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5. Consider the crank-slider mechanism (1-crank, 2- connecting rod). They were first used in antiquity, in Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The given crank rotates at a constant angular velocity $\omega=10$ ... | Solution. Let's write the equations of motion of point M in coordinate form. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,121 |
1. Solve the equation $9^{x}+4 \cdot 3^{x+1}=13$. | Solution: $3^{2 x}+4 \cdot 3^{x+1}=13, 3^{2 x}+12 \cdot 3^{x}-13=0$,
$\left(3^{x}\right)_{1,2}=\frac{-12 \pm \sqrt{144+52}}{2}=\frac{-12 \pm 14}{2}=\left[\begin{array}{l}1, \\ -13 .- \text { n.s. }\end{array} 3^{x}=1 \Rightarrow x=0\right.$. Answer: $x=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,127 |
3. Find all extrema of the function $y=x \lg 2 x$.
---
Note: The translation maintains the original format and line breaks as requested. | Solution: $y=x \lg 2 x, x>0, \quad y'=\lg 2 x+\frac{x \cdot 2}{2 x \ln 10}=\frac{\ln 2 x}{\ln 10}+\frac{1}{\ln 10}=0$, Answer: $x=\frac{e^{-1}}{2}, y=\frac{-1}{2 e \ln 10}$.
 | \frac{e^{-1}}{2},\frac{-1}{2e\ln10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,128 |
4. Calculate $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}$. | Solution: $(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})^{2}=6-2 \sqrt{9-5}=2, A=\sqrt{2}$. Answer: $\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,129 |
5. Solve the inequality $\frac{1-0.4 x^{2}}{5-x} \geq \frac{2 x}{5}$. | Solution: $\frac{1-0.4 x^{2}}{5-x} \geq \frac{2 x}{5}, \frac{1-\frac{2}{5} x^{2}}{5-x}-\frac{2 x}{5} \geq 0$,
$\frac{5-2 x^{2}-2 x(5-x)}{5(5-x)} \geq 0, \frac{5-10 x}{5(5-x)} \geq 0$ ;
An... | (-\infty;\frac{1}{2}]\cup(5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,130 |
6. Solve the equation $\sqrt{3 x^{2}+5 x+7}-\sqrt{3 x^{2}+5 x+2}=1$. | Solution: Domain of definition $3 x^{2}+5 x+2 \geq 0 . x_{1,2}=\frac{-5 \pm \sqrt{25-24}}{6}=\frac{-5 \pm 1}{6}=\left[\begin{array}{l}\frac{2}{3}, \\ -1\end{array}\right.$
$t=3 x^{2}+5 x+7, \quad \sqrt{t}=1+\sqrt{t-5} ; \quad t=1+2 \sqrt{t-5}+t-5$;
$4=2 \sqrt{t-5} ; t-5=4 ; t=9 ; 3 x^{2}+5 x+7=9 ; 3 x^{2}+5 x-2=0$;
... | -2;\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,131 |
8. Solve the system $\left\{\begin{array}{l}x+y-20=0 \\ \log _{4} x+\log _{4} y=1+\log _{4} 9 .\end{array}\right.$ | Solution: $\left\{\begin{array}{l}x+y=20, \\ x y=36,\end{array} x, y>0,\left\{\begin{array}{l}y=20-x, \\ x(20-x)=36\end{array},\left\{\begin{array}{l}20 x-x^{2}=36 \\ y=20-x,\end{array},\left\{\begin{array}{l}x^{2}-20+36=0, \\ y=20-x,\end{array}\right.\right.\right.\right.$ $\left\{\begin{array}{l}x=18, \\ y=2\end{arra... | {\begin{pmatrix}18,\\2\end{pmatrix}. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,132 |
9. Find the maximum value of the function $f(x)=3 \sin x+4 \cos x$. | Solution: $f(x)=3 \sin x+4 \cos x=\sqrt{3^{2}+4^{2}} \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)=5 \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)$.
Answer: The maximum value is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,133 |
10. The seller is distributing candies into gift boxes. When arranging them into boxes with a capacity of 10 candies, 6 candies are left over, and when arranging them into boxes with a capacity of 15 candies, 4 candies are missing. Find the number of candies, given that there were no fewer than 200 and no more than 250... | Solution: $\begin{array}{ll}a=10 x+6=10(x+1)-4, & a+4 \vdots 10, \\ a=15 y-4, & a+4 \vdots 15,\end{array} \Rightarrow a+4 \vdots 30, a+4=180,210,240,270$.
Answer: 206, 236.
## TICKET № 2 | 206,236 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,134 |
1. Solve the equation $9^{x-1}+3^{x+2}=90$. | Solution: $9^{x-1}+3^{x+2}=90,\left(3^{x-1}\right)^{2}+27 \cdot 3^{x-1}=90$, $3^{x-1}=\frac{-27 \pm \sqrt{3^{6}+4 \cdot 3^{2} \cdot 10}}{2}=\frac{-27 \pm 3 \sqrt{121}}{2}=\frac{-27 \pm 33}{2}=\left[\begin{array}{l}3, \\ -30 .\end{array} \quad x-1=1, x=2\right.$ Answer: $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,135 |
2. The first term of the geometric progression $b_{1}=\sqrt{3}$, the fifth term $b_{5}=\sqrt{243}$. Find the sixth term and the common ratio of the progression. | Solution: $b_{1}=\sqrt{3} ; b_{5}=\sqrt{243}, b_{6}=$ ? $q=$ ?
$b_{5}=b_{1} \cdot q^{4}=\sqrt{3} \cdot q^{4}=\sqrt{243} \Rightarrow q^{4}=\sqrt{\frac{243}{3}}=3^{2} ; q= \pm \sqrt{3}, b_{6}=\sqrt{243} \cdot q= \pm \sqrt{81 \cdot 3^{2}}= \pm 27$.
Answer: $b_{6}= \pm 27$ | b_{6}=\27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,136 |
3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[-1 ; 1]$. | Solution: $y=x^{3}-3 x^{2}+5$ on $[-1 ; 1], y^{\prime}=3 x^{2}-6 x=0,3 x(x-2)=0, x=0 ; x=2$, $y(-1)=1 ; y(1)=3 ; y(0)=5$. Answer: $5 ; 1$. | 5;1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,137 |
4. Calculate $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}$. | Solution: $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=A ; A^{2}=8-2 \sqrt{16-7}=2, A= \pm \sqrt{2} \Rightarrow A=\sqrt{2}$.
