problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln x}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln x}=\lim _{y \rightarrow 0} \frac{(y+1)^{2}-1}{\ln (y+1)}=$
Using the substitution of equivalent infinitesimals:
$\ln (1+y) \sim y$, as $y \rightarrow 0$
We get:
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{\sin ^{2} x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{e^{-x}\left(e^{2 x}-2 e^{x}+1\right)}{\sin ^{2} x}=$
$=\lim _{x \rightarrow 0} \frac{e^{-x}\left(e^{x}-1\right)^{2}}{\sin ^{2} x}=$
Using the substitution of equivalent infinitesimals:
$e^{x}-1 \sim x$, as $x \righ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)^{1+x}=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{\lim _{x \rightarrow 0} 1+x}=$
$=\left(\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{x}\right)\right)^{1}=\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=$
Using the substitution o... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}
$$ | ## Solution
Since $\operatorname{arctg}\left(\frac{1}{x}\right)_{\text { is bounded, then }}$
$$
x \cdot \operatorname{arctg}\left(\frac{1}{x}\right) \rightarrow 0 \underset{\text { as } x \rightarrow 0}{ }
$$
Then:
$\lim _{x \rightarrow 0} \sqrt{4 \cos 3 x+x \cdot \operatorname{arctg}\left(\frac{1}{x}\right)}=\sqr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$$
A(-3 ; -7 ; -5), B(0 ; -1 ; -2), C(2 ; 3 ; 0)
$$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(0-(-3) ;-1-(-7) ;-2-(-5))=(3 ; 6 ; 3) \\
& \overrightarrow{A C}=(2-(-3) ; 3-(-7) ; 0-(-5))=(5 ; 10 ; 5)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-q$
$b=p+3 q$
$|p|=3$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(2 p-q) \times(p+3 q)=2 \cdot p \times p+2 \cdot 3 \cdot p \times q-q \times p-3 ... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{4} ; \frac{1}{3} ; 1\right)$
a: $4 x-3 y+5 z-10=0$
$k=\frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate plane, the plane $a: A x + B y + C z + D = 0$ and the coefficient $k$ transitions to the plane $a^{\prime}: A x + B y + C z + k \cdot D = 0$. We find the image of the plane $a$:
$a^{\prime}: 4 x - 3 y + 5 z - 5 = 0$
Substitute t... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
x^{2} \cos ^{2} \frac{11}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((2 n+1)^{3}+(3 n+2)^{3}\right)}{\frac{1}{n^{3}}\left((2 n+3)^{3}-(n-7)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(2+\frac{1}{n}\ri... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{4 n^{2}-\sqrt[4]{n^{3}}}{\sqrt[3]{n^{6}+n^{3}+1}-5 n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{4 n^{2}-\sqrt[4]{n^{3}}}{\sqrt[3]{n^{6}+n^{3}+1}-5 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(4 n^{2}-\sqrt[4]{n^{3}}\right)}{\frac{1}{n^{2}}\left(\sqrt[3]{n^{6}+n^{3}+1}-5 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{4-\sqrt[4]... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2-5+4-7+\ldots+2 n-(2 n+3)}{n+3}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{2-5+4-7+\ldots+2 n-(2 n+3)}{n+3}=$
$=\{2-5=4-7=\ldots=2 n-(2 n+3)=-3\}=$
$=\lim _{n \rightarrow \infty} \frac{-3 \cdot n}{n+3}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-3) \cdot n}{\frac{1}{n}(n+3)}=$
$=\lim _{n \rightarrow \infty} \frac{-3}{1+\frac{3}{n}}=\f... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-3^{\sin ^{2} x}\right)^{\frac{1}{\ln (\cos x)}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(2-3^{\sin ^{2} x}\right)^{\frac{1}{\ln (\cos x)}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(2-3^{\sin ^{2} x}\right)}\right)^{\frac{1}{\ln (\cos x)}}= \\
& =\lim _{x \rightarrow 0} e^{\ln \left(2-3^{\sin ^{2} x}\right) / \ln (\cos x)}= \\
& =\lim _{x... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{11 x+8}{12 x+1}\right)^{\cos ^{2} x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{11 x+8}{12 x+1}\right)^{\cos ^{2} x}=\left(\frac{11 \cdot 0+8}{12 \cdot 0+1}\right)^{\cos ^{2} 0}=$
