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
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
39. Prove that for any values of $\alpha$ the inequality $4 \sin 3 \alpha+5 \geqslant 4 \cos 2 \alpha+5 \sin \alpha$ holds. | 39. Let's reduce all functions to the argument $\alpha$.
$4\left(3 \sin \alpha-4 \sin ^{3} \alpha\right)+5 \geqslant 4\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)+5 \sin \alpha,(1-\sin \alpha)(4 \sin \alpha+1)^{2} \geqslant 0$. The first factor is non-negative, and the second is positive for any $\alpha$. We obtain ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,841 |
40. Prove that $\cos \frac{2}{5} \pi+\cos \frac{4}{5} \pi=-\frac{1}{2}$. | 40. $\cos \frac{2}{5} \pi+\cos \frac{4}{5} \pi=\sin \frac{\pi}{10}-\sin \frac{3 \pi}{10}=-2 \sin \frac{\pi}{10} \cos \frac{\pi}{5}=$
$$
=-\frac{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}{\cos \frac{\pi}{10}} \cos \frac{\pi}{5}=-\frac{\sin \frac{\pi}{5} \cos \frac{\pi}{5}}{\cos \frac{\pi}{10}}=-\frac{\sin \frac{2}{5} \... | -\frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 34,842 |
41. Prove that
$\operatorname{tg} \alpha \operatorname{tg} 2 \alpha+\operatorname{tg} 2 \alpha \operatorname{tg} 3 \alpha+\ldots+\operatorname{tg}(n-1) \alpha \operatorname{tg} n \alpha=\frac{\operatorname{tg} n \alpha}{\operatorname{tg} \alpha}-n$. | 41. $\operatorname{tg} \alpha=\operatorname{tg}(2 \alpha-\alpha)=\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \alpha}{1+\operatorname{tg} \alpha \operatorname{tg} 2 \alpha}$. Hence, $\operatorname{tg} \alpha \operatorname{tg} 2 \alpha=\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \alpha}{\operatorname{tg} \a... | \frac{\operatorname{tg}n\alpha}{\operatorname{tg}\alpha}-n | Algebra | proof | Yes | Yes | olympiads | false | 34,843 |
4. $\log _{b} N=\frac{\log _{a} N}{\log _{a} b}(N>0, a>0, b>0, a \neq 1, b \neq 1)$.
This property is usually called the change of base rule.
We recommend proving these properties.
There are quite a number of problems where the variable $x$ or some function of it is under the logarithm sign. Before solving such prob... | 4. Instruction. Use the equality $\log _{b} a=\frac{1}{\log _{a} b}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 34,847 | |
8. Prove that if $\log _{k} x, \log _{m} x, \log _{n} x$, where $x>0, x \neq 1$, form an arithmetic progression, then $n^{2}=(n k)^{\log _{k} m}$.
In № $9-14$ calculate: | 8. According to the condition, we have $\div \frac{\log _{m} x}{\log _{m} k} ; \log _{m} x ; \frac{\log _{m} x}{\log _{m} n}$.
By the property of the terms of a progression, $2 \log _{m} x=\frac{\log _{m} x}{\log _{m} k}+\frac{\log _{m} x}{\log _{m} n}$, or $2 \log _{m} x \log _{m} k \log _{m} n=\left(\log _{m} n+\log... | n^{2}=(nk)^{\log_{k}} | Algebra | proof | Yes | Yes | olympiads | false | 34,850 |
9. $\log _{3} 6$, if $\log _{6} 2=a$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to restate it clearly:
9. $\log _{3} 6$, if... | 9. Answer: $\log _{3} 6=\frac{1}{1-a}$. | \log_{3}6=\frac{1}{1-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,851 |
10. $\log _{54} 168$, if $\log _{7} 12=a, \log _{12} 24=b$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. If you intended to have the note or instructio... | 10. $\log _{54} 168=\log _{54} 7+\log _{54} 24=\frac{1}{\log _{12} 54}\left(\log _{12} 7+b\right)=\frac{1}{\log _{12} 54}\left(\frac{1}{a}+b\right)$.
$$
\frac{1}{\log _{12} 54}=\log _{54} 3+\log _{54} 4=\frac{\log _{2} 3}{3 \log _{2} 3+1}+\frac{2}{3 \log _{2} 3+1}=\frac{2+\log _{2} 3}{3 \log _{2} 3+1}
$$
$\log _{12} ... | \frac{1+}{(8-5b)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,852 |
13. $\lg \sqrt[3]{25}$, if $\lg 64=a$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if you meant to translate the problem statement, here it i... | 13. $\lg 64=6 \lg 2 ; \lg 2=\frac{a}{6} ; \lg 5=\lg \frac{10}{2}=1-\lg 2 ; \lg \sqrt[3]{25}=\frac{6-a}{9}$. | \frac{6-}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,855 |
14. $\log _{\frac{\sqrt{3}}{3}}\left(\log _{8} \frac{\sqrt{2}}{2}-\log _{3} \frac{\sqrt{3}}{3}\right)$.
In № $15-27$ solve the equations: | 14. We should switch to the logarithm with base $\frac{\sqrt{3}}{3}$.
Then replace $\frac{1}{3}$ with $\left(\frac{\sqrt{3}}{3}\right)^{2}$. Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,856 |
16. $\log _{4}\left\{2 \log _{3}\left[1+\log _{2}\left(1+3 \log _{2} x\right)\right]\right\}=\frac{1}{2}$. | 16. Gradually potentiating, we arrive at the equation $1+3 \log _{2} x=4$. From which $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,858 |
17. $\log _{x} \sqrt{5}+\log _{x}(5 x)-2.25=\left(\log _{x} \sqrt{5}\right)^{2}$. | 17. Simple transformations lead to the equation
$\log _{x}^{2} 5-6 \log _{x} 5+5=0$. From which $x_{1}=\sqrt[5]{5}, x_{2}=5$. | x_{1}=\sqrt[5]{5},x_{2}=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,859 |
18. $3 \log _{a^{2} x} x+\frac{1}{2} \log _{\sqrt{a}} x=2$. | 18. Transitioning to logarithms with base $x$, we obtain the equation
$$
4 \log _{x}^{2} a-7 \log _{x} a+3=0 . \quad \text { Hence } x_{1}=a, x_{2}=a^{\frac{4}{3}}
$$ | x_{1}=,x_{2}=^{\frac{4}{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,860 |
19. $\log _{5} 120+(x-3)-2 \log _{5}\left(1-5^{x-3}\right)=-\log _{5}\left(0.2-5^{x-4}\right)$. | 19. $1-5^{x-3} \neq 0, x \neq 3,0,2-5^{x-4}=\frac{1}{5}\left(1-5^{x-3}\right)$.