Answer: $\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,138 |
5. Solve the inequality $\frac{4-x}{x-5}>\frac{1}{1-x}$. | Solution: $\frac{4-x}{x-5}+\frac{1}{x-1}>0 ; \frac{(4-x)(x-1)+x-5}{(x-5)(x-1)}>0$,
$\frac{-x^{2}+6 x-9}{(x-5)(x-1)}>0, \frac{-(x-3)^{2}}{(x-1)(x-5)}>0$. Answer: $(1 ; 3) \cup(3 ; 5)$. | (1;3)\cup(3;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,139 |
6. Solve the equation $\sqrt{\frac{x-3}{2 x+1}}+2=3 \sqrt{\frac{2 x+1}{x-3}}$. | Solution: $t=\frac{x-3}{2 x+1}>0 ; \sqrt{t}+2=\frac{3}{\sqrt{t}}, \sqrt{t}+2 \sqrt{t}-3=0, \sqrt{t}=\left[\begin{array}{l}1 \\ -3 \text {, but this is not valid }\end{array}\right.$
$\frac{x-3}{2 x+1}=1 ; x-3=2 x+1, x=-4, \sqrt{\frac{-7}{-7}}+2=3 \sqrt{\frac{-7}{-7}}$. Answer: $x=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,140 |
7. The area of an equilateral triangle inscribed in a circle is 81 cm $^{2}$. Find the radius of the circle. | Solution: $S_{\Delta}=81 c M^{2}, S_{\Delta}=\frac{1}{2} a \cdot h=\frac{1}{2} a \cdot a \cos 30^{\circ}=\frac{a^{2} \sqrt{3}}{4}, r=$ ?
$S_{\triangle}=3 S_{\triangle A O C}=3 \cdot \frac{1}{2} a \cdot r \cos 60^{\circ}=\frac{3}{4} a r=\frac{3}{2} r \cdot r \cdot \sin 120^{\circ}=\frac{3 r^{2} \sqrt{3}}{4}$,
$\frac{3... | 6\sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,141 |
9. Solve the equation $4 \sin x - 3 \cos x = 5$.
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Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
---
9. Solve the equation $4 \sin x - 3 \cos x = 5$. | Solution: $5=4 \sin x-3 \cos x ; \sqrt{4^{2}+3^{2}} \cdot \sin \left(x-\operatorname{arctg} \frac{3}{4}\right)=5$, $x-\operatorname{arctg} \frac{3}{4}=\frac{\pi}{2}+2 \pi k, x=\operatorname{arctg} \frac{3}{4}+\frac{\pi}{2}+2 \pi k$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,142 | |
2. Find the sum of the first fifteen terms of an arithmetic progression if its third term is -5, and the fifth term is 2.4. | Solution: $a_{3}=-5 ; a_{5}=2.4, S_{15}=$ ?
$\left\{\begin{array}{l}a_{1}+2 d=-5, \\ a_{1}+4 d=2.4 ;\end{array} 2 d=7.4, d=3.7 ; d_{1}=-5-2 d=-12.4\right.$,
$S_{15}=\frac{2 \cdot(-12.4)+3.7 \cdot 14}{2} \cdot 15=202.5$. Answer: $S_{15}=202.5$ | 202.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,143 |
3. Find all extrema of the function $y=\sin ^{2} 3 x$ on the interval $(0 ; 0.6)$. | Solution: $y=\sin ^{2} 3 x,(0 ; 0.6) ; y^{\prime}=2 \sin 3 x \cdot \cos 3 x=\sin 6 x=0 ; x=\frac{\pi k}{6}$ $x=0, y_{\max }\left(\frac{\pi}{6}\right)=1$
 | y_{\max}(\frac{\pi}{6})=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,144 |
4. Calculate $\sqrt{4+\sqrt{15}}-\sqrt{4-\sqrt{15}}$. | Solution: $\sqrt{4+\sqrt{15}}-\sqrt{4-\sqrt{15}}=A ; A^{2}=8-2 \sqrt{16-15}=1, A= \pm \sqrt{6}, A=\sqrt{6}$. | \sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,145 |
5. Solve the inequality $\frac{5 x^{2}}{x-1}>5 x+1$. | Solution: $\frac{5 x^{2}}{x-1}>5 x+1 ; \frac{5 x^{2}(5 x+1)(x-1)}{x-1}>0$,
$\frac{5 x^{2}-5 x^{2}+4 x+1}{x-1}>0$. Answer: $x \in\left(-\infty ;-\frac{1}{4}\right) \cup(1 ;+\infty)$. | x\in(-\infty;-\frac{1}{4})\cup(1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,146 |
6. Solve the equation $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12}$. | Solution: $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12} ; t=\frac{2 x+2}{x+2}>0, \sqrt{t}-\sqrt{\frac{1}{t}}=\frac{7}{12}$,
$\sqrt{t}=\frac{7 \pm \sqrt{49+4 \cdot 144}}{24}=\frac{7 \pm 25}{24}=\left[\begin{array}{l}\frac{32}{24}=\frac{4}{3} \\ -\frac{18}{24}=-\frac{3}{4}\end{array}, \frac{2 x+2}{x+2}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,147 |
7. A chord of length $10 \mathrm{~cm}$ is drawn in a circle. Through one of its ends, a tangent to the circle is drawn, and through the other end, a secant parallel to the tangent is drawn. The inner segment of the secant is 12 cm. Find the radius of the circle. | Solution: $\quad A B=10, \quad \sqrt{r^{2}}=\sqrt{O D^{2}+B D^{2}}=\sqrt{x^{2}+6^{2}}, \quad \triangle D O B$ $B C=12,(r+x)^{2}+6^{2}=10^{2}$,
$r=? \quad(\triangle A D B)$$\Rightarrow\left\{\begin{array}{l}r+x=\sqrt{64}=8, \\ r^{2}-x^{2}=36,\end{array}\right.$,
$x=8-r, \quad r^{2}-(8-r)^{2}=36, \quad r^{2}-64+16 r-r^{... | \frac{25}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,148 |
8. Solve the system $\left\{\begin{array}{l}3^{x} \cdot 2^{y}=972 ; \\ \log _{\sqrt{3}}(x-y)=2 .\end{array}\right.$ | Solution: $\left\{\begin{array}{l}3^{x} \cdot 2^{y}=972, \\ x-y=3,\end{array},\left\{\begin{array}{l}x=3+y, \\ 3^{3+y} \cdot 2^{y}=9 \cdot 9 \cdot 3 \cdot 4\end{array},\left\{\begin{array}{l}x=3+y, \\ 27 \cdot(3 \cdot 2)^{y}=9 \cdot 3 \cdot 9 \cdot 4,\end{array} 6^{y}=36,\left\{\begin{array}{l}x=5, \\ y=2\end{array}\ri... | {5;2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,149 |
9. Find the maximum value of the function $f(x)=6 \sin x+8 \cos x$. | Solution: $f(x)=6 \sin x+8 \cos x=\sqrt{6^{2}+8^{2}} \sin \left(x+\operatorname{arctg} \frac{8}{6}\right)$.
