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 |
|---|---|---|---|---|---|---|---|---|---|
6.352 Solve the equations $x^{4}-x^{3}-22 x^{2}+16 x+96=0$ and $x^{3}-2 x^{2}-3 x+10=0$, given that they have a common root. | Solution. Let $x_{0}$ be a common root,
$$
f(x)=x^{4}-x^{3}-22 x^{2}+16 x+96, g(x)=x^{3}-2 x^{2}-3 x+10
$$
- Represent $f(x)$ in the form (3). Then $r_{1}(x)=-17 x^{2}+9 x+86$.
$$
\text { Further, } \eta(x)=0 \Leftrightarrow\left[\begin{array}{l}
x_{0}=-2 \\
x_{0}=\frac{43}{17}
\end{array} . \text { By checking, we ... | x_{1}=-2,x_{2}=3,x_{3,4}=\4;-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,783 |
6.353 Find all values of $\lambda$ for which the equations
$\lambda x^{3}-x^{2}-x+(\lambda+1)=0$ and $\lambda x^{2}-x-(\lambda+1)=0$ have a common root, and find this root. | Solution. Let $x_{0}$ be a common root,
$$
f(x)=\lambda x^{3}-x^{2}-x+(\lambda+1), g(x)=\lambda x^{2}-x-(\lambda+1)
$$
Represent $f(x)$ in the form (3). Then $r_{1}(x)=\lambda x+(\lambda+1) \Rightarrow r_{1}(x)=0$ when $x_{0}=-\frac{\lambda+1}{\lambda}$. But then $g\left(-\frac{\lambda+1}{\lambda}\right)=0$, from whi... | 0when\lambda=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,784 |
6.354 Solve the equation $8 x^{3}+4 x^{2}-34 x+15=0$, given that two of its roots $x_{1}$ and $x_{2}$ satisfy the relation $2 x_{1}-4 x_{2}=1$. | ## Solution.
Let $f(x)=8 x^{3}+4 x^{2}-34 x+15$. Then $f\left(x_{1}\right)=f\left(x_{2}\right)=0 \Rightarrow$
$\Rightarrow f\left(x_{1}\right)-f\left(x_{2}\right)=0 \Leftrightarrow 4\left(x_{1}^{3}-x_{2}^{3}\right)+2\left(x_{1}^{2}-x_{2}^{2}\right)-34\left(x_{1}-x_{2}\right)=0 \Leftrightarrow$
$\Leftrightarrow\left(... | x_{1}=\frac{3}{2},x_{2}=\frac{1}{2},x_{3}=-\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,785 |
6.355 Prove that for any natural $n$
$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2}$
and using this equality, solve the equation
$$
(1+3+5+\ldots+(2 n+1)):\left(\frac{1}{2}+\frac{1}{6}+\ldots+\frac{1}{342}\right)=342
$$ | ## Solution.
$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{(n+1)(n+2)}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)+$ $+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=1-\frac{1}{n+2}=\frac{n+1}{n+2}$.
$1+3+5+\ld... | 17 | Algebra | proof | Yes | Yes | olympiads | false | 51,786 |
6.356 Solve the equation $x^{4}-6 x^{3}+7 x^{2}+6 x-2=0$, given that it has at least one pair of roots $x_{1}$ and $x_{2}$, the difference of which is 1. | Solution. Let $f(x)=x^{4}-6 x^{3}+7 x^{2}+6 x-2 ; x_{1}-x_{2}=1$, then $f\left(x_{1}\right)=f\left(x_{2}\right)=0 \Rightarrow f\left(x_{1}\right)-f\left(x_{2}\right)=0 \Leftrightarrow$ $\Leftrightarrow\left(x_{1}^{4}-x_{2}^{4}\right)-6\left(x_{1}^{3}-x_{2}^{3}\right)+7\left(x_{1}^{2}-x_{2}^{2}\right)+6\left(x_{1}-x_{2}... | x_{1}=1+\sqrt{3},x_{2}=2+\sqrt{3},x_{3}=1-\sqrt{3},x_{4}=2-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,787 |
6.357 Solve the equation $3 x^{3}+2 \sqrt{3} x^{2}-21 x+6 \sqrt{3}=0$, given that the product of two of its roots is 1. | Solution. Let $x_{1}, x_{2}, x_{3}$ be the roots of the original equation, $x_{1} x_{2}=1$. By Vieta's theorem, we have:
$\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-\frac{2 \sqrt{3}}{3}, \\ x_{1} x_{2} x_{3}=-2 \sqrt{3}\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}+x_{2}=\frac{4 \sqrt{3}}{3}, \\ x_{1} x_{2}=1, \\ ... | x_{1}=\sqrt{3},x_{2}=\frac{\sqrt{3}}{3},x_{3}=-2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,788 |
6.358 Solve the equations $x^{3}-7 x^{2}+12 x-10=0$ and $x^{3}-10 x^{2}-2 x+20=0$, given that one of the roots of the first equation is half of one of the roots of the second equation. | Solution. Let $x_{0}$ be the root of the first equation, $2 x_{0}$ be the root of the second equation, $f(x)=x^{3}-7 x^{2}+12 x-10, g(x)=x^{3}-10 x^{2}-2 x+20$. Then $x_{0}$ is the root of the equation $\tilde{f}(x)=0$, where $\tilde{f}(x)=f(2 x)=8 x^{3}-28 x^{2}+24 x-10$.
Represent $\tilde{f}(x)$ in the form (3). The... | 5,x_{1}=10,x_{2}=-\sqrt{2},x_{3}=\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,789 |
6.359 Solve the equation $a x^{3}+b x^{2}+c x+d=0$,
if $a d=b c \neq 0, b d<0$. | Solution. $\left\{\begin{array}{l}a x^{3}+b x^{2}+c x+d=0, \\ a d=b c\end{array} \Leftrightarrow\left\{\begin{array}{l}a x\left(x^{2}+\frac{c}{a}\right)+b\left(x^{2}+\frac{d}{b}\right)=0, \\ a d=b c .\end{array}\right.\right.$ $a x\left(x^{2}+\frac{d}{b}\right)+b\left(x^{2}+\frac{d}{b}\right)=0 \Leftrightarrow\left(x+\... | x_{1}=-\frac{b}{},x_{2,3}=\\sqrt{-\frac{}{b}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,790 |
6.361 Solve the equation $a x^{3}+b x^{2}+c x+d=0$, if its coefficients $a, b, c, d$ in the given order form a geometric progression with a given common ratio $q$. | Solution. The original equation can be written as:
$$
\begin{aligned}
& a x^{3}+a q x^{2}+a q^{2} x+a q^{3}=0 \Leftrightarrow x^{3}+q x^{2}+q^{2} x+q^{3}=0 \Leftrightarrow \\
& \Leftrightarrow(x+q)\left(x^{2}+q^{2}\right)=0 \Leftrightarrow x=-q .
\end{aligned}
$$
Answer: $x=-q$. | -q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,791 |
6.362 Prove that if the roots of the equation $x^{3}+a x^{2}+b x+c=0$ form a geometric progression, then one of them is equal to $-\sqrt[3]{c}$. | Solution. Let $a, a q, a q^{2}$ be the roots of the original equation. Then, by Vieta's theorem, $a^{3} q^{3}=-c \Rightarrow a q=-\sqrt[3]{c}$, which is what we needed to prove. | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,792 |
6.363 Solve the equation $64 x^{3}-24 x^{2}-6 x+1=0$, given that its roots form a geometric progression. | Solution. Let $a, a q, a q^{2}$ be the roots of the original equation. Then, by Vieta's theorem, $\left\{\begin{array}{l}a^{3} q^{3}=-\frac{1}{64}, \\ a+a q+a q^{2}=\frac{3}{8}\end{array} \Leftrightarrow\left\{\begin{array}{l}a q=-\frac{1}{4}, \\ 2 q^{2}+5 q+2=0\end{array} \Leftrightarrow\left[\begin{array}{l}q=-2, a=\... | \frac{1}{8};-\frac{1}{4};\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,793 |
6.364 1. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial $a x^{3}+b x^{2}+c x+d$.
## Use the identity
$a x^{3}+b x^{2}+c x+d=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)$ to obtain formulas relating the roots and coefficients of the given polynomial.
2. Find the roots $x_{1}, x_{2}, x_{3}$ ... | ## Solution.
1. The required formulas are obtained by expanding the brackets of the given identity and combining like terms:
$$
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=\frac{c}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a}
$$
These formulas are Vieta's theorem for cubic equations.
2. The new cubic ... | x_{1}=1.5;x_{2,3}=0.5\\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,794 |
6.365 Solve the equation $\left(x^{3}+x^{-3}\right)+\left(x^{2}+x^{-2}\right)+\left(x+x^{-1}\right)=6$. | Solution. $\left(x^{3}-2+\frac{1}{x^{3}}\right)+\left(x^{2}-2+\frac{1}{x^{2}}\right)+\left(x-2+\frac{1}{x}\right)=0 \Leftrightarrow \frac{\left(x^{3}-1\right)^{2}}{x^{3}}+\frac{\left(x^{2}-1\right)^{2}}{x^{2}}+\frac{(x-1)^{2}}{x}=0$ $(x-1)^{2}\left(\left(x^{2}+x+1\right)^{2}+x(x+1)^{2}+x^{2}\right)=0 \Leftrightarrow$ $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,795 |
6.366 Form an equation of the lowest degree with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt{3}$. | Solution. The equation $x^{2}-2 \sqrt{3} x+1=0$ has roots $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$.