Raising the entire expression, we get
$$
\frac{120 \cdot 5^{x-3}}{\left(1-5^{x-3}\right)^{2}}=\frac{5}{1-5^{x-3}} ; \quad 5^{x-3}=5^{-2} . \quad \text { Hence } x=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,861 |
20. $\frac{\log _{2} x}{\log _{2}^{2} a}-\frac{2 \log _{a} x}{\log _{\frac{1}{b}} a}=\log _{\sqrt[3]{a}} x \cdot \log _{a} x$. | 20. Transitioning to logarithms with base 2, we get
$$
3\left(\log _{2}:\right)^{2}-\log _{2} x\left(1+2 \log _{2} h\right)=0, \quad x>0 . \text { From which } x_{1}=1, \quad x_{2}=\sqrt[3]{2 b^{2}} .
$$ | x_1=1,\quadx_2=\sqrt[3]{2b^2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,862 |
21. $\sqrt{x^{\lg \sqrt{x}}}=10$. | 21. Taking the logarithm of both sides, we get $\frac{1}{4} \lg x \lg x=1$. From which $x_{1}=100 ; \quad x_{2}=0.01$. | x_{1}=100;\quadx_{2}=0.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,863 |
22. $x^{2+\lg x}=100 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
22. $x^{2+\lg x}=100 x$. | 22. After logarithmizing the left and right parts, we get
$$
(2+\lg x) \lg x=2+\lg x ; \quad x_{1}=0.01 ; \quad x_{2}=10
$$ | x_{1}=0.01;\quadx_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,864 |
25. $2 \log _{x} a+\log _{a x} a+3 \log _{a^{2} x} a=0$. | 25. Transitioning to logarithms with base $a(a \neq 1)$, we get
$$
\frac{2}{\log _{a} x}+\frac{1}{1+\log _{a} x}+\frac{3}{2+\log _{a} x}=0 . \text { From which } x_{1}=a^{-\frac{1}{2}} ; \quad x_{2}=a^{-\frac{4}{3}}
$$ | x_{1}=^{-\frac{1}{2}};\quadx_{2}=^{-\frac{4}{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,867 |
26. $5^{\lg x}-3^{\lg x-1}=3^{\lg x+1}-5^{\lg x-1}$. | 26. Grouping the powers with bases 5 and 3, we get
$$
5^{\lg x}\left(1+\frac{1}{5}\right)=3^{\lg x}\left(3+\frac{1}{3}\right) . \quad \text { Hence }\left(\frac{5}{3}\right)^{\lg \cdot x}=\left(\frac{5}{3}\right)^{2} ; \quad x=100
$$ | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,868 |
27. $\log _{\frac{1}{4}}\left[\log _{4}(\cos 2 x+\sin 2 x)\right]=1$.
In № $28-36$ solve the system of equations: | 27. Gradually potentiating, we get $\cos 2 x+\sin 2 x=\sqrt{2}$.
From where $\cos \left(\frac{\pi}{4}-2 x\right)=1 ; \quad x=\frac{\pi}{8}(1-8 k)$.
84 | \frac{\pi}{8}(1-8k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,869 |
28. $\left\{\begin{aligned} \log _{y} x-\log _{x} y & =\frac{8}{3} \\ x y & =16 .\end{aligned}\right.$
2 Order 5039 | 28. We should switch to logarithms with base $x(x>0, x \neq 1)$. We get
$$
\log _{x}^{2} y+\frac{8}{3} \log _{x} y-1=0 ; \quad \log _{x} y=\frac{1}{3} ; \log _{x} y=-3
$$
Considering these values together with the second equation, we find
$$
x_{1}=8 ; \quad y_{1}=2 ; \quad x_{2}=\frac{1}{4} ; \quad y_{2}=64
$$ | x_{1}=8,y_{1}=2;x_{2}=\frac{1}{4},y_{2}=64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,870 |
30. $\left\{\begin{array}{l}\log _{a} x+\log _{a} y=2 \\ \log _{b} x-\log _{b} y=4\end{array}\right.$ | 30. Potentiating, we get $x y=a^{2}, \frac{x}{y}=b^{4}, x= \pm a b^{2}, y= \pm \frac{a}{b^{2}}$. By the condition $x>0, y>0, a>0, b>0$. Answer: $x=a b^{2}, y=\frac{a}{b^{2}}$. | ^{2},\frac{}{b^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,872 |
32. $\left\{\begin{array}{l}\log _{y} \log _{y} x=\log _{x} \log _{x} y \\ \log _{a}^{2} x+\log _{a}^{2} y=8 .\end{array}\right.$ | 32. We will transform the left and right parts of the first equation to logarithms with base $a$. We get $\left(\log _{a} \log _{a} x-\log _{a} \log _{a} y\right)\left(\frac{1}{\log _{a} y}+\frac{1}{\log _{a} x}\right)=0 . \quad$ This system is equivalent to the following two:
a) $\left\{\begin{array}{l}\log _{a} \log ... | x_{1}=y_{1}=^{2},\quadx_{2}=y_{2}=^{-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,874 |
34. $\left\{\begin{array}{l}\log _{\sqrt{10}}\left(x^{2}+y^{2}\right)=2 \lg 2 a+2 \log _{100}\left(x^{2}-y^{2}\right) \\ x y=a^{2} .\end{array}\right.$ | 34. Let's move on to logarithms with base 10.
$\left(x^{2}+y^{2}\right)^{2}=4 a^{2}\left(x^{2}-y^{2}\right)$, but $y=\frac{a^{2}}{x}$, then $x^{8}+2 a^{4} x^{4}-4 a^{2} x^{6}+4 a^{\natural} x^{2}+a^{8}=0$,
$\left(x^{4}-2 a^{2} x^{2}-a^{4}\right)^{2}=0$. Answer: $x_{1,2}= \pm a \sqrt{\sqrt{2}+1}, y_{1,2}= \pm a \sqrt{... | x_{1,2}=\\sqrt{\sqrt{2}+1},y_{1,2}=\\sqrt{\sqrt{2}-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,876 |
36.
$$
\left\{\begin{array}{l}
x y=a^{2} \\
\lg ^{2} x+\lg ^{2} y=\frac{5}{2} \lg ^{2} a^{2}
\end{array}\right.
