v=-4
\end{array} \Leftrightarrow \sqrt{2 x+3}+\sqrt{x+1}=5 \Leftrightar... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,749 |
$6.279 \sqrt[4]{x+8}-\sqrt[4]{x-8}=2$ | Solution. Let $u=\sqrt[4]{x+8}, v=\sqrt[4]{x-8}$, then $\left\{\begin{array}{l}u-v=2, \\ u^{4}-v^{4}=16\end{array} \Rightarrow\right.$ $\Rightarrow(v+2)^{4}-v^{4}=16 \Leftrightarrow\left((v+2)^{2}-v^{2}\right)\left((v+2)^{2}+v^{2}\right)=16 \Leftrightarrow(v+1)(v+2 v+2)=2$.
Since $v \geq 0$, then $v+1 \geq 1, v^{2}+2 ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,750 |
$6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.280 \sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0$. | Solution. $\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4} \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ x+x+9+2 \sqrt{x(x+9)}=x+1+x+4+2 \sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ 2+\sqrt{x(x+9)}=\sqrt{(x+1)(x+4)}\end{array} \Leftrightarrow... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,751 |
$6.282 \sqrt{x^{2}-x-1}+\sqrt{x^{2}+x+3}=\sqrt{2 x^{2}+8}, x>0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$6.282 \sqrt{x^{2}-x-1}+\sqrt{x^{2}+x+3}=\sqrt{2 x^{2}+8}, x>0$. | Solution. Raise both sides to the square:
$\left\{\begin{array}{l}x^{2}-x-1 \geq 0, ; \\ 2 x^{2}+2+2 \sqrt{\left(x^{2}-x-1\right)\left(x^{2}+x+3\right)}=2 x^{2}+8, \\ x>0\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x^{2}-x-1 \geq 0, \\ \sqrt{\left(x^{2}-x-1\right)\left(x^{2}+x+3\right)} \\ x>0\end{array... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,752 |
$6.283 \frac{x \sqrt[5]{x}-1}{\sqrt[5]{x^{3}}-1}+\frac{\sqrt[5]{x^{3}}-1}{\sqrt[5]{x}-1}=16$ | ## Solution.
Let $y=\sqrt[5]{x}$, then $\frac{y^{6}-1}{y^{3}-1}+\frac{y^{3}-1}{y-1}=16 \Leftrightarrow y^{3}+y^{2}+y-14=0 \Leftrightarrow$ $\Leftrightarrow(y-2)\left(y^{2}+3 y+7\right)=0 \Leftrightarrow \sqrt[5]{x}=2 \Leftrightarrow x=32$.
Answer: $x=32$. | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,753 |
$6.284 \sqrt[4]{18+5 x}+\sqrt[4]{64-5 x}=4$ | Solution. Let $u=\sqrt[4]{18+5 x}, v=\sqrt[4]{64-5 x}$, then $\left\{\begin{array}{l}u+v=4, \\ u^{4}+v^{4}=82,\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ \left(u^{2}+v^{2}\right)^{2}-2 u^{2} v^{2}=82\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ \left((u+v)^{2}-2 u ... | x_{1}=\frac{63}{5},x_{2}=-\frac{17}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,754 |
$6.285 \frac{x^{2}}{\sqrt{5 x+4}}+\sqrt{5 x+4}=\frac{4}{3} x+2$. | Solution. Let's square both sides:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 5 x + 4 > 0 , } \\
{ 2 x + 3 > 0 , } \\
{ \frac { x ^ { 4 } } { 5 x + 4 } + 5 x + 4 + 2 x ^ { 2 } = \frac { 4 } { 9 } ( 2 x + 3 ) ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
5 x+4>0, \\
\frac{9 x^{4}}{5 x+4}+2 x^{2}... | x_{1}=0,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,755 |
$6.286 \sqrt{x^{3}+x^{2}-1}+\sqrt{x^{3}+x^{2}+2}=3$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.286 \sqrt{x^{3}+x^{2}-1}+\sqrt{x^{3}+x^{2}+2}=3$. | Solution. Let $u=\sqrt{x^{3}+x^{2}-1}, v=\sqrt{x^{3}+x^{2}+2}$, then $\left\{\begin{array}{l}u+v=3, \\ u^{2}-v^{2}=-3\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}u+v=3, \\ v-u=1\end{array} \Leftrightarrow\left\{\begin{array}{l}u=1, \\ v=2\end{array} \Leftrightarrow x^{3}+x^{2}-2=0 \Leftrig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,756 | |
$6.287 \frac{1}{\sqrt{x}+\sqrt[3]{x}}+\frac{1}{\sqrt{x}-\sqrt[3]{x}}=\frac{1}{3}$. | Solution. $\frac{\sqrt{x}-\sqrt[3]{x}+\sqrt{x}+\sqrt[3]{x}}{(\sqrt{x}+\sqrt[3]{x})(\sqrt{x}-\sqrt[3]{x})}=\frac{1}{3} \Leftrightarrow\left\{\begin{array}{l}6 \sqrt{x}=x-\sqrt[3]{x^{2}} \\ \sqrt{x} \neq \sqrt[3]{x} .\end{array}\right.$
Let $\sqrt[6]{x}=y$, then $\left\{\begin{array}{l}6 y^{3}=y^{6}-y^{4}, \\ y^{3}-y^{2... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,757 |
$6.288 \frac{x^{2}}{\sqrt{2 x+15}}+\sqrt{2 x+15}=2 x$. | Solution.