Therefore, the equation $\left(x^{2}-2 \sqrt{3} x+1\right)\left(x^{2}+2 \sqrt{3} x+1\right)=0$ has a root $\sqrt{3}+\sqrt{2}$. $\left(x^{2}-2 \sqrt{3} x+1\right)\left(x^{2}+2 \sqrt{3} x+1\right)=0 \Leftrightarrow x^{... | x^{4}-10x^{2}+1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,796 |
6.367 Prove that the roots of the equation $x+x^{-1}=2 \cos 40^{\circ}$ are roots of the equation $x^{4}+x^{-4}=2 \cos 160^{\circ}$. | ## Solution.
$x^{4}+\frac{1}{x^{4}}=2 \cos 160^{\circ} \Leftrightarrow\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=2\left(1+\cos 60^{\circ}\right) \Leftrightarrow\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=4 \cos ^{2} 80^{\circ} \Leftrightarrow$ $\Leftrightarrow x^{2}+\frac{1}{x^{2}}=2 \cos 80^{\circ} \Leftrightarrow\left(x+\fra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,797 |
6.368 Solve the equation $x^{4}-4 x^{3}+3 x^{2}+8 x-10=0$, given that two of its roots differ only in sign. | Solution. Let $x_{1}$ and $x_{2}$ be the roots of the original equation and $x_{1}=-x_{2}$. Then
$\left\{\begin{array}{l}x_{1}^{4}-4 x_{1}^{3}+3 x_{1}^{2}+8 x_{1}-10=0, \\ x_{1}^{4}+4 x_{1}^{3}+3 x_{1}^{2}-8 x_{1}-10=0\end{array} \Rightarrow 8 x_{1}^{3}-16 x_{1}=0 \Leftrightarrow x_{1}^{2}-2=0 \Rightarrow\right.$ $x_{1... | x_{12}=\\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,798 |
6.369 Solve the equation $2 x^{5}-x^{4}-2 x^{3}+x^{2}-4 x+2=0$, given that it has three roots, two of which differ only in sign. | Solution. Let $x_{1}, x_{2}, x_{3}$ be the roots, $x_{1}=-x_{2}$. Then
$\left\{\begin{array}{l}2 x_{1}^{5}-x_{1}^{4}-2 x_{1}^{3}+x_{1}^{2}-4 x_{1}+2=0, \\ -2 x_{1}^{5}-x_{1}^{4}+2 x_{1}^{3}+x_{1}^{2}+4 x_{1}+2=0\end{array} \Rightarrow 4 x_{1}^{5}-4 x_{1}^{3}-8 x_{1}=0 \Leftrightarrow\right.$ $x_{1}^{4}-x_{1}^{2}-2=0 \... | x_{1,2}=\\sqrt{2};x_{3}=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,799 |
6.370 Solve the equation $\sqrt[4]{x^{4}+x-2}+\sqrt{x^{4}+x-2}=6$ given that $\boldsymbol{x}>\mathbf{0}$. | Solution. Let $y=\sqrt[4]{x^{4}+x-2}$, then $y^{2}+y-6=0 \Leftrightarrow \sqrt[4]{x^{4}+x-2}=2 \Leftrightarrow$ $\Leftrightarrow x^{4}+x-18=0 \Leftrightarrow(x-2)\left(x^{3}+2 x^{2}+4 x+9\right)=0$.
Since $x>0$, then $x^{3}+2 x^{2}+4 x+9>0 \Rightarrow x=2$.
Answer: $x=2$.
## pRoGReSSiOnS | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,800 |
4.069 Find a three-digit number whose digits form a geometric progression. If 792 is subtracted from this number, the result is a number written with the same digits but in reverse order. If 4 is subtracted from the digit representing the hundreds, and the other digits of the desired number are left unchanged, the resu... | Solution. Let the desired number have the form:
$100 x+10 y+z \Rightarrow 100 x+10 y+z=792+100 z+10 y+x \Rightarrow x=8+z \Rightarrow\left[\begin{array}{l}z=0, x=8, \\ z=1, x=9 .\end{array}\right.$
Since the digits $x, y, z$ form a geometric progression, then $z \neq 0 \Rightarrow z=1$. The possible cases are:
1) $x... | 931 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 51,801 |
4.070 It is known that for any $n$ the sum of the first $n$ terms of a certain numerical sequence is expressed by the formula $S_{n}=2 n^{2}+3 n$. Find the tenth term of this sequence and prove that this sequence is an arithmetic progression. | Solution. Let $a_{1}, a_{2}, \ldots a_{n}$ be the given sequence. Then
$a_{n}=S_{n}-S_{n-1}=2 n^{3}+3 n-\left(2(n-1)^{2}+3(n-1)\right)=4 n+1$ for $n \geq 2$. Further, $a_{n+1}-a_{n}=(4(n+1)+1)-(4 n+1)=4$ for $n \geq 2$. It remains to check the difference $a_{2}-a_{1}: a_{2}=4 \cdot 2+1=9, a_{1}=S_{1}=2 \cdot 1^{2}+3 \... | 41 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,802 |
4.072 The lengths of the sides of a triangle are three consecutive terms of a geometric progression. Compare the common ratio of this progression with the number 2. | Solution. Let $b, b q, b q^{2}$ be the lengths of the sides of the triangle. By the triangle inequality, $b+b q>b q^{2} \Rightarrow q^{2}-q-1<0 \Rightarrow q<\frac{1+\sqrt{5}}{2}<2$.
Answer: less than 2. | than2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,803 |
4.073 Find the sum of the first four terms of a geometric progression that has the property that its first three terms, the sum of which is $\frac{148}{9}$, are simultaneously the first, fourth, and eighth terms of some arithmetic progression. | Solution. Let $b, b q, b q^{2}, b q^{3}$ be the given geometric progression, then
(if $q=1$, then $b q$ and $b q^{2}$ would not necessarily be the fourth and eighth terms of the arithmetic... | 25\frac{25}{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,804 |
4.074 The numbers $a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}$ form an arithmetic progression. Prove that $\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n} a_{n+1}}=\frac{n}{a_{1} a_{n+1}}$. | Solution. Let $d$ be the common difference of the arithmetic progression, then
$$
\begin{aligned}
& \frac{1}{a_{i} a_{i+1}}=\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right) \frac{1}{d} \Rightarrow \frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n} a_{n+1}}= \\
& =\frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,805 |
4.075 The sequence of numbers $1,8,22,43, \ldots$ has the property that the differences between two consecutive terms (the subsequent and the previous) form an arithmetic progression: $7,14,21, \ldots$. Find the index of the term in the sequence that equals 35351. | Solution. Let the numbers of the given sequence $1, 8, 22, \ldots$ be denoted as $a_{1}, a_{2}, a_{3}, \ldots$.
By the condition, $a_{2}-a_{1}=7 \cdot 1, a_{3}-a_{2}=7 \cdot 2, \ldots, a_{n}-a_{n-1}=7(n-1)$.
Adding these equations term by term, we have: $\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\ldots+$
$+\... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,806 |
4.077 The sum of four numbers forming a geometric progression is -40, and the sum of their squares is 3280. Find this progression. | Solution. Let $b, b q, b q^{2}, b q^{3}$ be the given numbers: Then
$\left\{\begin{array}{l}b\left(1+q+q^{2}+q^{3}\right)=-40, \\ b^{2}\left(1+q^{2}+q^{4}+q^{6}\right)=3280\end{array} \Leftrightarrow\left\{\begin{array}{l}b(1+q)\left(1+q^{2}\right)=-40, \\ b^{2}\left(1+q^{2}\right)\left(1+q^{4}\right)=3280 .\end{array}... | 2,-6,18,-54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,807 |
4.079 Find the sum $\Sigma=1+2 \cdot 3+3 \cdot 7+\ldots+n\left(2^{n}-1\right)$. | Solution. $\Sigma=1 \cdot\left(2^{1}-1\right)+2 \cdot\left(2^{2}-1\right)+\ldots+n \cdot\left(2^{n}-1\right)=2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n}-$ $-(1+2+3+\ldots+n)$
Let: $S_{n}=2^{i}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n} \Rightarrow$
$\Rightarrow S_{n-1}=2^{1}+2 \cdot 2^{2}+3 \cdot 2^{... | (n-1)2^{n+1}+2-\frac{n(n+1)}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,808 |
4.083 Prove the following statement: for three numbers $x, y$ and $z$ to form a geometric progression in the given order, it is necessary and sufficient that the equality $\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)=(x y+y z)^{2}$ holds. | ## Solution.