$$ | 36. We take the logarithm and square both sides of the first equation
$$
\begin{aligned}
& \lg ^{2} x+2 \lg x \lg y+\lg ^{2} y=\lg ^{2} a^{2} \\
& -\lg ^{2} x+\lg ^{2} y=\frac{5}{2} \lg ^{2} a^{2} \\
& 2 \lg x \lg y=-\frac{3}{2} \lg ^{2} a^{2}
\end{aligned}
$$
Consider the system $\left\{\begin{array}{l}\lg x \lg y=-... | x_{1}=^{3},y_{1}=\frac{1}{},x_{2}=\frac{1}{},y_{2}=^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,878 |
4. Find the sum $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}$. | 4. Use the formula from problem 3.
$$
1^{3}+2^{3}+3^{3}+\ldots+(2 n)^{3}=\left[\frac{2 n(2 n+1)}{2}\right]^{2}
$$
Isolate the sum of cubes of odd numbers
$$
\begin{gathered}
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+2^{3}+4^{3}+\ldots+(2 n)^{3}=\left[\frac{2 n(2 n+1)}{2}\right]^{2} \\
S+2^{3}\left(1^{3}+2^{3}+\ldots+n^{3... | n^{2}(2n^{2}-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,881 |
14. Calculate the product $p=1 \cdot 2^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdot 8^{\frac{1}{8}} \cdot 16^{\frac{1}{16}} \cdot \ldots$
In № $15-20$ find the sums: | 14. Let's write all powers as powers with base 2
$$
p=1 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{2}{4}} \cdot 2^{\frac{3}{8}} \cdot 2^{\frac{4}{16}} \ldots=2^{\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots}
$$
The problem has been reduced to finding the sum $S=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\ld... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,884 |
18. $\left(\frac{1}{a}-\frac{n}{x}\right)+\left(\frac{1}{a}-\frac{n-1}{x}\right)+\left(\frac{1}{a}-\frac{n-2}{x}\right)+\ldots+\left(\frac{1}{a}-\frac{1}{x}\right)$. | 18. Expanding the brackets, we get
$$
\begin{aligned}
& \frac{1}{a}-\frac{n}{x}+\frac{1}{a}-\frac{n-1}{x}+\frac{1}{a}-\frac{n-2}{x}+\ldots+\frac{1}{a}-\frac{1}{x}= \\
= & \frac{n}{a}-\frac{n+(n-1)+(n-2)+\ldots+1}{x}=\frac{n}{a}+\frac{n+1}{2 x} n=n\left(\frac{1}{a}-\frac{n+1}{2 x}\right)
\end{aligned}
$$ | n(\frac{1}{}-\frac{n+1}{2x}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,885 |
20. $\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}$. | 20. $S=\frac{n}{2(3 n+2)}$. | \frac{n}{2(3n+2)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,887 |
21. Find the sum $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{n}{(n+1)!}$ and compute its limit as $n \rightarrow \infty$. | 21. By forming the sequence of partial sums, we get
$$
S_{1}=\frac{2!-1}{2!}, \quad S_{2}=\frac{3!-1}{3!}, \ldots \quad \text { Hence } S_{n}=\frac{(n+1)!-1}{(n+1)!}, \quad S=\lim _{n \rightarrow \infty} S_{n}=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,888 |
22. Find the sum of $n$ numbers of the form $5,55,555,5555, \ldots$ | 22. Let's factor out the multiplier 5
$$
5(1+11+111+\ldots)=5\left(\frac{10-1}{9}+\frac{10^{2}-1}{9}+\ldots+\frac{10^{n}-1}{9}\right)=\frac{5}{9}\left(\frac{10^{n}-1}{9} 10-n\right)
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,889 | |
23. Calculate the sum $1 \cdot x+2 x^{2}+3 x^{3}+\ldots+n x^{n}$. | 23. Let $S=1 \cdot x+2 x^{2}+3 x^{3}+\ldots+n x^{n}$. (1)
Then $S x=1 \cdot x^{2}+2 x^{3}+3 x^{4}+\ldots+(n-1) x^{n}+n x^{n+1}$.
Subtract the second equation from the first, we get
$$
S(1-x)=x+x^{2}+x^{3}+\ldots+x^{n}-n x^{n+1} . \quad S=\frac{n x^{n+1}}{x-1}-\frac{x\left(x^{n}-1\right)}{(x-1)^{2}}
$$ | \frac{nx^{n+1}}{x-1}-\frac{x(x^{n}-1)}{(x-1)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,890 |
24. Calculate the sum $\left(x+\frac{1}{x}\right)^{2}+\left(x^{2}+\frac{1}{x^{2}}\right)^{2}+\ldots+\left(x^{n}+\frac{1}{x^{n}}\right)^{2}$. | 24. Expanding the brackets and then grouping the terms, we get
\[
\begin{gathered}
x^{2}+2+\frac{1}{x^{2}}+x^{4}+2+\frac{1}{x^{4}}+\ldots+x^{2 n}+2+\frac{1}{x^{2 n}}= \\
=2 n+\left(x^{2}+x^{4}+\ldots+x^{2 n}\right)+\left(\frac{1}{x^{2}}+\frac{1}{x^{4}}+\ldots+\frac{1}{x^{2 n}}\right)= \\
=2 n+\frac{x^{2}\left(1-x^{2 n... | 2n+\frac{1-x^{2n}}{1-x^{2}}\cdot\frac{x^{2n+2}+1}{x^{2n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,891 |
25. Calculate the sum of $n$ terms of the series $1+\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\ldots$ | 25. Each term of the sum can be represented as $\frac{2^{n-1}+1}{2^{n-1}}$.
Then $S=1+\frac{2+1}{2}+\frac{2^{2}+1}{2^{2}}+\ldots+\frac{2^{n-1}+1}{2^{n-1}}=n+1-\frac{1}{2^{n-1}}$. | n+1-\frac{1}{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,892 |
26. Find the sum $1+3+6+10+15+\ldots+\frac{n(n+1)}{2}$. | 26. Let's rewrite the given sum: $S=\frac{1 \cdot 2}{2}+\frac{2 \cdot 3}{2}+\ldots+\frac{n(n+1)}{2}$.
Then $S=\frac{1}{2}[(1+1)+2(2+1)+3(3+1)+\ldots+n(n+1)]=\frac{n(n+1)(n+2)}{6}$. | \frac{n(n+1)(n+2)}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,893 |
27. Find the sum $1-2+3-4+\ldots+(-1)^{n-1} n$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The note above is not part of the translation and should not be included in the final output. Here is the correct translation:
27. Find the sum $1-2+3-4+\ldots+(-1)^{n-1} n$. | 27. Let $n$ be an odd number, then
$1+3+5+\ldots+n-[2+4+\ldots+(n-1)]=\left(\frac{1+n}{2}\right)^{2}-\frac{2+(n-1)}{2} \cdot \frac{n-1}{2}=\frac{n+1}{2}$.