Answer: The maximum value is 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,150 |
10. For a sports parade, the coach decided to line up the children in rows of 8, but 5 children were left over. Then he lined them up in rows of 10, but 3 places were left empty. It is known that there were no fewer than 100 and no more than 150 children. How many children were there? | Solution: $a=8 x+5=8(x+1)-3, a+3 \vdots 8$,
$$
\begin{array}{lc}
a=8 x+5=8(x+1)-3, & a+3 \vdots 8 \\
a=10 y-3, & a+3 \vdots 10,
\end{array} \Rightarrow a+3 \vdots 40, a+3=\underline{120,160}
$$
Answer: 117.
TICKET № 4 | 117 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,151 |
1. Solve the equation $2 \cdot 4^{2 x}-17 \cdot 4^{x}+8=0$. | Solution: $2 \cdot 4^{2 x}-17 \cdot 4^{x}+8=0, 2\left(4^{x}\right)^{2}-17 \cdot 4^{x}+8=0$,
$\left(4^{x}\right)_{1,2}=\frac{17 \pm \sqrt{89-64}}{4}=\frac{17 \pm 15}{4}=\left[\begin{array}{l}\frac{32}{4} \\ \frac{1}{2}\end{array}=\left[\begin{array}{l}8 \\ \frac{1}{2}\end{array}, x=\left[\begin{array}{cc}\frac{3}{2} & ... | \frac{3}{2},-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,152 |
2. The second term of the geometric progression $b_{2}=37 \frac{1}{3}$, the sixth term $b_{6}=2 \frac{1}{3}$. Find the first term and the common ratio of the progression. | Solution: $b_{2}=37 \frac{1}{3}=\frac{112}{3} ; b_{6}=2 \frac{1}{3}=\frac{7}{3}, b_{1}=? q=$ ?
 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,153 | |
3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[1 ; 3]$. | Solution: $y=x^{3}-3 x^{2}+5,[1 ; 3] ; y^{\prime}=3 x^{2}-6 x=3 x(x-2) \Rightarrow x=0 ; 2$
$y(1)=3 ; y(3)=5 ; y(2)=1$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,154 |
4. Calculate $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}$. | Solution: $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}=A ; A^{2}=8-2 \sqrt{16-12}=4 ; \quad A= \pm 2$ Answer: $A=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,155 |
5. Solve the inequality $\frac{5 x+1}{x-3} \leq \frac{1}{3}$. | # Solution:
$\frac{5 x+1}{x-3} \leq \frac{1}{3}, \frac{5 x+1}{x-3}-\frac{1}{3} \leq 0, \frac{15 x+3-x+3}{3(x-3)} \leq 0, \frac{2(7 x+3)}{3(x-3)} \leq 0$
Answer: $\left[-\frac{3}{7} ; 3\right)$.
 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,156 |
6. Solve the equation $\sqrt{\frac{3 x-1}{x+4}}+3-4 \sqrt{\frac{x+4}{3 x-1}}=0$.
Solution: $t=\frac{3 x-1}{x+4}>0, \sqrt{t}+3-\frac{4}{\sqrt{t}}=0, \sqrt{t}^{2}+3 \sqrt{t}-4=0, \sqrt{t}=\left[\begin{array}{l}1 \\ -4 \text { (not valid) }\end{array}\right.$
$\frac{3 x-1}{x+4}=1,3 x-1=x+4,2 x=5, x=\frac{5}{2}$. Verifica... | Answer: $x=\frac{5}{2}$. | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,157 |
8. Solve the system $\left\{\begin{array}{l}x \log _{2} 3+y=\log _{2} 18 ; \\ 5^{x}=25^{y} .\end{array}\right.$
 | # Solution:
$\left\{\begin{array}{l}\log _{2} 3^{x}+y=\log _{2} 18, \\ 5^{x}=5^{2 y},\end{array}\left\{\begin{array}{l}2 y \log _{2} 3+y=18 \\ x=2 y,\end{array},\left\{\begin{array}{l}y\left(\log _{2} 3^{2}+1\right)=18, \\ x=2 y,\end{array}\left\{\begin{array}{l}y\left(\log _{2}\left(3^{2} \cdot 2\right)\right)=18, \\... | {2;1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,159 |
9. Solve the equation $8 \sin x - 6 \cos x = 10$.
---
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
---
9. Solve the equation $8 \sin x - 6 \cos x = 10$. | Solution: $8 \sin x-6 \cos x=10, \sqrt{8^{2}+6^{2}} \sin \left(x-\operatorname{arctg} \frac{6}{8}\right)=10, x-\operatorname{arctg} \frac{6}{8}=\frac{\pi}{2}+2 \pi k$.
Answer: $x=\operatorname{arctg} \frac{3}{4}+\frac{\pi}{2}+2 \pi k$. | \operatorname{arctg}\frac{3}{4}+\frac{\pi}{2}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,160 |
10. In the packaging workshop, there are 2 types of boxes: one for 20 parts and one for 27 parts. A batch of parts between 500 and 600 pieces has arrived for packaging. When the parts are packed in the first type of box, 13 parts are left unpacked, and when packed in the second type of box, 7 spaces are left unfilled. ... | Solution: ${ }^{a=20 x+13=20(x+1)-7,} a+7 \vdots 20 ; a+7 \vdots 27, \Rightarrow a+7 \vdots 540$.
$$
a=27 x-7
$$
Answer: $a=533$.