1) Given: $\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)=(x y+y z)^{2} \Rightarrow$
$\Rightarrow x^{2} y^{2}+x^{2} z^{2}+y^{4}+y^{2} z^{2}=x^{2} y^{2}+2 x y^{2} z+y^{2} z^{2} \Rightarrow$
$\Rightarrow y^{4}-2 y^{2} x z+x^{2} z^{2}=0 \Rightarrow\left(y^{2}-x z\right)^{2}=0 \Rightarrow y^{2} = x z$, i.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,809 |
4.084 In a volleyball competition, $n$ teams participated. Each team played with all the others once. For each game, the winning team was awarded one point, and no points were awarded for a loss, as there are no draws in volleyball. At the end of the competition, it turned out that the points scored by the teams formed... | Solution. Let the points scored by the teams form a non-decreasing arithmetic progression $a_{1}, a_{2}, \ldots, a_{n}$ with a common difference $d$. A total of $\frac{n(n-1)}{2}$ games were played. Therefore, the total number of points scored by all teams is $\frac{n(n-1)}{2}$, from which we have $a_{1}+a_{2}+\ldots+a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,810 |
4.085 In an angle containing $60^{\circ}$, five circles are inscribed such that each subsequent circle, starting from the second, touches the previous one; How many times greater is the sum of the areas of all five corresponding circles compared to the area of the smallest circle? | ## Solution.
$\angle O_{1} A B_{1}=30^{\circ} \Rightarrow A O_{1}=2 a, A O_{2}=2 b$. But $A O_{2}=b+a+A O_{1}=3 a+b \Rightarrow \frac{b}{a}=3$.
Therefore, if $r_{i}$ is the radius of the $... | 7381 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,811 |
7.296
$2 \log _{a}^{\frac{1}{2}} b \cdot\left(\left(\log _{a} \sqrt[4]{a b}+\log _{b} \sqrt[4]{a b}\right)^{\frac{1}{2}}-\left(\log _{a} \sqrt[4]{\frac{b}{a}}+\log _{b} \sqrt[4]{\frac{a}{b}}\right)^{\frac{1}{2}}\right), a, b>1$. | Solution.
1) $\left(\log _{a} \sqrt[4]{a b}+\log _{b} \sqrt[4]{a b}\right)^{\frac{1}{2}}=\left(\frac{1}{4}\left(\log _{a} b+1+\log _{b} a+1\right)\right)^{\frac{1}{2}}=$
$=\left(\frac{1}{4}\left(2+\log _{a} b+\frac{1}{\log _{a} b}\right)\right)^{\frac{1}{2}}=\frac{1}{2}\left(\frac{\left(\log _{a} b+1\right)^{2}}{\log... | 2, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,813 |
7.299 A=\frac{1-\log _{1 / a} \frac{1}{(a-b)^{2}}+\log _{a}^{2}(a-b)}{\left(1-\log _{\sqrt{a}}(a-b)+\log _{a}^{2}(a-b)\right)^{\frac{1}{2}}}
$$
7.299 A=\frac{1-\log _{1 / a} \frac{1}{(a-b)^{2}}+\log _{a}^{2}(a-b)}{\left(1-\log _{\sqrt{a}}(a-b)+\log _{a}^{2}(a-b)\right)^{\frac{1}{2}}}
$$ | Solution. $A=\frac{1-2 \log _{a}(a-b)+\log _{a}^{2}(a-b)}{\left(1-2 \log _{a}(a-b)+\log _{a}^{2}(a-b)\right)^{\frac{1}{2}}}=\frac{\left(1-\log _{a}(a-b)\right)^{2}}{\left|1-\log _{a}(a-b)\right|}$.
1) $\log _{a}(a-b)-1>0$, if $\left\{\begin{array}{l}a>1, \\ b1, \\ 01, \\ b 1 , } \\
{ 0 < b < a }
\end{array} \text { or... | \frac{(1-\log_{}(-b))^{2}}{|1-\log_{}(-b)|} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,816 |
### 7.300 Which number is greater: $\log _{135} 675$ or $\log _{45} 75$ ? | Solution. $\log _{135} 675=\frac{\log _{3} 675}{\log _{3} 135}=\frac{\log _{3}(9 \cdot 75)}{\log _{3}(3 \cdot 45)}=\frac{2+\log _{3} 75}{1+\log _{3} 45}=\frac{\left(2+\log _{3} 75\right) \log _{3} 45}{\left(1+\log _{3} 45\right) \log _{3} 45}$.
$
\log _{45} 75=\frac{\log _{3} 75}{\log _{3} 45}=\frac{\log _{3} 75 \cdot... | \log_{135}675>\log_{45}75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,817 |
### 7.301 The equation $4^{x}+10^{x}=25^{x}$ has a unique root.
Find it and compare it with 0 and 1. | ## Solution.
$$
\begin{aligned}
& 2^{2 x}+2^{x} \cdot 5^{x}=5^{2 x} \Leftrightarrow\left(\frac{2}{5}\right)^{2 x}+\left(\frac{2}{5}\right)^{x}-1=0, \Leftrightarrow\left(\frac{2}{5}\right)^{x}=\frac{-1+\sqrt{5}}{2} \Leftrightarrow \\
& \Leftrightarrow x=\log _{0.4} \frac{\sqrt{5}-1}{2} \text {. Since } \frac{2}{5}\log ... | \log_{0.4}\frac{\sqrt{5}-1}{2},0<x<1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,818 |
7.302 Prove that $\log _{3} 12=\log _{3} 7 \cdot \log _{7} 5 \cdot \log _{5} 4+1$. | Solution.
$$
\begin{aligned}
& \log _{3} 7 \cdot \log _{7} 5 \cdot \log _{5} 4+1=\log _{3} 7 \cdot \frac{\log _{3} 5}{\log _{3} 7} \cdot \frac{\log _{3} 4}{\log _{3} 5}+1=\log _{3} 4+1=\log _{3} 4+\log _{3} 3= \\
& =\log _{3} 12
\end{aligned}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,819 |
7.303 Expression
$T=\log _{m} A \cdot \log _{n} A+\log _{n} A \cdot \log _{p} A+\log _{p} A \cdot \log _{m} A$ represent in the form of a product. | Solution.
$$
\begin{aligned}
& T=\frac{1}{\log _{A} m \cdot \log _{A} n}+\frac{1}{\log _{A} n \cdot \log _{A} p}+\frac{1}{\log _{A} p \cdot \log _{A} m}=\frac{\log _{A} p+\log _{A} m+\log _{A} n}{\log _{A} m \cdot \log _{A} n \cdot \log _{A} p}= \\
& =\log _{A}(m \cdot n \cdot p) \cdot \log _{p} A \cdot \log _{n} A \c... | \log_{A}(\cdotn\cdotp)\cdot\log_{p}A\cdot\log_{n}A\cdot\log_{}A | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,820 |
7.304 Prove that
$$
\log _{3} 2 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{6} 5 \cdot \log _{7} 6 \cdot \log _{8} 7=\frac{1}{3}
$$ | Solution.
$$
\begin{aligned}
& \log _{3} 2 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{6} 5 \cdot \log _{7} 6 \cdot \log _{8} 7=\log _{3} 2 \cdot \frac{1}{\log _{3} 4} \cdot \frac{\log _{3} 4}{\log _{3} 5} \cdot \frac{\log _{3} 5}{\log _{3} 6} \times \\
& \times \frac{\log _{3} 6}{\log _{3} 7} \cdot \frac{\log _{... | \frac{1}{3} | Algebra | proof | Yes | Yes | olympiads | false | 51,821 |
7.305 Simplify the expression $A=\left(\left(\log _{b}^{4} a+\log _{a}^{4} b+2\right)^{\frac{1}{2}}-2\right)^{\frac{1}{2}}$ for $1<\boldsymbol{a}<\boldsymbol{b}$. | Solution.
$$
\begin{aligned}
& A=\left(\left(\log _{b}^{4} a+2 \log _{a}^{2} b \cdot \log _{b}^{2} a+\log _{a}^{4} b\right)^{\frac{1}{2}}-2\right)^{\frac{1}{2}}=\left(\log _{b}^{2} a+\log _{a}^{2} b-2\right)^{\frac{1}{2}}= \\
& =\left(\log _{b}^{2} a-2 \log _{a} b \cdot \log _{b} a+\log _{a}^{2} b\right)^{\frac{1}{2}}... | \log_{}b-\log_{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,822 |
7.307 For what values of a does the equation $2 \lg (x+3)=\lg (a x)$ have a unique solution? | Solution. The original equation is equivalent to the system:
$$
\left\{\begin{array} { c }
{ \operatorname { l g } ( x + 3 ) ^ { 2 } = \operatorname { l g } ( a x ) } \\
{ x + 3 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ ( x + 3 ) ^ { 2 } = a x , } \\
{ x + 3 > 0 }
\end{array} \Leftrightarrow \le... | \in(-\infty;0)\cup{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,823 |
$7.308 \quad(\sqrt[3]{0.5}+\sqrt[3]{4})^{x}=13.5$. | ## Solution.
$$
\left(\frac{1}{\sqrt[3]{2}}+\sqrt[3]{4}\right)^{x}=\frac{27}{2} \Leftrightarrow\left(\frac{3}{\sqrt[3]{2}}\right)^{x}=\frac{27}{2} \Leftrightarrow\left(\frac{27}{2}\right)^{\frac{x}{3}}=\left(\frac{27}{2}\right)^{1} \Leftrightarrow x=3
$$
Answer: $x=3$.