Let $n$ be an even number, then
$1+3+5+\ldots+(n-1)-(2+4+\ldots+n)=\left(\frac{n}{2}\right)^{2}+\frac{2+n}{2} \cdot \frac{n}{2}=-\frac{n}{2}$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,894 | |
28. Prove the identity
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2 n-1}-\frac{1}{2 n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}
$$ | 28. Group the members of the right-hand side in pairs and transform each pair, we have $n$ pairs in total.
$$
\begin{gathered}
1-\frac{1}{2}=\frac{1}{2} \\
\frac{1}{3}-\frac{1}{4}=\frac{1}{3}+\frac{1}{4}-\frac{1}{2} \\
\cdots \cdot \cdots \cdot \cdots \cdot \cdot \\
\frac{1}{2 n-1}-\frac{1}{2 n}=\frac{1}{2 n-1}-\frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,895 |
6. $\left|-\frac{3}{x-1}\right|>\left|-\frac{x-2}{2}\right|$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
6. $\left|-\frac{3}{x-1}\right|>\left|-\frac{x-2}{2}\right|$. | 6. This inequality can be rewritten as: $\frac{3}{|1-x|}>\frac{|2-x|}{2}$.
From this, $|1-x| \cdot|2-x|0 \\ x^{2}-3 x-40$, since $D<0$, the inequality holds for any $x$;
b) $x^{2}-3 x-4<0,-1<x<4$. Answer: $-1<x<4$. | -1<x<4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,901 |
7. $\left|\frac{2}{x+2}\right|<\left|\frac{4}{x^{2}+x-2}\right|$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7. $\left|\frac{2}{x+2}\right|<\left|\frac{4}{x^{2}+x-2}\right|$. | 7. The inequality can be written as: $\frac{2}{|x+2|}<\frac{4}{\left|x^{2}+x-2\right|}$,
$\frac{2}{|x+2|}<\frac{4}{|x+2||x-1|} \quad$ or $1<\frac{2}{|x-1|} ;|x-1|<2$
$-2<x-1<2,-1<x<3, x \neq 1$. | -1<x<3,x\neq1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,902 |
9. $|x-2|-|x|>0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
9. $|x-2|-|x|>0$. | 9. Roots of submodular expressions 2 and 0.
If $x \leqslant 0$, then $-x+2+x>0$, $x$ is any from the interval $x \leqslant 0$.
If $0 \leqslant x \leqslant 2$, then $-x+2-x>0$, $x0$, no solution.
Answer: $x<1$ | x<1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,904 |
10. $\left|\frac{-3}{x-2}\right|<\left|\frac{6}{x^{2}-3 x+2}\right|$.
Construct the graphs of the functions given in problems 11-19: | 10. The inequality can be written as follows:
$$
\begin{gathered}
\frac{3}{|x-2|}<\frac{6}{|x-2| \cdot|x-1|} \text { or } 1<\frac{2}{|x-1|} ; \quad|x-1|<2 \\
-2<x-1<2, \quad-1<x<3, \quad x \neq 1, \quad x \neq 2
\end{gathered}
$$ | -1<x<3,\quadx\neq1,\quadx\neq2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,905 |
11. $y=|x-1|-|x-2|+|x-3|-|x-4|$. | 11. (Fig. 11) Roots of the submodular expressions $1,2,3,4$.
For $x \leqslant 1, y=-2$.
For $1 \leqslant x \leqslant 2, \quad y=2 x-4$.
For $2 \leqslant x \leqslant 3, \quad y=0$.
For $3 \leqslant x \leqslant 4, \quad y=2 x-6$.
For $x \geqslant 4, \quad y=2$.
 The roots of the submodular expressions are $4,1,4$.
If $x \leqslant-4, \quad$ then $\quad y=\frac{1}{4}(3 x+12)$.
If $-4 \leqslant x \leqslant 1$, then $y=\frac{1}{4}(-3 x-12)$.
If $1 \leqslant x \leqslant 4, \quad$ then $\quad y=\frac{1}{4}\left(2 x^{2}+3 x-20\right)$.
If $x \geqslant 4, \quad$ then... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,907 |
14. $y=\frac{1}{2} x^{2}-\frac{3}{2}|x|+1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
14. $y=\frac{1}{2} x^{2}-\frac{3}{2}|x|+1$. | 14. (Fig. 14) The root of the submodular expression 0.
If $x \geqslant 0, \quad y=\frac{1}{2} x^{2}-\frac{3}{2} x+1$.
If $x < 0, \quad y=\frac{1}{2} x^{2}+\frac{3}{2} x+1$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,909 | |
15. $y=\frac{|x|-2}{|x|}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
15. $y=\frac{|x|-2}{|x|}$. | 15. (Fig. 15) The root of the submodular expression is 0.
For $x>0, \quad y=1-\frac{2}{x}$.
For $x<0, \quad y=1+\frac{2}{x}$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,910 |
16. $y=1+\log _{3}|x|$. | 16. (Fig. 16). The root of the submodular expression is 0.
For $x>0, \quad y=1+\log _{3} x$.
For $x<0, y=1+\log _{3}(-x)$.
Fig. 15
90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,911 | |
17. $y=2^{|x|}-2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
17. $y=2^{|x|}-2$. | 17. (Fig. 17) For $x \geqslant 0, y=2^{x}-2$. For $x \leqslant 0, \quad y=2^{-x}-2$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,912 |
20. Among the complex numbers $z$, find the numbers having the smallest and largest arguments, if $|z-5-5i|=5$. | 20. (Fig. 20) The desired number has the form \( z = x + yi \). Then \( |x \div yi - 5 - 5i| = 5 \) or \( \sqrt{(x-5)^2 + (y-5)^2} = 5 \). The smallest value of \( \varphi = 0 \). The largest \( \varphi = 90^\circ \). Answer: \( z_1 = 5, \quad z_2 = 5i \).
 Let the desired number be $\boldsymbol{z}=x+y i$.