$=(8)^{\left(1^{2}\right)}=8^{1}=8$
## Problem Kuznetsov Limits 18-19 | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}(\arcsin x)^{\tan \pi x}$ | ## Solution
$\lim _{x \rightarrow 1}(\arcsin x)^{\operatorname{tg} \pi x}=(\arcsin 1)^{\operatorname{tg} \pi}=\left(\frac{\pi}{2}\right)^{0}=1$
## Problem Kuznetsov Limits 20-19 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=(x-2)^{3}, y=4 x-8
$$ | ## Solution
From the graph, it can be seen that the area between the curves consists of two identical parts:
$$
S_{(0,4)}=2 S_{(2,4)}
$$
We will find the area of the part where \( x \in (2,4) \) as the difference of two integrals:
$$
\begin{aligned}
& S=\int_{2}^{4}(4 x-8) d x-\int_{2}^{4}(x-2)^{3} d x= \\
& =\left... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x \sqrt{9-x^{2}}, y=0, (0 \leq x \leq 3)
$$ | ## Solution
$$
\begin{aligned}
& S=\int_{0}^{3}\left(x \sqrt{9-x^{2}}\right) d x= \\
& =-\frac{1}{2} \int_{0}^{3}\left(9-x^{2}\right)^{\frac{1}{2}} d\left(9-x^{2}\right)= \\
& =-\left.\frac{... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=4-x^{2}, y=x^{2}-2 x
$$ | ## Solution
Find the abscissas of the points of intersection of the graphs of the functions:
$$
\begin{aligned}
& 4-x^{2}=x^{2}-2 x \\
& 2 x^{2}-2 x-4=0 \\
& x^{2}-x-2=0
\end{aligned}
$$
... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\frac{1}{x \sqrt{1+\ln x}}, y=0, x=1, x=e^{3}
$$ | ## Solution
Answer: 2
We construct the graphs:
We obtain a figure bounded above by the curve $\frac{1}{x \cdot \sqrt{1+\ln x}}$, on the left by the line $x=1$, on the right by the line ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\arccos x, y=0, x=0
$$ | ## Solution
$S=\int_{0}^{1} \arccos x d x$
We will use the integration by parts formula:
$$
\begin{aligned}
& \int_{a}^{b} u d v=\left.u \cdot v\right|_{a} ^{b}-\int_{a}^{b} v d u \\
& ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=2 x-x^{2}+3, y=x^{2}-4 x+3
$$ | ## Solution
Let's find the points of intersection of the graphs of the functions:
$$
2 x - x^{2} + 3 = x^{2} - 4 x + 3 \Rightarrow \left[\begin{array}{l}
x = 0 \\
x = 3
\end{array}\right]
$... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=x \sqrt{36-x^{2}}, y=0, (0 \leq x \leq 6)
$$ | ## Solution
Integrals 14-13
$S=\int_{0}^{6} x \sqrt{36-x^{2}} d x=$
$$
\begin{aligned}
& =\frac{1}{2} \int_{0}^{6} \sqrt{36-x^{2}} d\left(x^{2}\right)= \\
& =-\frac{1}{2} \int_{0}^{6}\l... | 72 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
y=\frac{1}{1+\cos x}, y=0, x=\frac{\pi}{2}, x=-\frac{\pi}{2}
$$ | ## Solution
$$
\begin{aligned}
S= & \int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x} d x= \\
& =\int_{-\pi / 2}^{\pi / 2} \frac{1}{2 \cos ^{2} \frac{x}{2}} d x= \\
& =\int_{-\pi / 2}^{\pi / 2} \frac{1}{\cos ^{2} \frac{x}{2}} d\left(\frac{x}{2}\right)= \\
& =\left.\operatorname{tg}_{\frac{1}{2}}\right|_{-\pi / 2} ^{\pi / 2... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=(y-2)^{3}, x=4 y-8
$$ | ## Solution
$$
\begin{aligned}
& (y-2)^{3}=4(y-2) \\
& (y-2)\left[4-(y-2)^{2}\right]=0 \\
& \begin{array}{l}
4-y^{2}+4 y-4=0 \\
y(y-4)=0
\end{array} \\
& \frac{S}{2}=\int_{2}^{4}(4 y-8) d y-\int_{2}^{4}(y-3)^{3} d y= \\
& =\left.2 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}-\int_{2}^{4}\left(y^{3}-6 y^{2}+12 y-8\r... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=4-y^{2}, x=y^{2}-2 y
$$ | ## Solution
Let's find the limits of integration:
$$
\begin{aligned}
& \left(y^{2}-2 y\right)-\left(4-y^{2}\right)=0 \\
& 2 y^{2}-2 y-4=0 \\
& y^{2}-y-2=0
\end{aligned}
$$
Then the area ... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the area of the figure bounded by the graphs of the functions:
$$
x=\frac{1}{y \sqrt{1+\ln y}}, x=0, y=1, y=e^{3}
$$ | ## Solution