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,824 |
7.3092 \log _{9}^{2} x=\log _{3} x \cdot \log _{3}(\sqrt{2 x+1}-1) | Solution. $\frac{\log _{3}^{2} x}{2}=\log _{3} x \cdot \log _{3}(\sqrt{2 x+1}-1) \Leftrightarrow \log _{3} x \cdot\left(\log _{3} x-2 \log _{3}(\sqrt{2 x+1}-1)\right)=0$
Answer: $x_{1}=1, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,825 |
$7.3113 \log _{x} 4+2 \log _{4 x} 4+3 \log _{16 x} 4=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.3113 \log _{x} 4+2 \log _{4 x} 4+3 \log _{16 x} 4=0$. | Solution. The original equation is equivalent to the following:
$\frac{3}{\log _{4} x}+\frac{2}{\log _{4} x+1}+\frac{3}{\log _{4} x+2}=0$. Let $y=\log _{4} x$, then
$\frac{3}{y}+\frac{2}{y+1}+\frac{3}{y+2}=0 \Leftrightarrow\left\{\begin{array}{l}3(y+1)(y+2)+2 y(y+2)+3 y(y+1)=0, \\ y \neq 0, y \neq-1, y \neq-2\end{arr... | x_1=\frac{1}{8},x_2=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,826 |
$7.313 \log _{x+1}\left(x^{3}-9 x+8\right) \cdot \log _{x-1}(x+1)=3$. | ## Solution.
$\log _{x+1}\left((x-1)\left(x^{2}+x-8\right)\right) \cdot \log _{x-1}(x+1)=3 \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}\log _{x+1}\left((x-1)\left(x^{2}+x-8\right)\right)=3 \log _{x+1}(x-1), \\ x \neq 2\end{array} \Leftrightarrow\left\{\begin{array}{l}x-1>0, \\ x \neq 2, \\ (x-1)\left(x^{2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,827 |
$7.314 \frac{2-4 \log _{12} 2}{\log _{12}(x+2)}-1=\frac{\log _{6}(8-x)}{\log _{6}(x+2)}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$7.314 \frac{2-4 \log _{12} 2}{\log _{12}(x+2)}-1=\frac{\log _{6}(8-x)}{\log _{6}(x+2)}$. | ## Solution.
$\frac{\log _{12} 144-\log _{12} 16-\log _{12}(x+2)}{\log _{12}(x+2)}=\frac{\log _{6}(8-x)}{\log _{6}(x+2)} \Leftrightarrow \frac{\log _{12} \frac{9}{x+2}}{\log _{12}(x+2)}=\frac{\log _{6}(8-x)}{\log _{6}(x+2)} \Leftrightarrow$
$\Leftrightarrow \log _{x+2}\left(\frac{9}{x+2}\right)=\log _{x+2}(8-x) \Left... | x_{1,2}=1\\sqrt{1-0.5\lgp}for1<p\leq100,nosolutionsforothervaluesofp | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,828 |
$7.317 \log _{4} x+\log _{x} 2-\log _{4} \sqrt{x}=1$ | Solution. $\frac{1}{2} \log _{2} x+\frac{1}{\log _{2} x}-\frac{1}{4} \log _{2} x=1 \Leftrightarrow \log _{2}^{2} x-4 \log _{2} x+4=0 \Leftrightarrow$ $\left(\log _{2} x-2\right)^{2}=0 \Leftrightarrow x=4$.
Answer: $x=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,829 |
$7.320 m^{1+\log _{3} x}+m^{1-\log _{3} x}=m^{2}+1, m>0, m \neq 1$.
The equation above is to be solved for $x$, given that $m>0$ and $m \neq 1$. | Solution. $m \cdot m^{\log _{3} x}+\frac{m}{m^{\log _{3} x}}=m^{2}+1$.
Let $y=m^{\log _{3} x}$, then $m y^{2}-\left(m^{2}+1\right) y+m=0 \Leftrightarrow y=\frac{m^{2}+1 \pm\left(m^{2}-1\right)}{2 m} \Leftrightarrow$ $\Leftrightarrow\left[\begin{array}{l}y=m, \\ y=\frac{1}{m}\end{array} \Leftrightarrow\left[\begin{arra... | 2if\neq1;x\in(0;6)if1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,831 |
7.326 $|x-3|^{x^{2}-x}=(x-3)^{2}$. | ## Solution.
1) $|x-3|=1 \Leftrightarrow\left[\begin{array}{l}x=4 \\ x=2\end{array}\right.$
2) $\left\{\begin{array}{l}|x-3| \neq 0, \\ x^{2}-x-2=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq 3, \\ {\left[\begin{array}{l}x=2, \\ x=-1\end{array}\right.}\end{array} \Leftrightarrow\left[\begin{array}{l}x=2, \... | x_{1}=-1,x_{2}=2,x_{3}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,832 |
$7.327 \log _{\sqrt{x}}(x+12)=8 \log _{x+12} x, x$-integer. | ## Solution.
$\log _{x}(x+12)=\frac{4}{\log _{x}(x+12)} \Leftrightarrow \log _{x}(x+12)= \pm 2 \Leftrightarrow\left\{\begin{array}{l}x \geq 2, x-\text { integer, } \\ {\left[\begin{array}{l}x+12=x^{2} \\ x+12=\frac{1}{x^{2}}\end{array}\right.}\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}x... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,833 |
$7.329 .\left|\log _{2}(3 x-1)-\log _{2} 3\right|=\left|\log _{2}(5-2 x)-1\right|$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$7.329 .\left|\log _{2}(3 x-1)-\log _{2} 3\right|=\left|\log _{2}(5-2 x)-1\right|$. | Solution. The original equation is equivalent to the following:
$\left[\begin{array}{l}\log _{2}(3 x-1)-\log _{2} 3=\log _{2}(5-2 x)-1, \\ \log _{2}(3 x-1)-\log _{2} 3=1-\log _{2}(5-2 x)\end{array} \Leftrightarrow\left[\begin{array}{l}\log _{2} \frac{3 x-1}{3}=\log _{2} \frac{5-2 x}{2}, \\ \log _{2} \frac{3 x-1}{3}=\l... | x_{1}=1,x_{2}=\frac{17}{12},x_{3}=\frac{11}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,834 |
$7.331 \frac{\log _{3} x-1}{\log _{3} \frac{x}{3}}-2 \log _{3} \sqrt{x}+\log _{3}^{2} x=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$7.331 \frac{\log _{3} x-1}{\log _{3} \frac{x}{3}}-2 \log _{3} \sqrt{x}+\log _{3}^{2} x=3$. | Solution.
$\frac{\log _{3} x-1}{\log _{3} x-1}-\log _{3} x+\log _{3}^{2}=3 \Leftrightarrow\left\{\begin{array}{l}x \neq 3, \\ \log _{3}^{2} x-\log _{3} x-2=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq 3, \\ {\left[\begin{array}{l}\log _{3} x=2, \\ \log _{3} x=-1\end{array}\right.}\end{array} \Leftrightarr... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,835 | |
$7.333 \sqrt{\log _{2}\left(2 x^{2}\right) \cdot \log _{4}(16 x)}=\log _{4} x^{3}$. | ## Solution.
$$
\text { So, } \sqrt{\left(1+2 \log _{2} x\right)\left(2+\frac{1}{2} \log _{2} x\right)}=\frac{3}{2} \log _{2} x
$$
Let $y=\log _{2} x$, then $\sqrt{(2+4 y)(4+y)}=3 y \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}y \geq 0, \\ (2+4 y)(y+4)=9 y^{2}\end{array} \Leftrightarrow\left\{\begin{array}... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,837 |
7.337
$$
\left\{\begin{array}{l}
x y=a^{2} \\
\lg ^{2} x+\lg ^{2} y=2.5 \lg ^{2} a^{2},
\end{array} a<0\right.
$$ | Solution. Taking the logarithm of the first equation:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \operatorname { l g } x + \operatorname { l g } y = 2 \operatorname { l g } ( - a ) } \\
{ \operatorname { l g } ^ { 2 } x + ( 2 \operatorname { l g } ( - a ) - \operatorname { l g } x ) ^ { 2 } = 1 0 \operatorname... | (0;0),(8;-8),(-4;-2),(3,\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,838 |
### 7.340
$$
\left\{\begin{array}{l}
x^{\log _{3} y}+2 y^{\log _{3} x}=27 \\
\log _{3} y-\log _{3} x=1
\end{array}\right.
$$ | ## Solution.
$$
\left\{\begin{array} { l }
{ x ^ { \log _ { 3 } y } + 2 x ^ { \log _ { 3 } y } = 27 , } \\
{ \log _ { 3 } y - \log _ { 3 } x = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{\log _{3} y}=9 \\
\log _{3} y=\log _{3} x+1
\end{array}\right.\right.
$$
Logarithmize the first equation:
$$
\begi... | (3;9),(\frac{1}{9};\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,839 |
$3.396 \frac{3-4 \cos 2 \alpha+\cos 4 \alpha}{3+4 \cos 2 \alpha+\cos 4 \alpha}=\tan^{4} \alpha$. | Solution. $L=\frac{2-4 \cos 2 \alpha+(1+\cos 4 \alpha)}{2+4 \cos 2 \alpha+(1+\cos 4 \alpha)}=\frac{1-2 \cos 2 \alpha+\cos ^{2} 2 \alpha}{1+2 \cos 2 \alpha+\cos ^{2} 2 \alpha}=\left(\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}\right)^{2}=$ $=\left(\tan^{2} \alpha\right)^{2}=\tan^{4} \alpha$. | \tan^{4}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,840 |
$3.3998 \cos ^{4} \alpha-4 \cos ^{3} \alpha-8 \cos ^{2} \alpha+3 \cos \alpha+1=-2 \sin \frac{7 \alpha}{2} \sin \frac{\alpha}{2}$. | ## Solution.