Then $|x+y i-10 i|=5 \sqrt{2}$ or $x^{2}+(y-10)^{2}=50$ The locus of points (GMP), corresponding to the numbers $z$ and $10 i$, is the circle $x^{2}+(y-10)^{2}=50$. The smallest value $\bigodot_{1}=45^{\circ}, \quad z_{1}=55 i$. The largest value $\varphi... | z_{1}=5+5i,z_{2}=-5+15i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,916 |
2. Prove the inequality $a b(a+b)+b c(b+c)+a c(a+c) \geqslant 6 a b c$, where $a>0, b>0, c>0$. | 2. We will transform the given inequality to the form $b(a-c)^{2}+a(b-c)^{2}+c(b-a)^{2} \geqslant 0$. This inequality is clearly visible. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,919 |
3. If $a, b, c, d$ are positive numbers and $a b c d=1$, then $a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d \geqslant 10$. Prove. | 3. For any positive numbers $x$ and $y$, the inequality $x+y \geqslant 2 \sqrt{x y}$ or $\frac{x+y}{\sqrt{x y}} \geqslant 2$ holds. Using these inequalities and considering that $a b c d=1$, we get
$$
\begin{gathered}
a b+c d \geqslant 2 \\
+a c+b d \geqslant 2 \\
a d+b c \geqslant 2 \\
\left(a^{2}+b^{2}\right)+\left(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,920 |
4. If $a>0, b>0, c>0$, then $8 a b c \leqslant(a+b)(b+c)(c+a)$. Prove.
| 4. $a+b \geqslant 2 \sqrt{a b}$
$b+c \geqslant 2 \sqrt{b c}$
$$
\begin{aligned}
& c+a \geqslant 2 \sqrt{a c} \\
& (a+b)(b+c)(c+a) \geqslant 8 a b c
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,921 |
5. Prove that for any positive real numbers $a, b, c, d$, the inequality $a^{4}+b^{4}+c^{4}+d^{4} \geqslant 4 a b c d$ holds.
6. If $a+b=1$, then $a^{4}+b^{4} \geqslant \frac{1}{8}$. Prove this. | 5. Let's write two inequalities $a^{4}+b^{4} \geqslant 2 \sqrt{a^{4} b^{4}}=2 a^{2} b^{2}, c^{4}+d^{4} \geqslant 2 c^{2} d^{2}$.
Adding them term by term, we get $a^{4}+b^{4}+c^{4}+d^{4} \geqslant 2\left(a^{2} b^{2}+c^{2} d^{2}\right) \geqslant 2 \cdot 2 \sqrt{a^{2} b^{2} c^{2} d^{2}}$ or $a^{4}+b^{4}+c^{4}+d^{4} \geq... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,922 |
7. Prove that $(a+b)^{4} \leqslant 8\left(a^{4}+b^{4}\right)$. | 7. Let's determine the sign of the difference $(a+b)^{4}-8\left(a^{4}+b^{4}\right)=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+$ $+b^{4}-8 a^{4}-8 b^{4}$. By grouping, we get
$-(b-a)^{2}\left[5(a+b)^{2}+2\left(a^{2}+b^{2}\right)\right] \leqslant 0$. Therefore, $(a+b)^{4} \leqslant 8\left(a^{4}+b^{4}\right)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,923 |
8. Prove that the inequality $x^{2}+y^{2}+z^{2}-x y-x z-y z \geqslant 0$ holds for any real $x, y, z$. | 8. Multiply both sides of the inequality by 2 and perform grouping
$$
x^{2}-2 x y+y^{2}+x^{2}-2 x z+z^{2}+y^{2}-2 y z+z^{2} \geqslant 0
$$
or $(x-y)^{2}+(x-z)^{2}+(y-z)^{2} \geqslant 0$, we obtain an obvious inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,924 |
9. Prove that the inequality $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{9}{a+b+c}$ holds for any positive $a, b, c$. | 9. In the left part of the inequality, we bring the fractions to a common denominator, obtaining $\frac{b c+a c+a b}{a b c} \geqslant \frac{9}{a+b+c}$. Multiplying both sides of the inequality by $a b c(a+b+c)$ and performing grouping, we obtain the obvious inequality: $c(a-b)^{2}+b(a-c)^{2}+a(b-c)^{2} \geqslant 0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,925 |
10. For positive numbers $a, b, c$, the inequality $a+b+c \geqslant \sqrt[m+n+k]{a^{m} b^{n} c^{k}}+\sqrt[m+n+k]{a^{n} b^{k} c^{m}}+\sqrt[m+n+k]{a^{k} b^{m} c^{n}}$ holds. Prove it. | 10. Generalizing the statement that the arithmetic mean of $m+n+k$ numbers is greater than or equal to their geometric mean, we can write
$$
\begin{gathered}
\sqrt{a b} \leqslant \frac{a+b}{2} ; \quad \sqrt[3]{a b c} \leqslant \frac{a+b+c}{3} ; \quad \sqrt[n]{\underbrace{a b c d \ldots l}_{n} \leqslant \frac{a+b+c+\ld... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,926 |
11. If $a>0, b>0, c>0, d>0$,
$$
\text { then } \quad \sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+\sqrt{c d} . \quad \text { Prove. }
$$
26 | 11. $(a+c)(b+d)=a b+c b+a d+c d$, but $c b+a d \geqslant 2 \sqrt{a b c d}$.
Then $(a+c)(b+d) \geqslant a b+c d+2 \sqrt{a b c d}=(\sqrt{a b}+\sqrt{c d})^{2}$
or $\sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+\sqrt{c d}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,927 |
13. For any numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ and $b_{1}, b_{2}, b_{3}, \ldots, b_{n}$, the statement $\left|a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\ldots+a_{n} b_{n}\right| \leqslant 1$ holds if $a_{1}{ }^{2}+a_{2}^{2}+a^{2}{ }_{3}+\ldots+a_{n}{ }^{2}=1$ and $b_{1}{ }^{2}+b_{2}{ }^{2}+b_{3}{ }^{2}+\ldots+b_{n}... | 13. For any $a$ and $b$, the inequality $|a b| \leqslant \frac{a^{2}+b^{2}}{2}$ holds.
Then $\left|a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}\right| \leqslant\left|a_{1} b_{1}\right|+\left|a_{2} b_{2}\right|+\ldots+\left|a_{n} b_{n}\right| \leqslant$
$$
<\frac{a_{1}^{2}+b_{1}^{2}}{2}+\frac{a_{2}^{2}+b_{2}^{2}}{2}+\ld... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,928 |
14. If $a_{1}, a_{2}, a_{3}, \ldots, a_{n} ; b_{1}, b_{2}, b_{3}, \ldots, b_{n}$ are any real numbers, then
$$
\left(a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}\right)^{2} \leqslant\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\ldots+b_{n}^{2}\right)
$$ | 14. Let $x$ be any real number. Consider the obvious inequality $\left(a_{1}-x b_{1}\right)^{2}+\left(a_{2}-x b_{2}\right)^{2}+\ldots+\left(a_{n}-x b_{n}\right)^{2} \geqslant 0$.
After expanding the brackets and grouping, we get
$x^{2}\left(b_{1}^{2}+b_{2}^{2}+\ldots+b_{n}^{2}\right)-2 x\left(a_{1} b_{1}+a_{2} b_{2}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,929 |
15. If $x$ and $y$ are any real numbers, $x \neq 0, y \neq 0$,
$$
\text { then } \quad 3\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right)-8\left(\frac{x}{y}+\frac{y}{x}\right)+10 \geqslant 0
$$ | 15. Let $\frac{x}{y}+\frac{y}{x}=u, \quad$ then $\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}=u^{2}-2$. The left-hand side takes the form $3 u^{2}-8 u+4$.