$$
\begin{aligned}
& \frac{x^{2}}{\sqrt{2 x+15}}+\sqrt{2 x+15}-2 x=0 \Leftrightarrow\left(\frac{x}{\sqrt[4]{2 x+15}}-\sqrt[4]{2 x+15}\right)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{x}{\sqrt[4]{2 x+15}}=\sqrt[4]{2 x+15} \Leftrightarrow x=\sqrt{2 x+15} \Leftrightarrow\left\{\begin{array}{l}
x \geq 0, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,758 |
6.289. $x^{\frac{4}{5}}-7 x^{-\frac{2}{5}}+6 x^{-1}=0$. | Solution. Let $y=\sqrt[5]{x}$, then
$y^{4}-\frac{7}{y^{2}}+\frac{6}{y^{5}}=0 \Leftrightarrow y^{9}-7 y^{3}+6=0 \Leftrightarrow$
$\Leftrightarrow\left(y^{9}-y^{3}\right)-6\left(y^{3}-1\right)=0 \Leftrightarrow\left(y^{3}-1\right)\left(y^{6}+y^{3}-6\right)=0 \Leftrightarrow$
$\Leftrightarrow\left[\begin{array}{l}y=1, ... | x_{1}=1,x_{2}=2\sqrt[3]{4},x_{3}=-3\sqrt[3]{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,759 |
$6.290 \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=x-1$. | Solution. $\sqrt{(x-1)+2 \sqrt{x-1}+1}+\sqrt{(x-1)-2 \sqrt{x-1}+1}=x-1 \Leftrightarrow$ $\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}}=x-1 \Leftrightarrow \sqrt{x-1}+1+|\sqrt{x-1}-1|=x-1 \Leftrightarrow$ $\Leftrightarrow \sqrt{x-1}+|\sqrt{x-1}-1|=x-2$.
Since $x \geq 2$, then $\sqrt{x-1} \geq 1$. ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,760 |
$6.292 \sqrt{2 x^{2}+8 x+6}+\sqrt{x^{2}-1}=2 x+2$. | Solution. $\sqrt{2(x+1)(x+3)}+\sqrt{(x-1)(x+1)}=2(x+1)$.
1) $x+1=0 \Leftrightarrow x=-1$.
2) $\sqrt{2(x+3)}+\sqrt{x-1}=2 \sqrt{x+1} \Leftrightarrow\left\{\begin{array}{l}x-1 \geq 0, \\ 2(x+3)+(x-1)+2 \sqrt{2(x+3)(x-1)}=4(x+1)\end{array} \Leftrightarrow\right.$
=x^{3} \cdot \sqrt[7]{x^{2}} \Leftrightarrow$ $\Leftrightarrow\left(x^{2}-2\right)^{8}=x^{23} \Leftrightarrow\left|x^{2}-2\right|=x^{\frac{23}{8}}$.
Consid... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,762 |
$6.294 \sqrt[3]{(2-x)^{2}}+\sqrt[3]{(7+x)^{2}}-\sqrt[3]{(7+x)(2-x)}=3$. | Solution. Let $u=\sqrt[3]{2-x}, v=\sqrt[3]{7+x}$, then $\left\{\begin{array}{l}u^{2}+v^{2}-u v \doteq 3, \\ u^{3}+v^{3}=9\end{array} \Leftrightarrow\right.$
$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array} { l }
{ u ^ { 2 } + v ^ { 2 } - u v = 3 } \\
{ ( u + v ) ( u ^ { 2 } + v ^ { 2 } - u v ) = 9 }
\end{arra... | x_{1}=1,x_{2}=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,763 |
$6.2955 \sqrt[3]{x \sqrt[5]{x}}+3 \sqrt[5]{x \sqrt[3]{x}}=8$ | Solution. $5 x^{\frac{1}{3}} \cdot x^{\frac{1}{15}}+3 x^{\frac{1}{5}} \cdot x^{\frac{1}{15}}=8 \Leftrightarrow 5 x^{\frac{6}{15}}+3 x^{\frac{4}{15}}-8=0$.