$L=8 \cos ^{2} \alpha \cdot\left(\cos ^{2} \alpha-1\right)-4 \cos \alpha \cdot\left(\cos ^{2} \alpha-1\right)-\cos \alpha+1=-8 \cos ^{2} \alpha \cdot \sin ^{2} \alpha+$ $+4 \cos \alpha \cdot \sin ^{2} \alpha-\cos \alpha+1=-2 \sin ^{2} 2 \alpha+2 \sin \alpha \cdot \sin 2 \alpha-\cos \alpha+1=$ $=-1+\cos 4 ... | -2\sin\frac{7\alpha}{2}\cdot\sin\frac{\alpha}{2} | Algebra | proof | Yes | Yes | olympiads | false | 51,842 |
$3.400 \cos (\alpha+\beta) \cdot \cos \gamma+\cos \alpha+\cos \beta+\cos \gamma-\sin (\alpha+\beta) \cdot \sin \gamma=$
$=4 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha+\gamma}{2} \cos \frac{\beta+\gamma}{2}$. | ## Solution.
$L=(\cos (\alpha+\beta) \cdot \cos \gamma-\sin (\alpha+\beta) \cdot \sin \gamma)+\cos \gamma+(\cos \alpha+\cos \beta)=$
$=\cos (\alpha+\beta+\gamma)+\cos \gamma+2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=2 \cos \frac{\alpha+\beta}{2} \cdot\left(\cos \left(\frac{\alpha+\beta}{2}+\gamm... | 4\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha+\gamma}{2}\cos\frac{\beta+\gamma}{2} | Algebra | proof | Yes | Yes | olympiads | false | 51,843 |
### 3.401
$$
\cos \left(\frac{5}{2} \pi-6 \alpha\right) \cdot \sin ^{3}(\pi-2 \alpha)-\cos (6 \alpha-\pi) \cdot \sin ^{3}\left(\frac{\pi}{2}-2 \alpha\right)=\cos ^{3} 4 \alpha
$$ | ## Solution.
$L=\sin 6 \alpha \cdot \sin ^{3} 2 \alpha+\cos 6 \alpha \cdot \cos ^{3} 2 \alpha=\sin 6 \alpha \cdot \sin 2 \alpha \cdot \frac{1}{2}(1-\cos 4 \alpha)+$ $+\cos 6 \alpha \cdot \cos 2 \alpha \cdot \frac{1}{2}(1+\cos 4 \alpha)=\frac{1}{2}[(\cos 6 \alpha \cdot \cos 2 \alpha+\sin 6 \alpha \cdot \sin 2 \alpha)+$... | \cos^{3}4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,844 |
$3.4028 \cos ^{4} \alpha+4 \cos ^{3} \alpha-8 \cos ^{2} \alpha-3 \cos \alpha+1=2 \cos \frac{7 \alpha}{2} \cdot \cos \frac{\alpha}{2}$. | ## Solution.
$L=8 \cos ^{2} \alpha \cdot\left(\cos ^{2} \alpha-1\right)+4 \cos \alpha\left(\cos ^{2} \alpha-1\right)+\cos \alpha+1=-8 \sin ^{2} \alpha \cdot \cos ^{2} \alpha-$
$-4 \sin ^{2} \alpha \cdot \cos \alpha+\cos \alpha+1=-2 \sin ^{2} 2 \alpha-2 \sin 2 \alpha \cdot \sin \alpha+\cos \alpha+1=$
$=-1+\cos 4 \alp... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,845 |
$3.405 \frac{\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)}{\cos \alpha+\cos \beta+\cos \gamma+\cos (\alpha+\beta+\gamma)}=\tan \frac{\alpha+\beta}{2} \cdot \tan \frac{\beta+\gamma}{2} \cdot \tan \frac{\gamma+\alpha}{2}$. | Solution.
$$
\begin{aligned}
& L=\frac{2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}-2 \sin \frac{\alpha+\beta}{2} \cdot \cos \left(\frac{\alpha+\beta}{2}+\gamma\right)}{2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}+2 \cos \frac{\alpha+\beta}{2} \cdot \cos \left(\frac{\alpha+\beta}{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,847 |
$3.406 \operatorname{ctg} \alpha-\operatorname{tg} \alpha-2 \operatorname{tg} 2 \alpha-\ldots-2^{n} \operatorname{tg} 2^{n} \alpha=2^{n+1} \operatorname{ctg} 2^{n+1} \alpha$,
## - natural. | Solution. Note that $2^{k} \operatorname{ctg} 2^{k} \alpha-2^{k} \operatorname{tg} 2^{k} \alpha=2^{k}\left(\frac{\cos 2^{k} \alpha}{\sin 2^{k} \alpha}-\frac{\sin 2^{k} \alpha}{\cos 2^{k} \alpha}\right)=$ $=2^{k}\left(\frac{\cos ^{2} 2^{k} \alpha-\sin ^{2} 2^{k} \alpha}{\sin 2^{k} \alpha \cdot \cos 2^{k} \alpha}\right)=... | 2^{n+1}\operatorname{ctg}2^{n+1}\alpha | Algebra | proof | Yes | Yes | olympiads | false | 51,848 |
$3.407 \cos \alpha+\cos 3 \alpha+\ldots+\cos (2 n-1) \alpha=\frac{\sin 2 n \alpha}{2 \sin \alpha}$. | Solution. $2 \sin \alpha \cdot L=2 \sin \alpha \cdot \cos \alpha+2 \sin \alpha \cdot \cos 3 \alpha+\ldots+2 \sin \alpha \cdot \cos (2 n-1) \alpha=$ $=\sin 2 \alpha+(-\sin 2 \alpha+\sin 4 \alpha)+(-\sin 4 \alpha+\sin 6 \alpha)+\ldots+(-\sin (2 n-2) \alpha+\sin 2 n \alpha)=$ $=\sin 2 n \alpha=2 \sin \alpha \cdot R$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,849 |
$3.408 \cos ^{2} \alpha+\cos ^{2} 2 \alpha+\ldots+\cos ^{2} n \alpha=\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha}+\frac{n}{2}$.
$3.408 \cos ^{2} \alpha+\cos ^{2} 2 \alpha+\ldots+\cos ^{2} n \alpha=\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha}+\frac{n}{2}$.
(Note: The equation provided i... | Solution. $4 \sin \alpha \cdot L=4 \sin \alpha \cdot\left(\frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 4 \alpha}{2}+\ldots+\frac{1+\cos 2 n \alpha}{2}\right)=$
$=2 \sin \alpha \cdot(n+\cos 2 \alpha+\cos 4 \alpha+\ldots+\cos 2 n \alpha)=$
$=2 n \cdot \sin \alpha+2 \sin \alpha \cdot \cos 2 \alpha+2 \sin \alpha \cdot \cos 4 \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,850 |
$3.409 \sin ^{2} \alpha+\sin ^{2} 2 \alpha+\ldots+\sin ^{2} n \alpha=\frac{n}{2}-\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha}$.
$3.409 \sin ^{2} \alpha+\sin ^{2} 2 \alpha+\ldots+\sin ^{2} n \alpha=\frac{n}{2}-\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha}$. | Solution. $n-L=\left(1-\sin ^{2} \alpha\right)+\ldots+\left(1-\sin ^{2} n \alpha\right)=\cos ^{2} \alpha+\ldots+\cos ^{2} n \alpha=$ (see problem 3.408) $=\frac{n}{2}+\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha} \Rightarrow L=\frac{n}{2}-\frac{\cos (n+1) \alpha \cdot \sin n \alpha}{2 \sin \alpha}$.
## S... | \frac{n}{2}-\frac{\cos(n+1)\alpha\cdot\sinn\alpha}{2\sin\alpha} | Algebra | proof | Yes | Yes | olympiads | false | 51,851 |
$3.410 \sin ^{3} 2 \alpha \cdot \cos 6 \alpha+\cos ^{3} 2 \alpha \cdot \sin 6 \alpha$. | ## Solution.