## TICKET № 5 | 533 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,161 |
1. Solve the equation $5^{2 x}-7 \cdot 5^{x}+10=0$. | Solution: $5^{2 x}-7 \cdot 5^{x}+10=0,\left(5^{x}\right)^{2}-7 \cdot\left(5^{x}\right)+10=0,5^{x}=\frac{7 \pm \sqrt{49-40}}{2}=\left[\begin{array}{l}5 \\ 2\end{array}\left[\begin{array}{c}x=1 \\ x=\log _{5} 2\end{array}\right.\right.$.
Answer: $x=1, x=\log _{5} 2$. | 1,\log_{5}2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,162 |
2. The third term of an arithmetic progression $a_{3}=3$, the eleventh term $a_{11}=15$.
Find the first term and the common difference of the progression. | Solution: $a_{3}=3 ; a_{11}=15$. Find $a_{1}, d$.
$\left\{\begin{array}{l}a_{1}+2 d=3, \\ a_{1}+10 d=15 ;\end{array}\left\{\begin{array}{l}8 d=12, \\ a_{1}=3-2 d ;\end{array}\left\{\begin{array}{l}d=\frac{3}{2}, \\ a_{1}=0 .\end{array}\right.\right.\right.$ | a_{1}=0,=\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,163 |
3. Find all extrema of the function $y=\frac{2}{3} \cos \left(3 x-\frac{\pi}{6}\right)$ on the interval $(0 ; \pi / 2)$. | Solution: $y=\frac{2}{3} \cos \left(3 x-\frac{\pi}{6}\right) ; y^{\prime}=\frac{2}{3}\left(-\sin \left(3 x-\frac{\pi}{6}\right)\right) \cdot 3=-2 \sin \left(3 x-\frac{\pi}{6}\right)$
$\left(0 ; \frac{\pi}{2}\right) \cdot 3 x-\frac{\pi}{6}=\pi k, x=\pi k+\frac{\pi}{6}=\frac{\pi(6 k+1)}{18}, k=1,0$
$y_{\max }\left(\fra... | y_{\max}(\frac{\pi}{18})=\frac{2}{3},y_{\}(\frac{7\pi}{18})=-\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,164 |
4. Calculate $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}$. | Solution: $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}=A ; A^{2}=6-2 \sqrt{9-8}=4, \quad A= \pm 2, \quad A=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,165 |
5. Solve the inequality $\frac{x+2}{x-1}>\frac{x+4}{x-3}$. | Solution: $\frac{x+2}{x-1}>\frac{x+4}{x-3}, \frac{x+2}{x-1}-\frac{x+4}{x-3}>0, \frac{(x+2)(x-3)-(x+4)(x-1)}{(x-1)(x-3)}>0$, $\frac{x^{2}-x-6-x^{2}-3 x+4}{(x-1)(x-3)}>0, \frac{-4 x-2}{(x-1)(x-3)}>0$.
Answer: $\left(-\infty ;-\frac{1}{2}\right) \cup(1 ; 3)$. | (-\infty;-\frac{1}{2})\cup(1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,166 |
6. Solve the equation $\sqrt{2 x^{2}+3 x-5}-\sqrt{2 x^{2}+3 x-8}=1$. | Solution: $\sqrt{2 x^{2}+3 x-5}-\sqrt{2 x^{2}+3 x-8}=1,\left\{\begin{array}{l}2 x^{2}+3 x-5 \geq 0, \\ 2 x^{2}+3 x-8 \geq 0,\end{array}\left\{\begin{array}{l}x \leq-\frac{3-\sqrt{73}}{4}, \\ x \geq-\frac{3+\sqrt{73}}{4}\end{array}\right.\right.$,
$\sqrt{t}=1+\sqrt{t-3} ; t=1+2 \sqrt{t-3}+t-3 ; \sqrt{t-3}=1 ; t-3=1, t=4... | \frac{3}{2},-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,167 |
7. The distance from the point of intersection of the diameter of a circle with radius $11 \mathrm{~cm}$ and a chord of length 18 cm to the center of the circle is 7 cm. In what ratio does the point of intersection divide the chord? | Solution: $R=11, AB=18, PO=7, \frac{AP}{PB}=$ ?
$AP \cdot PB=DP \cdot PC, DP=DO-PO=11-7=4, PC=11+7=18$,
$\left\{\begin{array}{l}AP \cdot PB=4 \cdot 18 \\ AP+PB=18\end{array}, AP(18-AP)=72,18AP-AP^{2}=72\right.$,
=5 \sin x+12 \cos x$. | Solution: $f(x)=5 \sin x+12 \cos x=\sqrt{5^{2}+12^{2}}\left(\sin \left(x+\operatorname{arctg} \frac{12}{5}\right)\right)=\sqrt{13^{2} \sin x}$.
Answer: 13 . | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,170 |
10. A florist received between 300 and 400 roses for a celebration. When he arranged them in vases with 21 roses in each, 13 roses were left. But when arranging them in vases with 15 roses in each, 8 roses were missing. How many roses were there in total? | Solution: $\left\{\begin{array}{ll}a=21 x+13=21(x+1)-8, & a+8 \vdots 21, \\ a=15 y-8, & a+8 \vdots 15,\end{array}\right\} \Rightarrow a+8 \vdots 105$. Answer: $a=307$.
$$
a+8=105,210, \underline{315,} 420
$$
## TICKET № 6 | 307 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,171 |
1. Solve the equation $3 \cdot 9^{x}+2 \cdot 3^{x}=1$. | Solution: $3 \cdot 9^{x}+2 \cdot 3^{x}=1,3 \cdot\left(3^{x}\right)^{2}+2 \cdot 3^{x}-1=0,3^{x}=\frac{-2 \pm \sqrt{4+12}}{6}=\left[\begin{array}{l}\frac{1}{3}, \\ -1-\text { not valid }\end{array} \quad x=-1\right.$.