If $x$ and $y$ have different signs $x0(x>0, y0$
If $x$ and $y$ have the same sign $x>0, y>0(x0$.
$\frac{x}{y}+\frac{y}{x}=\frac{x^{2}+y^{2}}{x y} \geqslant 2 . \quad$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,930 |
16. Given positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{100}$. It is known that $a_{1}>a_{0}, a_{2}=3 a_{1}-2 a_{0}, a_{3}=3 a_{2}-2 a_{1}, \ldots a_{100}=3 a_{99}-2 a_{98}$. Prove that $a_{100}>2^{99}$. | 16. Expressing the numbers through $a_{0}$ and $a_{1}$, we get
$$
a_{2}=3 a_{1}-2 a_{0}, \quad a_{3}=3 a_{2}-2 a_{1}=7 a_{1}-6 a_{0}, \quad a_{4}=3 a_{3}-2 a_{2}=15 a_{1}-14 a_{0}
$$
In general, $a_{n}=a_{1}\left(2^{n}-1\right)-\left(2^{n}-2\right) a_{0}$.
$$
a_{n}=a_{1}\left(2^{n}-1\right)-\left(2^{n}-1\right) a_{0... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,931 |
17. For the numbers $a_{0}, a_{1}, a_{2}, \ldots, a_{n-1}, a_{n}$, it is known that $a_{0}=a_{n}=0$ and that $a_{k-1}-2 a_{k}+a_{k+1} \geqslant 0$ for all $k(k=1,2,3, \ldots, n-1)$. Prove that all numbers $a$ are non-positive. | 17. Assume the opposite. Let $a_{r}$ be the first positive number, i.e., $a_{r-1} \leqslant 0, a_{r}>0$, then $a_{r}-a_{r-1}>0$. According to the problem's condition, $a_{k+1}-a_{k} \geqslant \geqslant a_{k}-a_{k-1}$ for all $k(k=1,2,3, \ldots, n-1)$. Starting from $k=r$, we sequentially obtain from the inequality $a_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,932 |
18. Prove that for any real $x, y, z$ the inequality $4 x(x+y)(x+z)(x+y+z)+y^{2} z^{2} \geqslant 0$ holds. | 18. Multiplying $(x+y)(x+z)$ and $x(x+y+z)$, we transform the left side of the inequality to the form $\left[2\left(x^{2}+x y+x z\right)+y z\right]^{2} \geqslant 0$. This proves the validity of the given inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,933 |
19. Show that for any $n \geqslant 1$ the inequality holds
$$
\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2}
$$ | 19. Consider the first $n-1$ terms
$$
\frac{1}{n+1}>\frac{1}{n+2}>\frac{1}{n+3}>\ldots>\frac{1}{2 n-1}
$$
Then $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n-1}>(n-1) \cdot \frac{1}{2 n}$. Adding $\frac{1}{2 n}$ to both sides, we get $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,934 |
20. Prove that for any natural number $n$ the following inequality holds
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots+\frac{1}{n^{2}}<\frac{n-1}{n}
$$ | 20. Let's write down the obvious inequalities
\[
\begin{aligned}
& \frac{1}{2^{2}}<\frac{1}{1 \cdot 2}=1-\frac{1}{2} \\
& \frac{1}{3^{2}}<\frac{1}{2 \cdot 3}=\frac{1}{2}-\frac{1}{3} \\
& \cdots \cdots \cdots \cdots \cdot \cdots \\
& \frac{1}{n^{2}}<\frac{1}{(n-1) n}=\frac{1}{n-1}-\frac{1}{n}
\end{aligned}
\]
Adding t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,935 |
21. Prove that for $n>2,(n!)^{2}>n^{n}(n!=1 \cdot 2 \cdot 3 \cdots n)$. | 21. Let's rewrite parts of the inequality as follows:
$n^{n}=n \cdot n \cdot n \ldots n$. Each factor in the square brackets is not less than $n$. We will prove that $(n-k+1) k \geqslant ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,936 |
23. Prove that
$$
\left(a_{1}+a_{2}+a_{3}+\ldots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n}}\right) \geqslant n
$$
if each term is positive. | 23. In the left part of the inequality, we expand the brackets
$$
\begin{aligned}
\left(a_{1}+\right. & \left.a_{2}+\ldots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)= \\
& =n+\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{1}}+\ldots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{n-1}}
\end{aligned}
$$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,938 |
24. Find all real numbers $x$ that satisfy the inequality $\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}$. | 24. The arithmetic values of radicals are meant.
From the system of inequalities $\left\{\begin{array}{l}3-x \geqslant 0 \\ x+1 \geqslant 0\end{array}\right.$ it follows that $-1 \leqslant x \leqslant 3$.
The inequality is satisfied for $x < 1+\frac{\sqrt{31}}{8}$.
Considering condition (3), we finally obtain $-1 \l... | -1\leqslantx\leqslant1-\frac{\sqrt{31}}{8} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,939 |
25. Solve the inequality $\frac{-5}{2 x^{2}-11 x+12}<\frac{-5}{2 x^{2}-9 x+7}$. | 25. The inequality makes no sense at $x=1 ; \frac{3}{2} ; \frac{7}{2} ; 4$.
a)
$$
\left\{\left.\begin{array}{l}
2 x^{2}-11 x+12>0, \quad x4 \\
2 x^{2}-9 x+7>0, \quad x\frac{7}{2} \\
2 x^{2}-9 x+7>2 x^{2}-11 x+12, x>\frac{5}{2}
\end{array} \right\rvert\, x>4\right.
$$
b)
$$
\left\{\left.\begin{array}{l}
2 x^{2}-11 x+... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,940 |
26. For what values of $x$ is the inequality
$$
\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9 ?
$$ | 26. The inequality makes sense for $x \geqslant-\frac{1}{2}$, but $x \neq 0$.
By getting rid of the denominator and the radical, we obtain the inequality $8 x^{3}-45 x^{2}<0$, from which $x<5 \frac{5}{8}$. Answer: $-\frac{1}{2} \leqslant x<0,0<x<5 \frac{5}{8}$. | -\frac{1}{2}\leqslantx<0,0<x<5\frac{5}{8} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,941 |
27. For what values of $\boldsymbol{a}$ is the inequality $-3<\frac{x^{2}+a x-2}{x^{2}-x+1}<2$ true for all values of $x$?