$(\sin 6 \alpha \cdot \cos 2 \alpha) \cdot \cos ^{2} 2 \alpha+(\cos 6 \alpha \cdot \sin 2 \alpha) \cdot \sin ^{2} 2 \alpha=\sin 6 \alpha \cdot \cos 2 \alpha \cdot \frac{1+\cos 4 \alpha}{2}+$ $+\cos 6 \alpha \cdot \sin 2 \alpha \cdot \frac{1-\cos 4 \alpha}{2}=\frac{1}{2}[(\sin 6 \alpha \cdot \cos 2 \alpha+... | \frac{3}{4}\sin8\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,852 |
$3.4113 \sin \alpha \cdot \cos 3 \alpha+9 \sin \alpha \cdot \cos \alpha-\sin 3 \alpha \cdot \cos 3 \alpha-3 \sin 3 \alpha \cdot \cos \alpha$. | Solution. $3(\sin \alpha \cdot \cos 3 \alpha-\cos \alpha \cdot \sin 3 \alpha)+\frac{9}{2} \sin 2 \alpha-\frac{\sin 6 \alpha}{2}=$ $=-3 \sin 2 \alpha+\frac{9}{2} \sin 2 \alpha-\frac{3 \sin 2 \alpha-4 \sin ^{3} 2 \alpha}{2}=2 \sin ^{3} 2 \alpha$.
$3.4124\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)-4\left(\sin ^{6} \al... | -\cos^{2}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,853 |
$3.416 \cos \frac{\pi}{33} \cdot \cos \frac{2 \pi}{33} \cdot \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cdot \cos \frac{16 \pi}{33}=\frac{1}{32}$. | Solution.
$32 \sin \frac{\pi}{33} \cdot L=16 \sin \frac{2 \pi}{33} \cdot \cos \frac{2 \pi}{33} \cdot \cos \frac{4 \pi}{33} \cdot \cos \frac{8 \pi}{33} \cdot \cos \frac{16 \pi}{33}=8 \sin \frac{4 \pi}{33} \cdot \cos \frac{4 \pi}{33} \times$ $\times \cos \frac{8 \pi}{33} \cdot \cos \frac{16 \pi}{33}=4 \sin \frac{8 \pi}{... | \frac{1}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,856 |
$3.4173 \sin \frac{2 \pi}{17}+\sin \frac{4 \pi}{17}-\sin \frac{6 \pi}{17}-\frac{1}{2} \sin \frac{8 \pi}{17}=8 \sin ^{3} \frac{2 \pi}{17} \cdot \cos ^{2} \frac{\pi}{17}$. | Solution.
$L=3 \sin \frac{2 \pi}{17}+\sin \frac{4 \pi}{17}-\left(3 \sin \frac{2 \pi}{17}-4 \sin ^{3} \frac{2 \pi}{17}\right)-\sin \frac{4 \pi}{17} \cdot \cos \frac{4 \pi}{17}=4 \sin ^{3} \frac{2 \pi}{17}+$
$+\sin \frac{4 \pi}{17} \cdot\left(1-\cos \frac{4 \pi}{17}\right)=4 \sin ^{3} \frac{2 \pi}{17}+2 \sin \frac{4 \p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,857 |
$3.418 \cos \frac{2 \pi}{31} \cdot \cos \frac{4 \pi}{31} \cdot \cos \frac{8 \pi}{31} \cdot \cos \frac{16 \pi}{31} \cdot \cos \frac{32 \pi}{31}=\frac{1}{32}$.
| ## Решение.
$L \cdot 32 \sin \frac{2 \pi}{31}=16 \sin \frac{4 \pi}{31} \cdot \cos \frac{4 \pi}{31} \cdot \cos \frac{8 \pi}{31} \cdot \cos \frac{16 \pi}{31} \cdot \cos \frac{32 \pi}{31}=8 \sin \frac{8 \pi}{31} \cdot \cos \frac{8 \pi}{31} \times$ $\times \cos \frac{16 \pi}{31} \cdot \cos \frac{32 \pi}{31}=4 \sin \frac{1... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,858 | |
$3.420 \sin 10^{\circ} \cdot \sin 20^{\circ} \cdot \sin 30^{\circ} \cdot \sin 40^{\circ} \cdot \sin 50^{\circ} \cdot \sin 60^{\circ} \cdot \sin 70^{\circ} \cdot \sin 80^{\circ}=\frac{3}{256} \cdot$ | ## Solution.
$L=\frac{\sqrt{3}}{4} \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot\left(\sin 10^{\circ} \cdot \cos 10^{\circ} \cdot \cos 20^{\circ} \cdot \cos 40^{\circ}\right)=\frac{\sqrt{3}}{32} \sin 20^{\circ} \cdot \sin 40^{\circ} \times$ $\times \sin 80^{\circ}=\frac{\sqrt{3}}{64} \sin 20^{\circ} \cdot\left(\cos 40^{... | \frac{3}{256} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,859 |
$3.421 \cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \ldots \cdot \cos \frac{14 \pi}{15}=-\frac{1}{2^{14}}$. | ## Solution.
$$
L=-\cos ^{2} \frac{\pi}{15} \cdot \cos ^{2} \frac{2 \pi}{15} \cdot \cos ^{2} \frac{3 \pi}{15} \cdot \cos ^{2} \frac{4 \pi}{15} \cdot \cos ^{2} \frac{5 \pi}{15} \cdot \cos ^{2} \frac{6 \pi}{15} \cdot \cos ^{2} \frac{7 \pi}{15}=-\frac{1}{2^{14}}
$$
The last part follows from problem 3.422. | -\frac{1}{2^{14}} | Algebra | proof | Yes | Yes | olympiads | false | 51,860 |
$3.422 \cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{5 \pi}{15} \cdot \cos \frac{6 \pi}{15} \cdot \cos \frac{7 \pi}{15}=\frac{1}{2^{7}} \cdot$
| ## Решение.
$$
2^{7} \sin \frac{\pi}{15} \cdot \sin \frac{3 \pi}{15} \cdot L=2^{7}\left(\sin \frac{\pi}{15} \cdot \cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15}\right) \times
$$
$\times\left(\sin \frac{3 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{6 \pi}{15}\right) \cdot \cos \f... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,861 |
$3.423 \sin 10^{\circ}+\sin 20^{\circ}+\sin 30^{\circ}+\sin 40^{\circ}+\sin 50^{\circ}=\frac{\sin 25^{\circ}}{2 \sin 5^{\circ}}$. | ## Solution.
$L \cdot 2 \sin 5^{\circ}=2 \sin 5^{\circ} \cdot\left(\cos 80^{\circ}+\cos 70^{\circ}+\cos 60^{\circ}+\cos 50^{\circ}+\cos 40^{\circ}\right)=$ $=\left(-\sin 75^{\circ}+\sin 85^{\circ}\right)+\left(-\sin 65^{\circ}+\sin 75^{\circ}\right)+\left(-\sin 55^{\circ}+\sin 65^{\circ}\right)+\left(-\sin 45^{\circ}+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,862 |
$3.427 \cos 50^{\circ}+8 \cos 200^{\circ} \cdot \cos 220^{\circ} \cdot \cos 80^{\circ}=2 \sin ^{2} 65^{\circ}$. | Solution. $\cos 50^{\circ}+8 \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ}=1-\cos 130^{\circ} \Leftrightarrow$
$\Leftrightarrow 8 \sin 20^{\circ} \cdot \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ}=\sin 20^{\circ} \Leftrightarrow 4 \sin 40^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ}... | \sin160=\sin20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,863 |
$3.428 \sin 18^{\circ} \cdot \sin 54^{\circ}=\frac{1}{4}$.
| ## Решение.
$\cos 72^{\circ} \cdot \cos 36^{\circ}=\frac{1}{4} \Leftrightarrow 4 \cos 72^{\circ} \cdot \cos 36^{\circ} \cdot \sin 36^{\circ}=\sin 36^{\circ} \Leftrightarrow$
$\Leftrightarrow 2 \sin 72^{\circ} \cdot \cos 72^{\circ}=\sin 36^{\circ} \Leftrightarrow \sin 144^{\circ}=\sin 36^{\circ}$.
| proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,864 |
$3.429 \sin ^{2}\left(\operatorname{arctan} 3-\operatorname{arccot}\left(-\frac{1}{2}\right)\right)=\frac{1}{2}$. | Solution. Let $\alpha=\operatorname{arctg} 3, \beta=\operatorname{arcctg}\left(-\frac{1}{2}\right)$.
Then $\operatorname{tg} \alpha=3, 0<\alpha<\frac{\pi}{2}, \operatorname{ctg} \beta=-\frac{1}{2}, \frac{\pi}{2}<\beta<\pi \Rightarrow$
$\Rightarrow \cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\... | \sin^{2}(\alpha-\beta)=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,865 |
$3.430 \sin ^{2}\left(\operatorname{arcctg} \frac{1}{2}-\operatorname{arcctg}\left(-\frac{1}{3}\right)\right)=\frac{1}{2}$. | Solution. $\operatorname{arcctg} \frac{1}{2}=\pi-\operatorname{arcctg}\left(-\frac{1}{2}\right), \operatorname{arcctg}\left(-\frac{1}{3}\right)=\pi-\operatorname{arctg} 3 \Rightarrow$
$\Rightarrow \operatorname{arcctg} \frac{1}{2}-\operatorname{arcctg}\left(-\frac{1}{3}\right)=\operatorname{arctg} 3-\operatorname{arcc... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,866 |
$3.431 \sin \left(2 \operatorname{arctg} \frac{1}{2}\right)+\operatorname{tg}\left(\frac{1}{2} \arcsin \frac{15}{17}\right)=\frac{7}{5}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$3.431 \sin \left(2 \operatorname{arctan} \f... | ## Solution.
$$
\begin{aligned}
& \text { Let } \alpha=\operatorname{arctg} \frac{1}{2}, \beta=\arcsin \frac{15}{17}, \text { then } \operatorname{tg} \alpha=\frac{1}{2}, 0<\alpha<\frac{\pi}{2}, \\
& \sin \beta=\frac{15}{17}, \cos \beta=\frac{8}{17} \Rightarrow \sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operat... | \frac{7}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,867 |
$3.433 \cos (2 \operatorname{arctg} 2)-\sin (4 \operatorname{arctg} 3)=\frac{9}{25}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$3.433 \cos (2 \operatorname{arctan} 2)-\sin (4 \operatorname{arctan} 3)=\frac{9}{25}$. | ## Solution.