The desired area $S$ is:
$S=\int_{1}^{e^{3}} \frac{1}{y \sqrt{\ln y+1}} d y$
We make a substitution of variables:
$t=\ln y$, hence
$d t=\frac{d y}{y}$
When $t=\ln 1 \Rightarrow t=0$ and when
$t=\ln e^{3} \Rightarrow t=3$
Then we get
^{2}, x=y^{2}-4 y+3
$$ | ## Solution
Find the ordinates of the points of intersection of the graphs of the functions $x=4-(y-1)^{2}$
, $x=y^{2}-4 y+3:$
$$
4-(y-1)^{2}=y^{2}-4 y+3
$$
$$
\begin{aligned}
& 4-y^{2}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=5(t-\sin t) \\
y=5(1-\cos t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
Let's find the derivatives with respect to $t$:
$$
\begin{aligned}
& x_{t}^{\prime}=5(1-\cos t) ; \quad y_{t}^{\prime}=5 \sin t \\
& L=5 \int_{0}^{\pi} \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t=5 \int_{0}^{\pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2} t} d t= \\
& =5 \int_{0}^{\pi} \sqrt{2-2 \cos t} d t=5 \int_... | 20 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(2 \cos t-\cos 2 t) \\
y=3(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=3(2 \cos t-\cos 2... | 48 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=10 \cos ^{3} t \\
y=10 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{2}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=10 \cos ^{3}(t) ; x_{t}^{\prime}=10 \cdot... | 15 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=3(t-\sin t) \\
y=3(1-\cos t)
\end{array}\right. \\
& \pi \leq t \leq 2 \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=3(t-\sin t) ; x... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8 \cos ^{3} t \\
y=8 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{6}
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$
From the equations of the curve, we find:
$$
\begin{aligned}
& x=8 \cos ^{3} t ; x_{t}^{\prime}=8 \cdot 3 ... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the parametric equations.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=4(2 \cos t-\cos 2 t) \\
y=4(2 \sin t-\sin 2 t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$ | ## Solution
The length of the arc of a curve defined by parametric equations is determined by the formula
$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$
Let's find the derivatives with respect to $t$ for the given curve:
$$
\begin{aligned}
& x=4(2 \cos t-\cos 2... | 32 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
e^{x \sin 5 x}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 0 ; 4), B(-3 ;-6 ; 1), C(-5 ;-10 ;-1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-3-0 ;-6-0 ; 1-4)=(-3 ;-6 ;-3)$
$\overrightarrow{A C}=(-5-0 ;-10-0 ;-1-4)=(-5 ;-10 ;-5)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overri... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(2 ; 3 ; 1) \)
\( A_{2}(4 ; 1 ;-2) \)
\( A_{3}(6 ; 3 ; 7) \)
\( A_{4}(7 ; 5 ;-3) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{4-2 ; 1-3 ;-2-1\}=\{2 ;-2 ;-3\} \\
& \vec{A}_{1} A_{3}=\{6-2 ; 3-3 ; 7-1\}=\{4 ; 0 ; 6\} \\
& \overrightarrow{A_{1} A_{4}}=\{7-2 ; 5-3 ;-3-1\}=\{5 ; 2 ;-4\}
\end{aligned}
$$
According to the geometric meaning of the... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2 ; 3 ;-2)$
$a: 3 x-2 y+4 z-6=0$
$k=-\frac{4}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-2 y+4 z+8=0$
Substitute the coordinates of point $A$ into the equ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(1, -2, 3), B(0, -1, 2), C(3, -4, 5)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-1 ;-1-(-2) ; 2-3)=(-1 ; 1 ;-1)$
$\overrightarrow{A C}=(3-1 ;-4-(-2) ; 5-3)=(2 ;-2 ; 2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{alig... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
\[
\begin{aligned}
& a=p+2 q \\
& b=3 p-q \\
& |p|=1 \\
& |q|=2 \\
& (\widehat{p, q})=\frac{\pi}{6}
\end{aligned}
\] | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$$
S=|a \times b|
$$