In № $28-32$ solve the inequalities. | 27. $x^{2}-x+1>0$ for any $x$.
Then $\left\{\begin{array}{l}-3\left(x^{2}-x+1\right)<0 \\ x^{2}+a x-2<0 .\end{array}\right.$
Both inequalities are valid for any $x$ if the discriminants of the quadratic polynomials are negative, i.e.
$$
\begin{aligned}
& (a-3)^{2}-16<0, \quad|a-3|<4,-1<a<7 \\
& (a+2)^{2}-16<0, \quad... | -1<<2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,942 |
29. $5^{2 x+1}+6^{x+1}>30+5^{x} 30^{x}$. | 29. The solution is similar to the previous example.
Answer: $\frac{1}{2} \log _{5} 6<x<\log _{6} 5$. | \frac{1}{2}\log_{5}6<x<\log_{6}5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,944 |
31. $\frac{4^{x}-2^{x+1}+8}{2^{1-x}}<8^{x}$. | 31. In all members of the inequality, we transition to base 2. By moving all terms to the right side of the inequality, we get $2^{x} \cdot\left(2^{2 x}+2 \cdot 2^{x}-2^{3}\right)>0$.
Since $2^{x}>0$ for any $x$, the inequality holds under the condition that $2^{2 x}+2 \cdot 2^{x}-2^{3}>0$, i.e., $2^{x}>2, x>1$. | x>1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,946 |
33. For what values of $r$ is the inequality
$$
\left(r^{2}-1\right) x^{2}+2(r-1) x+1>0
$$
satisfied for all $x$? | 33. A quadratic trinomial is positive for any values of $x$ if the coefficient of $x^{2}$ is positive, and the discriminant is negative, i.e., $r^{2}-1>0 (r>0, r>1$.
Answer: $r>1$. | r>1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,948 |
34. Find all values of $a$ for which the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+2 x \leqslant 1 \\
x-y=-a
\end{array}\right. \text { has a unique solution. }
$$
Solve the inequalities given in № $35-41$. | 34. Let's write down the equivalent system for the given one $\left\{\begin{array}{l}x+a=y \\ x^{2}+(x+a)^{2}+2 x \leqslant 1 .\end{array}\right.$
The unique solution will occur if the discriminant of the quadratic polynomial is zero, i.e., $(a+1)^{2}-2\left(a^{2}-1\right)=0$.
From this, $a=3, a=-1$. | =3,=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,949 |
35. $\log _{\frac{1}{3}} \log _{5}\left(\sqrt{x^{2}+1}+x\right)<\log _{3} \log _{\frac{1}{3}}\left(\sqrt{x^{2}+1}-x\right)$. | 35. In both parts of the inequality, we switch to the base 3 and 5.
The inequality will take the form $2 \log _{3} \log _{5}\left(\sqrt{x^{2}+1}+x\right)>0$.
From which $\log _{5}\left(\sqrt{x^{2}+1}+x\right)>1, \sqrt{x^{2}+1}+x>5$. Answer: $x>\frac{12}{5}$. | x>\frac{12}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,950 |
36. $\log _{2 x+3} x^{2}<1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
36. $\log _{2 x+3} x^{2}<1$. | 36. If $0<2 x+3$, i.e. $x>-\frac{3}{2}$.
The inequality is satisfied when $-\frac{3}{2}<x<1$, i.e. $2 x>-2$, then $x^{2}<2 x+3$, i.e. $-1<x<3$.
Answer: $-\frac{3}{2}<x<3, x \neq 0, x \neq-1$. | -\frac{3}{2}<x<3,x\neq0,x\neq-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,951 |
43. Solve the inequality $4 \sin 3 \alpha+5 \geqslant 4 \cos 2 \alpha+5 \sin \alpha$. | 43. The functions $\sin 3 \alpha$ and $\cos 2 \alpha$ expressed through the functions of the angle $\alpha$.
The inequality will take the form $(1-\sin \alpha)(4 \sin \alpha+1)^{2} \geqslant 0$.
We obtained an inequality that is valid for any $\alpha$. | (1-\sin\alpha)(4\sin\alpha+1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,958 |
44. Solve the inequality $4 \log _{16} \cos 2 x+2 \log _{4} \sin x+\log _{2} \cos x+$ $+3<0$. | 44. We transition to logarithms with base 2. The given inequality reduces to the inequality $\log _{2} 2 \sin 4 x<0$, i.e., $0<\sin 4 x<\frac{1}{2}$.
But $0<x<\frac{\pi}{4}$. Therefore, the inequality holds under the condition $0<4 x<\frac{\pi}{6}$, i.e., $0<x<\frac{\pi}{24}$ and $\frac{5}{6} \pi<4 x<\pi$, i.e., $\fra... | 0<x<\frac{\pi}{24} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,959 |
1. Given an angle and a point $M$ inside it. Draw a straight line through point $M$ such that the segment cut off by the sides of the angle is bisected at point $M$. | 1. Draw a line through point $M$ parallel to one of the sides of the angle; transfer the segment obtained on this side of the angle to the same side again. Draw a line through the end of the second segment and point $M$. This line will be the desired one.
98 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,960 |
2. Prove that the sum of the distances from a point lying on the base of an isosceles triangle to its lateral sides is equal to the length of the height dropped from the vertex to the lateral side of the triangle.