Answer: $x=-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,172 |
2. The fourth term of a geometric progression $b_{4}=\frac{1}{25}$, the fifth term $b_{5}=\frac{1}{125}$. Find the sum of the first five terms. | Solution: $b_{4}=\frac{1}{25} ; \quad b_{5}=\frac{1}{125}, S_{5}=$ ?
$$
\left\{\begin{array}{l}
b_{1} \cdot q^{3}=\frac{1}{25}, \quad q=\frac{1}{125} \cdot \frac{25}{1}=\frac{1}{5}, \quad S_{5}=\frac{5\left(1-\left(\frac{1}{5}\right)^{5}\right)}{1-\frac{1}{5}}=\frac{781}{125} \\
b_{1} \cdot q^{4}=\frac{1}{125} ; \quad... | \frac{781}{125} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,173 |
3. Find the greatest and least values of the function $y=3 x^{4}+4 x^{3}+34$ on the interval $[-2 ; 1]$. | Solution: $y=3 x^{4}+4 x^{3}+34 ; y^{\prime}=12 x^{3}+12 x^{2}=12 x^{2}(x+1)$
$x=0 ; x=-1 . y(-2)=50 ; y(1)=41 ; y(-1)=33$. | y(-2)=50,y(1)=41,y(-1)=33 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,174 |
4. Calculate $\sqrt{5+\sqrt{24}}-\sqrt{5-\sqrt{24}}$. | Solution: $\sqrt{5+\sqrt{24}}-\sqrt{5-\sqrt{24}}=A ; A^{2}=10-2 \sqrt{25-24}=8, A=\sqrt{8}$. | \sqrt{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,175 |
5. Solve the inequality $2 x-1>\frac{3}{2 x+1}$. | Solution:
$$
2 x-1>\frac{3}{2 x+1} ; 2 x-1-\frac{3}{2 x+1}>0
$$
$$
\frac{4 x^{2}-1-3}{2 x+1}>0 ; \frac{4(x-1)(x+1)}{2 x+1}>0
$$
Answer: $\left(-1 ;-\frac{1}{2}\right) \cup(1 ;+\infty)$. | (-1;-\frac{1}{2})\cup(1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,176 |
6. Solve the equation $\sqrt{3 x^{2}-5 x-4}+\sqrt{3 x^{2}-5 x-8}=2$. | Solution: $\sqrt{3 x^{2}-5 x-4}+\sqrt{3 x^{2}-5 x-8}=2$, Domain of Definition (ODZ) $3 x^{2}-5 x-8 \geq 0$.
$t=3 x^{2}-5 x-4, \quad \sqrt{t}+\sqrt{t-4}=2, \quad t+t-4+2 \sqrt{t(t-4)}=4, \quad 2 \sqrt{t(t-4)}=8-2 t$, $\sqrt{t-(t-4)}=4-t, \quad(t \leq 4) . \quad t^{2}-4 t=16-8 t+t^{2}, \quad t=\frac{8}{2}=4 . \quad 3 x^{... | \frac{8}{3};-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,177 |
7. Two vertices of a square with an area of $256 \mathrm{~cm}^{2}$ lie on a circle, while the other two vertices lie on a tangent to this circle. Find the radius of the circle. | Solution: $S_{A B C D}=256 \, \text{cm}^{2} \Rightarrow a=16 \, \text{cm}=x$,
$E F=2 R-x, F O=R-E F=R-(2 R-x)=x-R$,
$\triangle F C O: F O^{2}=R^{2}-F C^{2} \Rightarrow(x-R)^{2}=R^{2}-8^{2}$,
$x^{2}-2 R x+R^{2}=R^{2}-8^{2}, 2 R x=16^{2}+8^{2} \Rightarrow R=\frac{16^{2}+8^{2}}{2 \cdot 16}=10$. Answer:
$R=10$.
=13, x-\operatorname{arctg} \frac{5}{12}=\frac{\pi}{2}+2 \pi k$. | x-\operatorname{arctg}\frac{5}{12}=\frac{\pi}{2}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,179 |
10. Koschei is counting his gold coins. When he counts them by tens, there are 7 coins left, and he is 3 coins short of a whole number of dozens. Koschei's wealth is estimated at $300-400$ coins. How many coins does Koschei have? | Solution: $\left\{\begin{array}{ll}a=10 x+7=10(x+1)-3, & a+3 \mid 10, \\ a=12 y-3, & a+3 \mid 12,\end{array}\right\} \Rightarrow a+3 \mid 60$. Answer: $a=357$.
$$
a+3=300,360,420
$$
## TICKET № 7 | 357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,180 |
1. Solve the equation $3 \cdot 2^{x+1}+5 \cdot 2^{x}-2^{x+2}=21$. | Solution: $3 \cdot 2^{x+1}+5 \cdot 2^{x}-2^{x+2}=21,(6+5-4) 2^{x}=21,2^{x}=\frac{21}{7}=3, x=\log _{2} 3$. | \log_{2}3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,181 |
2. Find the eleventh term of the arithmetic progression if the sum of the first seven terms $S_{7}=77$, and the first term $a_{1}=5$. | Solution: $S_{7}=77 ; a_{1}=5, a_{11}=$ ?
$S_{7}=\frac{2 a_{1}+d(7-1)}{2} \cdot 7=\frac{10+6 d}{2} \cdot 7=(5+3 d) 7=77 \Rightarrow d=2, a_{11}=a_{1}+10 d=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,182 |
3. Find all extrema of the function $y=\frac{x}{8}+\frac{2}{x}$ on the interval $(-5 ; 10)$. | Solution: $y=\frac{x}{8}+\frac{2}{x},(-5 ; 10)$, Domain of Definition $x \neq 0$,
$y^{\prime}=\frac{1}{8}-\frac{2}{x^{2}}=\frac{x^{2}-16}{8 x^{2}}=\frac{(x-4)(x+4)}{8 x^{2}}$.
Answer: $y_... | y_{\max}(-4)=-1,y_{\}(4)=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,183 |
4. Calculate $\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}$. | Solution: $\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}=A, A^{2}=12-2 \sqrt{36-11}=2, A= \pm \sqrt{2}, A=\sqrt{2}$. | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,184 |
5. Solve the inequality $\frac{2 x^{2}-6 x+5}{x^{2}-4 x+3}<1$.
# | # Solution:
$$
\frac{2 x^{2}-6 x+5}{x^{2}-4 x+3}<1, \frac{2 x^{2}-6 x+5-x^{2}+4 x-3}{(x-1)(x-3)}<0, \frac{2 x^{2}-2 x+2}{(x-1)(x-3)}<0
$$
Answer: $(1 ; 3)$. | (1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,185 |
6. Solve the equation $\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}$. | Solution:
$$
\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}
$$
$\frac{x+3}{3 x-5}>0$
$t=\frac{x+3}{3 x-5}, \sqrt{t}+1=\frac{2}{\sqrt{t}}, \frac{t+\sqrt{t}-2}{\sqrt{t}}=0, \sqrt{t}=\frac{-1 \pm \sqrt{1+8}}{2}=\frac{-1 \pm 3}{2}=\left[\begin{array}{l}-2 \\ 1\end{array}\right.$
$t=1, \frac{x+3}{3 x-5}=1,3 x-5=x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,186 |
7. The distance from the point of intersection of the diameter of a circle with a chord of length 18 cm to the center of the circle is $7 \mathrm{~cm}$. This point divides the chord in the ratio $2: 1$. Find the radius.