We compute \(a \times b\) using the properties of the vector product:
$$
\begin{aligned}
& a \times b=(p+2 q) \times(3 p-q)=3 \cdot p \times p-p \times q+2 \... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(4 ; 3 ; 1)$
$a: 3x - 4y + 5z - 6 = 0$
$k = \frac{5}{6}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-4 y+5 z-5=0$
Substitute the coordinates of point $A$ into the equati... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{4 n^{4}+1}-\sqrt[3]{n^{4}-1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{4 n^{4}+1}-\sqrt[3]{n^{1}-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt{n^{2}+2}\right)}{\frac{1}{n}\left(\sqrt[4]{4 n^{\frac{1}{4}+1}}-\sqrt[3]{n^{4}-1}\right)}= \\
& =\lim _{n \rightarro... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x+x^{5}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{: 5}+3 \cdot 1^{2} \cdot x^{2}+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x\left(1+x^{4}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x\left(1+x^{4}\... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{\sin ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\frac{\sin ^{2} x \cdot \cos ^{2} x}{\cos ^{2} x}-\tan ^{2} x}{(x-\pi)^{4}}=$
$=\lim _{x \rightarrow \pi} \frac{\tan ^{2} x \cdot \cos ^{2} x-\tan ^{2} x}{(x-\pi)^{4}}=\lim _{x \rightarrow \pi} \frac{\ta... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}$ | ## Solution
Substitution:
$x=y+3 \Rightarrow y=x-3$
$x \rightarrow 3 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \left(\sqrt{2(y+3)^{2}-3(y+3)-5}-\sqrt... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\lim _{x \rightarrow 0} \frac{x}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\frac{0}{0+2}}=\left(\lim _{x \rightarrow 0} \... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}=\lim _{x \rightarrow 3}\left(e^{\ln \left(2-\frac{x}{3}\right)}\right)^{\sin (\pi x)}= \\
& =\lim _{x \rightarrow 3} e^{\sin (\pi x) \cdot \ln \left(2-\frac{x}{3}\right)}=\exp \left\{\lim _{x \rightarrow 3} \sin (\pi x) \... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-4 ; 3 ; 0), B(0 ; 1 ; 3), C(-2 ; 4 ;-2)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-4) ; 1-3 ; 3-0)=(4 ;-2 ; 3)$
$\overrightarrow{A C}=(-2-(-4) ; 4-3 ;-2-0)=(2 ; 1 ;-2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; 1 ; 4)$
$M_{2}(3 ; 5 ;-2)$
$M_{3}(-7 ;-3 ; 2)$
$M_{0}(-3 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
3-2 & 5-1 & -2-4 \\
-7-2 & -3-1 & 2-4
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
1 & 4 & -6 \\
-9 & -4 ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
e^{\sin \left(x^{\frac{3}{2}} \sin \frac{2}{x}\right)}-1+x^{2}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\sqrt{1+\ln \left(1+x^{2} \sin \frac{1}{x}\right)}-1, x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
2.160. $\frac{\left(a^{2} b \sqrt{b}-6 a^{5 / 3} b^{5 / 4}+12 a b \sqrt[3]{a}-8 a b^{3 / 4}\right)^{2 / 3}}{a b \sqrt[3]{a}-4 a b^{3 / 4}+4 a^{2 / 3} \sqrt{b}}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}a^{1 / 3} b^{1 / 4} \neq 2, \\ a \neq 0, \\ b \neq 0 .\end{array}\right.$
$$
\begin{aligned}
& \frac{\left(a^{2} b \sqrt{b}-6 a^{5 / 3} b^{5 / 4}+12 a b \sqrt[3]{a}-8 a b^{3 / 4}\right)^{2 / 3}}{a b \sqrt[3]{a}-4 a b^{3 / 4}+4 a^{2 / 3} \sqrt{b}}= \\
& =\frac{\l... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.196. $\frac{\left|x^{2}-1\right|+x^{2}}{2 x^{2}-1}-\frac{|x-1|}{x-1}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq \pm \frac{\sqrt{2}}{2}, \\ x \neq 1 .\end{array}\right.$
Expanding the absolute values with consideration of the domain of definition, we consider three cases:
1) $\left\{\begin{array}{l}x \in(-\infty ;-1), \\ \frac{x^{2}-1+x^{2}}{2 x^{2}-1}+\frac{x-1}{x... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.204. $\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$. | Solution.