untranslated text:
2. Доказать, что сумма расстояний от точки, лежащей на основании равнобедренного треу... | 2. From a point lying on the base of an isosceles triangle, we drop perpendiculars to the lateral sides; from the same point, we draw a line parallel to one of the lateral sides. This line will intersect the height dropped to the other lateral side, dividing it into segments whose lengths are equal to the previously co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,961 |
4. Construct a quadrilateral $A B C D$ given four sides, knowing that the diagonal $A C$ bisects angle $D A B$.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | 4. Suppose that the quadrilateral $A B C D$ is constructed and satisfies the conditions of the problem (Fig. 23). On side $A D$, from point $A$, we lay off a segment $A E$ equal to $A B$, and connect point $C$ with point $E$. In the quadrilateral $A B C E$ (a kite), side $B C$ is equal to $C E$. Triangle $C E D$ can be... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,963 | |
5. The sum of the angles at the base of a trapezoid is $90^{\circ}$. Prove that the length of the segment connecting the midpoints of the bases is equal to their semi-difference. | 5. Let the trapezoid be $ABCD$ and let $\angle A + \angle D = 90^{\circ}$. Denote the midpoint of the upper base as $E$, and the midpoint of the lower base as $F$. It is required to prove that $EF = \frac{AD - BC}{2}$
 will represent the geometric locus of the vertices of acute angles at vertex $C$ of triangle $A B C$. However, the angles at vertices $A$ and $B$ can be either acute, right, or obtuse. ... | Other | math-word-problem | Yes | Yes | olympiads | false | 34,967 | |
9. Show that two triangles, each having 3 equal angles and two equal sides, are not necessarily equal to each other. | 9. Triangles having three equal angles are always similar, and two similar triangles can have two equal sides. For example, let the sides of one triangle be equal to 27, 28, and 12, and the sides of the other be respectively equal to 18, 12, and 8. Such triangles are similar but not equal. | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,968 |
11. In triangle $ABC$, a circle is inscribed, touching its sides at points $D, E, F$. Prove that triangle $DEF$ is always acute-angled. | 11. Each angle of triangle $D E F$ (for example, at vertex $D$) is equal to half the central angle formed by the two radii of the inscribed circle that are drawn to the other two vertices of triangle $D E F$. Each such central angle will be at most obtuse, therefore, the angle at vertex $D$ will be acute. We will also ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,970 |
12. Prove that in a triangle, the bisector divides in half the angle between the altitude and the radius of the circumscribed circle, drawn from the same vertex. | 12. This task is equivalent to the following: to prove that the height of a triangle and the radius of the circumscribed circle, drawn to the vertex, form equal angles with the lateral sides of the triangle. To solve this problem, one needs to drop a perpendicular from the center of the circle to the larger of the late... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,971 |
13. Restore a triangle given its centroid and the midpoints of two of its medians. | 13. The segment connecting the centroid of a triangle with the midpoint of its midline constitutes $\frac{1}{6}$ of the corresponding median. We can construct two medians, find two vertices of the triangle and the midpoints of the sides opposite to them. From them, it is easy to construct the third vertex of the triang... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,972 |
14. On the sides of an arbitrary triangle, outside it, equilateral triangles are constructed. Prove that their centers themselves are the vertices of an equilateral triangle. | 14. Let the sides of triangle $A B C$ be denoted as $a, b, c$. Suppose the centers of the equilateral triangles $A C_{1} B, B A_{1} C$, and $C B_{1} A$, constructed on the sides of triangle $A B C$, are points $O_{1}, O_{2}, O_{3}$. Then we have: $\angle O_{3} A O_{1} = \angle A + 60^{\circ} ; O_{3} A = \frac{b \sqrt{3... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,973 |
15. Construct a triangle, given a line on which its base lies and two points that are the feet of the altitudes dropped from the lateral sides. | 15. By connecting two points - the bases of the altitudes - with a segment, draw a line perpendicular to it through its midpoint until it intersects with the given line. The point of intersection obtained is the center of the circle passing through the bases of the altitudes, and the diameter of this circle lying on th... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,974 |
18. Prove that a quadrilateral is a parallelogram if and only if the sum of the squares of the diagonals is equal to the sum of the squares of all its four sides. | 18. It should be preliminarily proved that for any quadrilateral, there is a relationship: $M N^{2}=\frac{1}{4}\left(a^{2}+b^{2}+c^{2}+d^{2}-e^{2}-f^{2}\right)$, where $M N-$ is the length of the segment connecting the midpoints of the diagonals of the quadrilateral; $a, b$, $\boldsymbol{c}, d$ are its sides; $e$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,977 |
19. Construct a triangle given the median, altitude, and angle bisector drawn from the same vertex. | 19. Let $m_{a}, \beta_{a}$, and $h_{a}$ be the median, bisector, and altitude of the desired triangle $ABC$, drawn from vertex $A$. We construct a right triangle $ADC$ with hypotenuse $AE = \beta_{a}$ and leg $AD = h_{a}$, with a right angle at vertex $D$. Next, from vertex $A$ as the center and with radius equal to $m... | m_{}\geqslant\beta_{}\geqslanth_{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,978 |
21. If the midpoints of the sides of a quadrilateral are taken as the vertices of a new one, then a parallelogram will be formed. Under what conditions will it be 1) a rectangle, 2) a rhombus, 3) a square? | 21. We will get a rectangle if the diagonals of a quadrilateral are perpendicular to each other; we will get a rhombus if the diagonals of a quadrilateral are equal (for example, in an isosceles trapezoid); we will get a square if the diagonals of a quadrilateral are both perpendicular and equal to each other. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,980 |
22. Construct a triangle given the lengths of all its three medians. | 22. Let the bases of the medians of a triangle be points $A_{1}, B_{1}$, and $C_{1}$; construct triangle $A_{1} B_{1} C_{1}$. Then, through vertex $A_{1}$, draw a line parallel to $B_{1} C_{1}$; through vertex $B_{1}$, draw a line parallel to $C_{1} A_{1}$; and through vertex $C_{1}$, draw a line parallel to $A_{1} B_{... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,981 |
24. Construct a pentagon using the midpoints of all its sides. | 24. Let $A_{1} A_{2} A_{3} A_{4} A_{5}$ be the desired pentagon, and the midpoints of its sides be points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$. These points are known: $B_{1}$ is the midpoint of side $A_{1} A_{2} ; B_{2}$ is the midpoint of side $A_{2} A_{3} ; B_{3}$ is the midpoint of $A_{3} A_{4} ; B_{4}$ is the midpo... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,983 |
25. Prove that the intersection of the bisectors of the interior angles of a parallelogram forms a rectangle, the diagonal of which is equal to the difference of the adjacent sides of the parallelogram. | 25. By constructing the bisectors of all the internal angles of a parallelogram, it is easy to prove that they intersect to form a rectangle, since the bisectors of the angles adjacent to one side of the parallelogram are mutually perpendicular. By choosing a diagonal of the rectangle parallel to the larger side of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,984 |
26. Restore the parallelogram if the midpoints $M, N, P$ of three of its sides are preserved, with $M$ and $N$ being the midpoints of two of its opposite sides. | 26. We construct triangle $M N P$. We divide side $M N$ in half and, marking the midpoint of this side as $Q$, connect point $Q$ with $P$ - the vertex of triangle $M N P$. Then, on the extension of ray $P Q$, by laying off segment $Q R=P Q$, we obtain the midpoint of the fourth side of the parallelogram. Now it is easy... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,985 |
28. On the sides of an arbitrary parallelogram $A B C D$, outside it, squares are constructed. Prove that the centers of these squares themselves are the vertices of a square. | 28. Let $O_{1}, O_{2}, O_{3}$ and $O_{4}$ be the centers of squares constructed on the sides $A B, B C, C D$ and $A D$ of parallelogram $A B C D$. Then it can be verified that when rotated by $90^{\circ}$ around point $O_{2}$ or around point $O_{4}$, point $O_{1}$ coincides with point $O_{3}$; from this it follows that... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,987 |
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