$$
A B=18, E O=7, A E=2 B E, R=?
$$ | Solution: $2 B E \cdot B E=(R-7)(7+R)$
$$
A E \cdot B E=D E \cdot E C, \quad A E+B E=18, \quad B E=6
$$
$$
2 \cdot 6 \cdot 6=\left(R^{2}-7^{2}\right), R^{2}=72+49=121=11^{2}
$$
Answer: $R... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,187 |
8. Solve the system $\left\{\begin{array}{l}\log _{4} x-\log _{2} y=0 \\ x^{2}-5 y^{2}+4=0 .\end{array}\right.$ | Solution: $\left\{\begin{array}{l}\frac{1}{2} \log _{2} x-\log _{2} y=0 ; \log _{2} \frac{\sqrt{x}}{y}=\log _{2} 1, \sqrt{x}=y, \\ x^{2}-5 y^{2}+4=0, \quad x^{2}-5 x+4=0,\end{array} \Rightarrow \begin{array}{l}x=1 ; 4, \\ y=4 ; 2 .\end{array}\right.$
Answer: $\{1 ; 1\},\{4 ; 2\}$. | {1;1},{4;2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,188 |
9. Find the maximum value of the function $f(x)=9 \sin x+12 \cos x$. | Solution: $f(x)=9 \sin x+12 \cos x=\sqrt{9^{2}+12^{2}} \sin \left(x-\operatorname{arctg} \frac{12}{9}\right)$. Answer: 15 . | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,189 |
10. There are between 200 and 300 pencils that can be placed in red boxes of 10 or in blue boxes of 12 pencils each. When all the pencils were packed in red boxes, 7 pencils were left, and when packed in blue boxes, 3 pencils were short. How many pencils were intended for packing | Solution: $\left\{\begin{array}{ll}a=10 x+7=10(x+1)-3, & a+3 \mid 10, \\ a=12 y-3, & a+3 \mid 12,\end{array}\right\} \Rightarrow a+3 \mid 60$
Answer: $a=237, a=297$.
## TICKET № 8 | 237,297 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,190 |
1. Solve the equation $7^{-x}-3 \cdot 7^{1+x}=4$. | Solution: $7^{-x}-3 \cdot 7^{1+x}=4, \frac{1}{7^{x}}-21 \cdot 7^{x}=4,-21\left(7^{x}\right)^{2}-4 \cdot 7^{x}+1=0$,
$\left(7^{x}\right)=\frac{4 \pm \sqrt{16+84}}{-42}=\frac{4 \pm 10}{-42}=\left[\begin{array}{l}-\frac{14}{42}=-\frac{1}{3} \text {, not valid } \\ \frac{1}{7}\end{array}, x=-1\right.$. Answer: $x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,191 |
3. Find the greatest and least values of the function $y=3 x^{4}-6 x^{2}+4$ on the interval $[-1 ; 3]$. | Solution: $\quad y=3 x^{4}-6 x^{2}+4 \quad$ on $\quad[-1 ; 3], \quad y^{\prime}=12 x^{3}-12 x=12 x\left(x^{2}-1\right), \quad x=0, \pm 1$, $y(-1)=1, y(3)=193, y(0)=4, y(1)=1$. Answer: $y(-1)=y(1)=1, y(3)=193$. | y(-1)=y(1)=1,y(3)=193 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,193 |
4. Calculate $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}$. | Solution: $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}=A, A^{2}=12-2 \sqrt{36-20}=4, A= \pm \sqrt{4}=2$. Answer: 2 . | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,194 |
5. Solve the inequality $\frac{x+2}{3 x+1} \leq \frac{x-2}{2 x-1}$. | Solution: $\frac{x+2}{3 x+1} \leq \frac{x-2}{2 x-1} ; \frac{x+2}{3 x+1}-\frac{x-2}{2 x-1} \leq 0, \frac{(x+2)(2 x-1)-(x-2)(3 x-1)}{(3 x+1)(2 x-1)} \leq 0$, $\frac{2 x^{2}+3 x-2-3 x^{2}+5 x+2}{(3 x+1)(2 x-1)} \leq 0, \frac{-8 x^{2}+8 x}{(3 x+1)(2 x-1)} \leq 0, \frac{-x(x-8)}{(3 x+1)(2 x-1)} \leq 0$
 | Solution: $\left\{\begin{array}{l}x+y=13, \\ \log _{4}(x y)=\log _{4} 40,\end{array}\left\{\begin{array}{l}x+y=13, \\ x y=40,\end{array}\left\{\begin{array}{l}x=5, y=8 \\ x=8, y=5\end{array}\right.\right.\right.$
Answer: $\{5,8\},\{8,5\}$ | {5,8},{8,5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,198 |
9. Solve the equation $9 \sin x - 12 \cos x = 15$. | Solution:
$$
9 \sin x-12 \cos x=15, \sqrt{9^{2}+12^{2}} \sin \left(x-\operatorname{arctg} \frac{4}{3}\right)=15
$$
$$
\sin \left(x-\operatorname{arctg} \frac{4}{3}\right)=1, x-\operatorname{arctg} \frac{4}{3}=\frac{\pi}{2}+2 \pi k, x=\operatorname{arctg} \frac{4}{3}+\frac{\pi}{2}+2 \pi k
$$ | \operatorname{arctg}\frac{4}{3}+\frac{\pi}{2}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,199 |
2. Find the sum of the first twelve terms of an arithmetic progression if its fifth term $a_{5}=1$, and the seventeenth term $a_{17}=18$. | Solution: $a_{5}=1 ; a_{17}=18$. Find $S_{12}$.