Let $y=x^{\frac{2}{15}}$, then $5 y^{3}+3 y^{2}-8=0 \Leftrightarrow\left(5 y^{3}-5\right)+\left(3 y^{2}-3\right)=0 \Leftrightarrow$ $\Leftrightarrow(y-1)\left(5 y^{... | x_{1}=1,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,764 |
$6.296 \frac{(34-x) \sqrt[3]{x+1}-(x+1) \sqrt[3]{34-x}}{\sqrt[3]{34-x}-\sqrt[3]{x+1}}=30$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.296 \frac{(34-x) \sqrt[3]{x+1}-(x+1) \sqrt[3]{34-x}}{\sqrt[3]{34-x}-\sqrt[3]{x+1}}=... | Solution. Let $u=\sqrt[3]{34-x}, v=\sqrt[3]{x+1}$, then
$$
\left\{\begin{array} { l }
{ \frac { u ^ { 3 } v - u v ^ { 3 } } { u - v } = 3 0 , } \\
{ u ^ { 3 } + v ^ { 3 } = 3 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u v(u+v)=30 \\
(u+v)\left((u+v)^{2}-3 u v\right)=35 \\
u \neq v
\end{array}\right.\righ... | x_{1}=7,x_{2}=26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,765 |
$6.297 \sqrt{x^{2}-19 x+204}-\sqrt{x^{2}-25 x-150}=3 \sqrt{\frac{x+5}{x-30}}$. | Solution. $\sqrt{x^{2}-19 x+204}=3 \sqrt{\frac{x+5}{x-30}}+\sqrt{(x+5)(x-30)} \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}\frac{x+5}{x-30} \geq 0, \\ x^{2}-19 x+204=\frac{9(x+5)}{x-30}+(x+5)(x-30)+6|x+5|\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}x \in(-\infty,-5] \cup(30,+\in... | x_{1}=31,x_{2}=\frac{-5-\sqrt{61705}}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,766 |
$6.298 \frac{\left(\sqrt[3]{(15-x)^{2}}+\sqrt[3]{(15-x)(x-6)}+\sqrt[3]{(x-6)^{2}}\right)^{2}}{\sqrt[3]{15-x}+\sqrt[3]{x-6}}=\frac{49}{3}$. | Solution. Let $u=\sqrt[3]{15-x}, v=\sqrt[3]{x-6}$, then
$$
\left\{\begin{array} { l }
{ \frac { ( u ^ { 2 } + u v + v ^ { 2 } ) ^ { 2 } } { u + v } = \frac { 4 9 } { 3 } , } \\
{ u ^ { 3 } + v ^ { 3 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{\left.(u+v)^{2}-u v\right)^{2}}{u+v}=\frac{49}{3} \\
... | x_{1}=7,x_{2}=14,x_{3,4}=\frac{21}{2}\\frac{7\sqrt{141}}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,767 |
$6.299 \frac{2}{19}\left(\sqrt{x^{2}+37 x+336}-\sqrt{x^{2}+18 x+32}\right)=\sqrt{\frac{21+x}{16+x}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$6.299 \frac{2}{19}\left(\sqrt{x^{2}+37 x+336}-\sqrt{x^{2}+18 x+32}\right)=... | Solution. Note that $x^{2}+37 x+336 \geq x^{2}+18 x+32$, i.e., $x \geq-16$.
Therefore, the original equation is equivalent to
$$
\begin{aligned}
& \left\{\begin{array}{l}
\frac{2}{19} \sqrt{x+16}(\sqrt{x+21}-\sqrt{x+2})=\frac{\sqrt{x+21}}{\sqrt{x+16}}, \Leftrightarrow \\
x>-16
\end{array}\right. \\
& \Leftrightarrow\... | 79 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,768 |
6.300 \sqrt{x-2}+\sqrt{4-x}=x^{2}-6 x+11 | Solution. $\sqrt{(x-3)+1}+\sqrt{1-(x-3)}=\left(x^{2}-6 x+9\right)+2$.
Let $y=x-3$, then $\sqrt{y+1}+\sqrt{1-y}=y^{2}+2 \Leftrightarrow$ $y^{4}+4 y^{2}+2\left(1-\sqrt{1-y^{2}}\right)=0$.