Let $\alpha=\operatorname{arctg} 2, \beta=\operatorname{arctg} 3$, then $\operatorname{tg} \alpha=2, \operatorname{tg} \beta=3 \Rightarrow$ $\Rightarrow \cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=-\frac{3}{5}$, $\sin 4 \beta=2 \sin 2 \beta \cdot \cos 2 \beta=\frac{... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,868 | |
$3.434 \arccos \frac{36}{85}-\arccos \frac{15}{17}=\frac{\pi}{2}-\arcsin \frac{4}{5}$. | ## Solution.
$\frac{36}{85}\arccos \frac{15}{17} \Rightarrow 0<L<\frac{\pi}{2}, \quad 0<R<\frac{\pi}{2}$. Therefore, from the equality of cosines of the left $L$ and right $R$ parts, it will follow that $L=R$.
$\cos \left(\arccos \frac{36}{85}-\arccos \frac{15}{17}\right)=\cos \left(\frac{\pi}{2}-\arcsin \frac{4}{5}\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 51,869 |
$3.435 \frac{\pi}{2}+\arccos \frac{36}{85}=\arccos \frac{15}{17}+\arccos \left(-\frac{3}{5}\right)$ | ## Solution.
$\arccos \left(-\frac{3}{5}\right)=\pi-\arccos \frac{3}{5}=\pi-\arcsin \frac{4}{5}$. Therefore, the original identity can be written as: $\arccos \frac{36}{85}-\arccos \frac{15}{17}=\frac{\pi}{2}-\arcsin \frac{4}{5}$.
The last expression matches problem 3.434.
## $3.436 \cos (2 \operatorname{arcctg} 7)=... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,870 |
$3.437 \cos \frac{11 \pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}$.
| ## Решение.
$$
\begin{aligned}
& 2 \cos \frac{\pi}{10} \cdot L=2 \cos \frac{\pi}{10} \cdot \cos \frac{\pi}{5}-2 \cos \frac{\pi}{10} \cdot \cos \frac{2 \pi}{5}= \\
& =\left(\cos \frac{\pi}{10}+\cos \frac{3 \pi}{10}\right)-\left(\cos \frac{3 \pi}{10}+\cos \frac{\pi}{2}\right)=\cos \frac{\pi}{10}=2 \cos \frac{\pi}{10} \c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,871 |
$3.438 \sin 84^{\circ} \cdot \sin 24^{\circ} \cdot \sin 48^{\circ} \cdot \sin 12^{\circ}=\frac{1}{16}$. | ## Solution.
$\left(\sin 84^{\circ} \cdot \sin 24^{\circ}\right) \cdot\left(\sin 48^{\circ} \cdot \sin 12^{\circ}\right)=\frac{1}{4}\left(\frac{1}{2}-\cos 108^{\circ}\right) \cdot\left(\cos 36^{\circ}-\frac{1}{2}\right)=$
$=\frac{1}{8}\left(\cos 36^{\circ}-\cos 72^{\circ}-2 \cos 108^{\circ} \cdot \cos 36^{\circ}-\fra... | \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,872 |
$3.439 \operatorname{tg} 830^{\circ}+\operatorname{tg} 770^{\circ}+\operatorname{tg} 740^{\circ}=\operatorname{tg} 470^{\circ} \cdot \operatorname{tg} 410^{\circ} \cdot \operatorname{tg} 380^{\circ}$. | ## Solution.
$\operatorname{tg} 470^{\circ}+\operatorname{tg} 410^{\circ}=\operatorname{tg} 470^{\circ} \cdot \operatorname{tg} 410^{\circ} \cdot \operatorname{tg} 380^{\circ}-\operatorname{tg} 380^{\circ} \Leftrightarrow$
$\Leftrightarrow \frac{\operatorname{tg} 470^{\circ}+\operatorname{tg} 410^{\circ}}{1-\operator... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,873 |
3.441 A=\operatorname{ctg}\left(\frac{11 \pi}{4}+\frac{1}{2} \arccos \frac{2 b}{a}\right)+\operatorname{ctg}\left(\frac{11 \pi}{4}-\frac{1}{2} \arccos \frac{2 b}{a}\right)
$$
3.441 A=\operatorname{ctg}\left(\frac{11 \pi}{4}+\frac{1}{2} \arccos \frac{2 b}{a}\right)+\operatorname{ctg}\left(\frac{11 \pi}{4}-\frac{1}{2} ... | Solution.
Let $\alpha=\arccos \frac{2b}{a}$, then $\cos \alpha=\frac{2b}{a}$,
$$
\begin{aligned}
& A=-\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)-\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=-\frac{1+\tan \alpha / 2}{1-\tan \alpha / 2}-\frac{1-\tan \alpha / 2}{1+\tan \alpha / 2}=-2 \frac{1+\tan^{2} \alpha / 2}... | -\frac{}{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,874 |
$3.442 A=\operatorname{tg}\left(\frac{7 \pi}{4}+\frac{1}{2} \arccos \frac{2 a}{b}\right)+\operatorname{tg}\left(\frac{7 \pi}{4}-\frac{1}{2} \arccos \frac{2 a}{b}\right)$
$3.442 A=\tan\left(\frac{7 \pi}{4}+\frac{1}{2} \arccos \frac{2 a}{b}\right)+\tan\left(\frac{7 \pi}{4}-\frac{1}{2} \arccos \frac{2 a}{b}\right)$ | ## Solution.
$$
\text { Let } \alpha=\arccos \frac{2 a}{b}, \text { then } \mathrm{A}=-\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)-\operatorname{tg}\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)
$$
Further, problem 3441.
Answer: $A=-\frac{b}{a}$. | -\frac{b}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,875 |
$3.443 A=\operatorname{ctg} \frac{5 \pi}{4}-2 \sin ^{2}\left(\frac{5 \pi}{2}+\frac{1}{2} \arcsin \frac{2 \sqrt{2}-1}{3}\right)$.
$3.443 A=\cot \frac{5 \pi}{4}-2 \sin ^{2}\left(\frac{5 \pi}{2}+\frac{1}{2} \arcsin \frac{2 \sqrt{2}-1}{3}\right)$. | ## Solution.
Let $\alpha=\arcsin \frac{2 \sqrt{2}-1}{3}$, then $0<\alpha<\frac{\pi}{2}, \sin \alpha=\frac{2 \sqrt{2}-1}{3}$, $\cos \alpha=\frac{2 \sqrt[4]{2}}{3} \Rightarrow A=1-2 \sin ^{2}\left(\frac{5 \pi}{2}+\frac{\alpha}{2}\right)=\cos (5 \pi+\alpha)=-\cos \alpha=-\frac{2 \sqrt[4]{2}}{3}$.
Answer: $A=-\frac{2 \sq... | -\frac{2\sqrt[4]{2}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,876 |
$3.444 A=\cos ^{6}\left(\frac{3 \pi}{2}-\frac{1}{2} \arcsin \frac{3}{5}\right)-\cos ^{6}\left(\frac{5 \pi}{2}+\frac{1}{2} \arcsin \frac{4}{5}\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$3.444 A=\cos ^{6}\left(\... | ## Solution.
Let $\alpha=\arcsin \frac{3}{5}, \beta=\arcsin \frac{4}{5}$, then $\sin \alpha=\cos \beta=\frac{3}{5} \Rightarrow$ $\Rightarrow A=\sin ^{6} \frac{\alpha}{2}-\sin ^{6} \frac{\beta}{2}=\left(\frac{1-\cos \alpha}{2}\right)^{3}-\left(\frac{1-\cos \beta}{2}\right)^{3}=-0.007$.
Answer: $A=-0.007$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,877 | |
3.445
$$
A=\frac{1}{4}-\cos ^{4}\left(\frac{5 \pi}{2}+\frac{1}{2} \arccos \frac{4}{5}\right)
$$ | Solution. Let $\alpha=\arccos \frac{4}{5}$, then $\cos \alpha=\frac{4}{5} \Rightarrow$
$\Rightarrow A=\frac{1}{4}-\sin ^{4} \frac{\alpha}{2}=\frac{1}{4}-\left(\frac{1-\cos \alpha}{2}\right)^{2}=\frac{6}{25}$.
Answer: $A=\frac{6}{25}$. | \frac{6}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,878 |
$3.448 A=\arcsin (\cos (2 \operatorname{arcctg}(\sqrt{2}-1)))$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$3.448 A=\arcsin (\cos (2 \operatorname{arcctg}(\sqrt{2}-1)))$. | ## Solution.