$$
\left\{\begin{array} { l }
{ a _ { 1 } + 4 d = 1 , } \\
{ a _ { 1 } + 1 6 d = 1 8 ; }
\end{array} \left\{\begin{array}{l}
12 d=17, \\
a_{1}=1-4 \frac{17}{12}=-\frac{14}{3} ;
\end{array} \quad S_{12}=\frac{2 \cdot\left(-\frac{14}{3}\right)+\frac{17}{12} \cdot 11}{2} \... | 37.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,201 |
3. Find all the extrema of the function $y=2^{x^{2}-2 x+3}$. | Solution: $y=2^{x^{2}-2 x+3}, y^{\prime}=2^{x^{2}-2 x+3} \ln 2(2 x-2)=0 \Rightarrow x=1$
Answer: $y_{\min }(1)=2^{2}=4$. | y_{\}(1)=4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,202 |
4. Calculate $\sqrt{6+\sqrt{32}}-\sqrt{6-\sqrt{32}}$. | Solution: $\sqrt{6+\sqrt{32}}-\sqrt{6-\sqrt{32}}=A, A^{2}=12-2 \sqrt{36-32}=8, A= \pm 2 \sqrt{2}, A=2 \sqrt{2}$ | 2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,203 |
5. Solve the inequality $\frac{1}{2-x}+\frac{5}{2+x}<1$.
# | # Solution:
$\frac{1}{2-x}+\frac{5}{2+x}<1, \frac{2+x+5(2-x)-1(2-x)(2+x)}{(2-x)(2+x)}<0$,
$\frac{2+x+10-5x+x^2-4}{(2-x)(2+x)}<0, \frac{x^2-4x+8}{(2-x)(2+x)}<0$
Answer: $(-\infty, -2) \cu... | (-\infty,-2)\cup(2,+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,204 |
6. Solve the equation $\sqrt{\frac{x+2}{x-3}}-6 \sqrt{\frac{x-3}{x+2}}+1=0$. | Solution: $\sqrt{\frac{x+2}{x-3}}-6 \sqrt{\frac{x-3}{x+2}}+1=0$, Domain of Definition $\frac{x+2}{x-3}>0, \frac{x+2}{x-3}=t>0$,
$\sqrt{t}-6 \frac{1}{\sqrt{t}}+1=0,(\sqrt{t})^{2}+\sqrt{t}-6=0,(\sqrt{t})_{1,2}=\frac{-1 \pm \sqrt{1+24}}{2}=\frac{-1 \pm 5}{2}=\left[\begin{array}{l}2 \\ -3-\text { not valid }\end{array}\ri... | \frac{14}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,205 |
7. Tangents are drawn from point $A$ to a circle with a radius of 10 cm, touching the circle at points $B$ and $C$ such that triangle $A B C$ is equilateral. Find its area. | Solution: $\triangle A B C$ is equilateral, $A C, A B$ are tangents, $R=10, A B=$ ?
$\angle A C B=60^{\circ}, \angle A O C=90^{\circ}(A C \perp O C), \angle O C B=30^{\circ}$,
$B C=R \cos 30^{\circ} \cdot 2=R \sqrt{3}=10 \sqrt{3}, A B=B C=A C=10 \sqrt{3}$,
$S_{\triangle A B C}=\frac{A B^{2} \sqrt{3}}{4}=\frac{100 \c... | 75\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,206 |
8. Solve the system $\left\{\begin{array}{l}3^{y} \cdot 81=9^{x^{2}} ; \\ \lg y=\lg x-\lg 0.5 .\end{array}\right.$ | Solution: $\left\{\begin{array}{l}3^{y} \cdot 81=9^{x^{2}}, \\ \lg y=\lg 2 x,\end{array} 3^{2 x} \cdot 9^{2}=9^{x^{2}}, x+2=x^{2}, x^{2}-x-2=0\right.$, $\left\{\begin{array}{l}x=-1-\text { n.s., } \\ x=2,\end{array} \Rightarrow x=2 \Rightarrow y=4\right.$. Verification: $\left\{\begin{array}{l}3^{4} \cdot 81=9^{4}, \\ ... | 2,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,207 |
9. Find the maximum value of the function $f(x)=8 \sin x+15 \cos x$. | Solution: $f(x)=8 \sin x+15 \cos x=\sqrt{8^{2}+15^{2}} \sin \left(x+\operatorname{arctg} \frac{15}{8}\right)=17 \sin \left(x+\operatorname{arctg} \frac{15}{8}\right)$
Answer: The maximum value of $f(x)=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,208 |
10. Between 200 and 300 children enrolled in the first grade of school. It was decided to form classes of 25 students, but it turned out that ten would not have a place. Then they formed classes of 30 students, but in one of the classes, there were 15 fewer students. How many children enrolled in the first grade? | Solution: $\left\{\begin{array}{ll}a=25 R+10=25(R+1)-15, & a+15: 25, \\ a=30 l-15, & a+15: 30,\end{array}\right\} \Rightarrow a+15 \vdots 150$. Answer: $a=285$.
$$
a+15=150, \underline{300,} 450
$$ | 285 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,209 |
1. Solve the equation $9 \cdot 3^{2 x-1}+3^{x}-30=0$. | Solution: $9 \cdot 3^{2 x-1}+3^{x}-30=0,3 \cdot 3^{2 x}+3^{x}-30=0$, $3^{x}=\frac{-1 \pm \sqrt{1+360}}{6}=\frac{-1 \pm 19}{6}=\left[\begin{array}{l}3 \\ -\frac{10}{3} \text {, not valid }\end{array}, x=1\right.$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,210 |
3. Find the greatest and least values of the function $y=e^{2 x^{2}-4 x-6}$ on the interval $[0 ; 3]$. | Solution: $y=e^{2 x^{2}-4 x-6}$ on $[0 ; 3] . y^{\prime}=e^{2 x^{2}-4 x-6}(4 x-4)=0, x=1$, $y(0)=e^{6}, y(1)=e^{-4}, y(3)=e^{12}$. | y(0)=e^{6},y(1)=e^{-4},y(3)=e^{12} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,212 |
4. Calculate $\sqrt{6+\sqrt{27}}-\sqrt{6-\sqrt{27}}$. | Solution: $(\sqrt{6+\sqrt{27}}-\sqrt{6-\sqrt{27}})^{2}=12-2 \sqrt{36-27}=6 \Rightarrow \pm \sqrt{6} \Rightarrow \sqrt{6}$ | \sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,213 |
5. Solve the inequality $\frac{3-x}{2 x+1}<-\frac{3}{4}$.
# | # Solution:
$\frac{3-x}{2 x+1}<-\frac{3}{4}, \frac{3-x}{2 x+1}+\frac{3}{4}<0, \frac{12-4 x+6 x+3}{4(2 x+1)}<0, \frac{2 x+15}{4(2 x+1)}<0$.