Since $1 \geq \sqrt{1-y^{2}}$, the last equation is equivalent to the system:
$$
\left\{\begin{array}{l}
y^{4}+4 y^{2}=0 \\
1=\sqrt... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,769 |
$6.3016 \sqrt[3]{x-3}+\sqrt[3]{x-2}=5 \sqrt[6]{(x-2)(x-3)}$. | Solution. There are two cases:
54
1) $\left\{\begin{array}{l}x-2 \geq 0, \\ x-3 \geq 0,\end{array} \Rightarrow x \geq 3\right.$. Let $u=\sqrt[6]{x-2}, v=\sqrt[6]{x-3}$, then
$6 v^{2}+u^{2}=5 u v \Leftrightarrow u^{2}-5 u v+6 v^{2}=0 \Leftrightarrow u=\frac{5 v \pm \sqrt{25 v^{2}-24 v^{2}}}{2} \Leftrightarrow$
$\Lef... | x_{1}=\frac{190}{63},x_{2}=\frac{2185}{728} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,770 |
6.312 \{ $\left\{\begin{array}{l}x-y+z=6, \\ x^{2}+y^{2}+z^{2}=14, \\ x^{3}-y^{3}+z^{3}=36 .\end{array}\right.$ | Solution. $\left\{\begin{array}{l}x+z=6+y, \\ x^{2}+z^{2}=14-y^{2} \\ (x+z)\left(x^{2}-x z+z^{2}\right)=36+y^{3} .\end{array}\right.$
Square the first equation and subtract the second:
$x z=y^{2}+6 y+11 \Rightarrow(6+y)\left(14-y^{2}-\left(y^{2}+6 y+11\right)\right)=36+y^{3} \Leftrightarrow$
$\Leftrightarrow y^{3}+6 ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,771 |
6.319
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=19 \\
(xy+8)(x+y)=2
\end{array}\right.
$$ | Solution. $\left\{\begin{array}{l}(x+y)\left((x+y)^{2}-3 x y\right)=19, \\ (x y+8)(x+y)=2 .\end{array}\right.$ Let $u=x+y, v=x y$, then
$$
\left\{\begin{array} { l }
{ u ^ { 3 } - 3 u v = 1 9 , } \\
{ u v + 8 u = 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u^{3}-3 u v=19, \\
24 u+3 u v=6
\end{array} \Righ... | (9,6),(6,9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,772 |
6.326
$$
\left\{\begin{array}{l}
9\left(u^{4}+v^{4}\right)=17(u+v)^{2} \\
3 u v=-2(u+v)
\end{array}\right.
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,773 | ||
6.342 Let $S_{n}=\alpha^{n}+\beta^{n}$, where $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$. Find the relationship between $S_{n}, S_{n+1}, S_{n+2}$. | ## Solution.
$$
\begin{gathered}
S_{n+1} \cdot S_{1}=\left(\alpha^{n+1}+\beta^{n+1}\right)(\alpha+\beta)=\left(\alpha^{n+2}+\beta^{n+2}\right)+\alpha \beta\left(\alpha^{n}+\beta^{n}\right)=S_{n+2}+\alpha \beta \cdot S_{n} \\
\text { Since } S_{1}=-\frac{b}{a}, \alpha \beta=\frac{c}{a} \Rightarrow S_{n+1} \cdot\left(-\... | S_{n+2}=\frac{-(b\cdotS_{n+1}+\cdotS_{n})}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,774 |
6.343 Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}+p x^{2}+q x+r=0$.
1) Form the equation with roots $x_{1} x_{2}, x_{2} x_{3}, x_{1} x_{3}$;
2) use the result from part 1 to find the roots of the equation $x^{3}-3 \sqrt{2} x^{2}+7 x-3 \sqrt{2}=0$. | Solution. 1) By Vieta's theorem, we have: $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-p, \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=q, \\ x_{1} \cdot x_{2} \cdot x_{3}=-r .\end{array}\right.$
By the inverse of Vieta's theorem, the desired equation is:
$y^{3}-\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) y^{2}+\left(x_{... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,775 |
6.344 Find the coefficients $a$ and $b$ of the equation $x^{4}+x^{3}-18 x^{2}+a x+b=0$, if among its roots there are three equal integer roots. | Solution. Let $x_{1}=x_{2}=x_{3}, x_{4}$ be the roots of the original equation. By Vieta's theorem,
we have: $\left\{\begin{array}{l}3 x_{1}+x_{4}=-1, \\ 3 x_{1}^{2}+3 x_{1} x_{4}=-18, \\ x_{1}^{3}+3 x_{1}^{2} x_{4}=-a, \\ x_{1}^{3} x_{4}=b\end{array} \Rightarrow x_{1}^{2}+x_{1}\left(-1-3 x_{1}\right)=-6 \Rightarrow x... | =-52,b=-40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,776 |
### 6.345 Find the coefficients $p$ and $q$ of the equation
$x^{4}-10 x^{3}+37 x^{2}+p x+q=0$, if among its roots there are two pairs of equal numbers. | Solution.