Let $\alpha=\operatorname{arcctg}(\sqrt{2}-1)$, then $\operatorname{tg} \alpha=\sqrt{2}+1$, $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=-\frac{1}{\sqrt{2}} \Rightarrow A=\arcsin \left(-\frac{1}{\sqrt{2}}\right)=-\frac{\pi}{4}$.
Answer: $\quad A=-\frac{\pi}{4}$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,880 | |
3.449 $A=\operatorname{tg}\left(\arccos \frac{1}{\sqrt{1+a^{2}}}+\arccos \frac{a}{\sqrt{1+a^{2}}}\right) a<0$.
3.449 $A=\tan\left(\arccos \frac{1}{\sqrt{1+a^{2}}}+\arccos \frac{a}{\sqrt{1+a^{2}}}\right) a<0$. | Solution.
$$
\begin{aligned}
& \text { Let } \alpha=\arccos \frac{1}{\sqrt{1+a^{2}}}, \beta=\arccos \frac{a}{\sqrt{1+a^{2}}}, \text { then } \cos \alpha=\frac{1}{\sqrt{1+a^{2}}} \\
& 0<\alpha<\frac{\pi}{2}, \quad \tan \alpha=\sqrt{\frac{1}{\cos ^{2} \beta}-1}=|a|=-a, \cos \beta=\frac{a}{\sqrt{1+a^{2}}}, \frac{\pi}{2}<... | \frac{1-^{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,881 |
$3.450 A=\cos ^{6}\left(\frac{5 \pi}{2}+\frac{1}{2} \arcsin \frac{3}{5}\right)+\cos ^{6}\left(\frac{7 \pi}{2}-\frac{1}{2} \arcsin \frac{4}{5}\right)$
$3.450 A=\cos ^{6}\left(\frac{5 \pi}{2}+\frac{1}{2} \arcsin \frac{3}{5}\right)+\cos ^{6}\left(\frac{7 \pi}{2}-\frac{1}{2} \arcsin \frac{4}{5}\right)$
The text above is ... | ## Solution.
Let $\alpha=\arcsin \frac{3}{5}, \beta=\arcsin \frac{4}{5}$, then $\cos \alpha=\frac{4}{5}, \cos \beta=\frac{3}{5} \Rightarrow$ $\Rightarrow A=\sin ^{6} \frac{\alpha}{2}+\sin ^{6} \frac{\beta}{2}=\left(\frac{1-\cos \alpha}{2}\right)^{3}+\left(\frac{1-\cos \beta}{2}\right)^{3}=0.009$.
Answer: $A=0.009$.
... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,882 |
$3.452 A=\operatorname{tg}\left(\frac{3 \pi}{4}-\frac{1}{4} \arcsin \left(-\frac{4}{5}\right)\right)$
$3.452 A=\tan\left(\frac{3 \pi}{4}-\frac{1}{4} \arcsin \left(-\frac{4}{5}\right)\right)$ | Solution. Let $\alpha=\arcsin \frac{4}{5}$, then $A=\operatorname{tg}\left(\frac{3 \pi}{4}+\frac{\alpha}{4}\right)=\frac{-\sin \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)}{1+\cos \left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)}=$ $=-\frac{\cos \frac{\alpha}{2}}{1+\sin \frac{\alpha}{2}}$. Since $\sin \alpha=\frac{4}{5},... | \frac{1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,883 |
3.453
$$
A=\operatorname{ctg}\left(\frac{5 \pi}{4}+\frac{1}{4} \arccos \left(-\frac{4}{5}\right)\right)
$$ | ## Solution.
Let $\quad \alpha=\arccos \frac{4}{5}, \quad A=\operatorname{ctg}\left(\frac{5 \pi}{4}+\frac{\pi-\alpha}{4}\right)=\operatorname{tg} \frac{\alpha}{4}=\frac{\sin \alpha / 2}{1+\cos \alpha / 2}$.
Since $\cos \alpha=\frac{4}{5}$, then $\sin \frac{\alpha}{2}=\frac{1}{\sqrt{10}}, \cos \frac{\alpha}{2}=\frac{3... | \sqrt{10}-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,884 |
$3.454 A=\sin ^{2}\left(\operatorname{arcctg} \frac{1}{2}-\operatorname{arctg}\left(-\frac{1}{3}\right)\right)$ | ## Solution.
Let $\alpha=\operatorname{arcctg} \frac{1}{2}, \beta=\operatorname{arctg} \frac{1}{3} \Rightarrow \operatorname{ctg} \alpha=\frac{1}{2}, \operatorname{tg} \beta=\frac{1}{3}, 0<\alpha<90^{\circ}$,
$0<\beta<90^{\circ} \Rightarrow \sin \alpha=\frac{1}{\sqrt{1+\operatorname{ctg}^{2} \alpha}}=\frac{2}{\sqrt{5... | 0.98 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,885 |
$3.455 A=\operatorname{tg}\left(2 \arccos \frac{5}{\sqrt{26}}-\arcsin \frac{12}{13}\right)$.
$3.455 A=\tan\left(2 \arccos \frac{5}{\sqrt{26}}-\arcsin \frac{12}{13}\right)$. | ## Solution.
Let $\alpha=\arccos \frac{5}{\sqrt{26}}, \beta=\arcsin \frac{12}{13} \Rightarrow \cos \alpha=\frac{5}{\sqrt{26}}, \operatorname{tg} \alpha=\frac{1}{5}$,
$\sin \beta=\frac{12}{13}, \operatorname{tg} \beta=\frac{12}{5} \Rightarrow A=\operatorname{tg}(2 \alpha-\beta)=\frac{\operatorname{tg} 2 \alpha-\operat... | -\frac{119}{120} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,886 |
$3.456 A=\sin ^{2}\left(\frac{1}{2} \arcsin \frac{4}{5}-2 \operatorname{arctg}(-2)\right)$
$3.456 A=\sin ^{2}\left(\frac{1}{2} \arcsin \frac{4}{5}-2 \arctan(-2)\right)$ | Solution. Let $\alpha=\arcsin \frac{4}{5}, \beta=\operatorname{arctg} 2 \Rightarrow \sin \alpha=\frac{4}{5}, \cos \alpha=\frac{3}{5}$, $\sin \frac{\alpha}{2}=\sqrt{\frac{1-\cos \alpha}{2}}=\frac{1}{\sqrt{5}}, \quad \cos \frac{\alpha}{2}=\frac{2}{\sqrt{5}}, \operatorname{tg} \beta=2, \sin 2 \beta=\frac{2 \operatorname{t... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,887 |
$3.457 A=\operatorname{ctg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$
$3.457 A=\cot\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$ | Solution. Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}\left(-\frac{1}{2}\right) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$, $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=2, \operatorname{ctg} \beta=-\frac{1}{2}, \operatorname{ctg} 2 \beta=\frac{\operatorname{c... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,888 |
$3.458 A=\operatorname{tg}\left(\frac{1}{2} \arccos \frac{3}{5}-3 \operatorname{arcctg}(-2)\right)$
$3.458 A=\tan\left(\frac{1}{2} \arccos \frac{3}{5}-3 \operatorname{arcctg}(-2)\right)$ | ## Solution.
Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}(-2) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$,
$\operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}=\frac{1}{2}, \operatorname{ctg} \beta=-2, \operatorname{tg} \beta=-\frac{1}{2}, \operatorname{tg} 3 \beta=\... | -\frac{24}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,889 |
$3.460 A=\cos \left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arctg}(-2)\right)$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$3.460 A=\cos \left(\frac{1}{2} \arccos \frac{3}{5}-2 \arctan(-2)\right)$. | Solution.
Since $\arccos \frac{3}{5}=\arcsin \frac{4}{5}, \operatorname{arctg}(-2)=\operatorname{arcctg}\left(-\frac{1}{2}\right)-\pi$, this problem is analogous to problem 3.459.
Answer: $A=-\frac{2 \sqrt{5}}{5}$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,891 | |
$3.461 A=\operatorname{tg}\left(\frac{5 \pi}{4}+\frac{1}{2} \arccos \frac{b}{a}\right)+\operatorname{tg}\left(\frac{5 \pi}{4}-\frac{1}{2} \arccos \frac{b}{a}\right)$
$3.461 A=\tan\left(\frac{5 \pi}{4}+\frac{1}{2} \arccos \frac{b}{a}\right)+\tan\left(\frac{5 \pi}{4}-\frac{1}{2} \arccos \frac{b}{a}\right)$ | ## Solution.
Let $\alpha=\arccos \frac{b}{a} \Rightarrow \cos \alpha=\frac{b}{a} \Rightarrow A=\frac{1+\operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg} \frac{\alpha}{2}}+\frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=2 \frac{1+\operatorname{tg}^{2} \frac{\alpha}{2}}{1-\operato... | \frac{2}{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,892 |
$3.462 A=\operatorname{tg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arctg}(-2)\right)$
$3.462 A=\tan\left(\frac{1}{2} \arccos \frac{3}{5}-2 \arctan(-2)\right)$ | Solution. Since $\operatorname{arctg}(-2)=\operatorname{arcctg}\left(-\frac{1}{2}\right)-\pi$, this problem is analogous to problem 3.457.
Answer: $A=-\frac{1}{2}$. | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,893 |
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