Answer: $\left(-\frac{15}{2} ;-\frac{1}{2}\right)$
 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,214 |
6. Solve the equation $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$. | Solution: $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$, Domain of Definition $\frac{x^{2}-16}{x-3} \geq 0, x>3 \Rightarrow x \geq 4$,
$\sqrt{x^{2}-16}+\sqrt{x^{2}-9}=7, x^{2}-16=t, \sqrt{t}+\sqrt{t+7}=7, t+2 \sqrt{t(t+7)}+t+7=49$,
$2 \sqrt{t(t+7)}=42-2 t, \sqrt{t^{2}+7 t}=21-t, t \leq 21, t^{2}+7 t=21... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,215 |
7. Two vertices of a square lie on a circle with a radius of $5 \mathrm{~cm}$, while the other two lie on a tangent to this circle. Find the area of the square. | Solution: $R=5, A B C D$-square, $S_{A B C D}=$ ?
$A B=x, E F=2 R-x, E O=R-E F=x-R$
$\Delta E O C: E C^{2}=O C^{2}-E O^{2}=R^{2}-(x-R)^{2} \Rightarrow\left(\frac{x}{2}\right)^{2}=R^{2}-x^{2}+2 R x-R^{2}$,
=17
$$
$$
\sin \left(x-\operatorname{arctg} \frac{8}{15}\right)=1 \Rightarrow x-\operatorname{arctg} \frac{8}{15}=\frac{\pi}{2}+2 \pi k
$$
Answer: $x=\operatorname{arctg} \frac{8}{15}+\frac{\pi}{2}+2 \pi k$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,217 | |
10. To pack books when moving a school library, you can buy small boxes that hold 12 books or large ones that hold 25 books. If all the books are packed in small boxes, 7 books will be left, and if all the books are packed in large boxes, there will be room for 5 more books. The library's collection contains between 50... | Solution: $\left\{\begin{array}{ll}a=12 R+7=12(R+1)-5, & a+5 \vdots: 12, \\ a=25 l-5, & a+5 \vdots 25,\end{array}\right\} a+5: 300$. Answer: $a=595$.
$$
a+5=300, \underline{600,} 900
$$ | 595 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,218 |
5. Consider the crank-slider mechanism. (1-crank, 2- connecting rod). They were first applied in antiquity, on Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank rotates at a constant angular velocity \(\omega=10 \m... | Solution. Let's write the equations of motion of point M in coordinate form. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,220 | |
5. Consider the crank-slider mechanism (1-crank, 2- connecting rod). They were first used in antiquity, in Roman sawmills. There, the rotation of the wheel, driven by the force of falling water, was converted into reciprocating motion of the saw blade. The crank rotates at a constant angular velocity $\omega=10$ $rad/s... | Solution. Let's write the equations of motion of point M in coordinate form. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 19,221 |
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,236 |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle \(\alpha=45^{\circ}\). (15 points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,237 |
8. Solve the system $\left\{\begin{array}{l}x \log _{2} 3+y=\log _{2} 18 ; \\ 5^{x}=25^{y} .\end{array}\right.$
 | # Solution:
$\left\{\begin{array}{l}\log _{2} 3^{x}+y=\log _{2} 18, \\ 5^{x}=5^{2 y},\end{array}\left\{\begin{array}{l}2 y \log _{2} 3+y=18 \\ x=2 y,\end{array},\left\{\begin{array}{l}y\left(\log _{2} 3^{2}+1\right)=18, \\ x=2 y,\end{array}\left\{\begin{array}{l}y\left(\log _{2}\left(3^{2} \cdot 2\right)\right)=18, \\... | {2;1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,238 |
8. Solve the system $\left\{\begin{array}{l}x+y=13 \\ \log _{4} x+\log _{4} y=1+\log _{4} 10 .\end{array}\right.$
 | Solution: $\left\{\begin{array}{l}x+y=13, \\ \log _{4}(x y)=\log _{4} 40,\end{array}\left\{\begin{array}{l}x+y=13, \\ x y=40,\end{array}\left\{\begin{array}{l}x=5, y=8 \\ x=8, y=5\end{array}\right.\right.\right.$
Answer: $\{5,8\},\{8,5\}$ | {5,8},{8,5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,239 |
10. To pack books when moving a school library, you can buy small boxes that hold 12 books or large ones that hold 25 books. If all the books are packed in small boxes, 7 books will remain, and if all the books are packed in large boxes, there will be room for 5 more books. The library's collection contains between 500... | Solution: $\left\{\begin{array}{ll}a=12 R+7=12(R+1)-5, & a+5 \vdots: 12, \\ a=25 l-5, & a+5 \vdots 25,\end{array}\right\} a+5: 300$. Answer: $a=595$.
$$
a+5=300, \underline{600,} 900
$$ | 595 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,240 |
1. Solve the equation $\left(\sin ^{2} x+\frac{1}{\sin ^{2} x}\right)^{3}+\left(\cos ^{2} y+\frac{1}{\cos ^{2} y}\right)^{3}=16 \cos z$. ( 5 points $)$ | Solution. The domain of the inequality is limited by the conditions $\sin x \neq 0, \cos y \neq 0$. Note that $u+\frac{1}{u} \geq 2$ for any positive $u$, as this inequality is equivalent to
$$
\frac{u^{2}+1}{u} \geq 2 \Leftrightarrow u^{2}+1 \geq 2 u \Leftrightarrow u^{2}+1-2 u \geq 0 \Leftrightarrow(u-1)^{2} \geq 0
... | \frac{\pi}{2}+\pi,\pik,2\pi, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,246 |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\}$. (10 points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equat... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,249 |
5. Find the area of the shaded figure described by a semicircle rotated about one of its ends by an angle $\alpha=30^{\circ}$. (15 points)
 | Solution. Let the area of the semicircle $S_{0}=\frac{\pi R^{2}}{2}$, the area of the lune $A B_{1} C=x$, the lune $C B_{1} B=y$ and the sector $A C B=a$. Then, if the area of the segment $A C$ is $b$, we have: $x+b=S_{0}=b+a \Rightarrow x=a$.
The entire shaded area is equal to $x+y=a+y$, but $a+y$ is the area of the ... | \frac{\piR^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,252 |
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