Let $x_{1}=x_{2}, x_{3}=x_{4}$ be the roots of the original equation. By Vieta's theorem, we have:
$\left\{\begin{array}{l}x_{1}+x_{3}=5, \\ x_{1}^{2}+4 x_{1} x_{3}+x_{3}^{2}=37, \\ 2 x_{1} x_{3}\left(x_{1}+x_{3}\right)=-\dot{p}, \\ \left(x_{1} x_{3}\right)^{2}=q\end{array} \Leftrightarrow\left\{\begin{arra... | p=-60,q=36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,777 |
6.346 For the equation $x^{3}+a x^{2}+b x+1=0$, express the product of the sum of its roots by the sum of their reciprocals in terms of the coefficients $a$ and $b$. | Solution. By Vieta's theorem $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-a, \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=b, \\ x_{1} \cdot x_{2} \cdot x_{3}=-1 .\end{array}\right.$
Therefore $\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-a \frac{x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3... | b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,778 |
### 6.347 Prove that the equality $a b=c$ is a necessary and
sufficient condition for the fact that among the roots of the equation $x^{3}+a x^{2}+b x+c=0$ there are two numbers whose sum is zero. | ## Solution.
1) $a b=c \Rightarrow x^{3}+a x^{2}+b x+a b=0 \Leftrightarrow\left(x^{2}+b\right)(x+a)=0 . \Rightarrow$
$\Rightarrow x_{1}=-a, x_{2}=-\sqrt{-b}, x_{3}=\sqrt{-b} \Rightarrow x_{2}+x_{3}=0 \quad$ (if $b>0$, then $x_{2}$ and $x_{3}$ are complex roots).
2) Let $x_{1}, x_{2}, x_{3}$ be the roots and $x_{2}+x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,779 |
6.349 Solve the equations $2 x^{3}-5 x^{2}+6 x-2=0$ and $6 x^{3}-3 x^{2}-2 x+1=0$, given that they have a common root. | Solution. Let $f(x)=6 x^{3}-3 x^{2}-2 x+1, g(x)=2 x^{3}-5 x^{2}+6 x-2$.
Represent $f(x)$ in the form (3). Then $r_{1}(x)=12 x^{2}-20 x+7$.
Next, $g(x)=r_{1}(x) \cdot q_{2}(x)+r_{2}(x)$, where $r_{2}(\dot{x})=\frac{37}{18} x-\frac{37}{36} \Rightarrow r_{2}(x)=0$ when $x_{0}=\frac{1}{2}$.
Thus, $g(x)=(2 x-1)\left(x^{2... | \frac{1}{2},x_{2,3}=\\frac{1}{\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,780 |
6.350 Form the equation of the third degree from its roots $x_{1}^{2}, x_{1} x_{2}$ and $x_{2}^{2}$, if $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$. | Solution. By the inverse Vieta's theorem, the desired equation has the form:
$y^{3}-\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right) y^{2}+\left(x_{1}^{3} x_{2}+x_{1} x_{2}^{3}+x_{1}^{2} x_{2}^{2}\right) y-x_{1}^{3} x_{2}^{3}=0$.
But by Vieta's theorem for the equation $x^{2}+p x+q=0$, we have:
$x_{1} x_{2}=q, x_{1}+x_{... | y^{3}-(p^{2}-q)y^{2}+(p^{2}q-q^{2})y-q^{3}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,781 |
6.351 Solve the equations $x^{3}-6 x^{2}-39 x-10=0$ and $x^{3}+x^{2}-20 x-50=0$, given that one of the roots of the first equation is twice one of the roots of the second equation. | Solution. Let $x_{0}$ be the root of the second equation, $2 x_{0}$ be the root of the first equation, $f(x)=x^{3}-6 x^{2}-39 x-10, g(x)=x^{3}+x^{2}-20 x-50$. Then $x_{0}$ is the root of the equation $\tilde{f}(x)=0$, where $\tilde{f}(x)=f(2 x)=8 x^{3}-24 x^{2}-78 x-10$. Represent $\tilde{f}(x)$ in the form (3). Then $... | x_{1}=10,x_{2,3}=-2\\sqrt{3};5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,782 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.