$$
\begin{aligned}
& \sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{(2+\sqrt{2+\sqrt{2+\sqrt{3}}})(2-\sqrt{2+\sqrt{2+\sqrt{3}}})}= \\
& =\sqrt{2+\sqrt{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.208. $\frac{\left((\sqrt[4]{m}+\sqrt[4]{n})^{2}-(\sqrt[4]{m}-\sqrt[4]{n})^{2}\right)^{2}-(16 m+4 n)}{4 m-n}+\frac{10 \sqrt{m}-3 \sqrt{n}}{\sqrt{n}+2 \sqrt{m}}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}n \neq 4 m, \\ m>0, \\ n>0 .\end{array}\right.$
$$
\frac{\left((\sqrt[4]{m}+\sqrt[4]{n})^{2}-(\sqrt[4]{m}-\sqrt[4]{n})^{2}\right)^{2}-(16 m+4 n)}{4 m-n}+\frac{10 \sqrt{m}-3 \sqrt{n}}{\sqrt{n}+2 \sqrt{m}}=
$$
$=\frac{(\sqrt{m}+2 \sqrt[4]{m n}+\sqrt{n}-\sqrt{m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.219. $\left(\frac{9}{a+8}-\frac{a^{1 / 3}+2}{a^{2 / 3}-2 a^{1 / 3}+4}\right) \cdot \frac{a^{4 / 3}+8 a^{1 / 3}}{1-a^{2 / 3}}+\frac{5-a^{2 / 3}}{1+a^{1 / 3}}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}a \neq-8, \\ a \neq \pm 1 .\end{array}\right.$
$$
\begin{aligned}
& \left(\frac{9}{a+8}-\frac{a^{1 / 3}+2}{a^{2 / 3}-2 a^{1 / 3}+4}\right) \cdot \frac{a^{4 / 3}+8 a^{1 / 3}}{1-a^{2 / 3}}+\frac{5-a^{2 / 3}}{1+a^{1 / 3}}= \\
& =\left(\frac{9}{\left(a^{1 / 3}+2\rig... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.230. $\frac{\sqrt{x-2 \sqrt{2}}}{\sqrt{x^{2}-4 x \sqrt{2}+8}}-\frac{\sqrt{x+2 \sqrt{2}}}{\sqrt{x^{2}+4 x \sqrt{2}+8}} ; x=3$. | ## Solution.
$$
\begin{aligned}
& \frac{\sqrt{x-2 \sqrt{2}}}{\sqrt{x^{2}-4 x \sqrt{2}+8}}-\frac{\sqrt{x+2 \sqrt{2}}}{\sqrt{x^{2}+4 x \sqrt{2}+8}}=\frac{\sqrt{x-2 \sqrt{2}}}{\sqrt{(x-2 \sqrt{2})^{2}}}-\frac{\sqrt{x+2 \sqrt{2}}}{\sqrt{(x+2 \sqrt{2})^{2}}}= \\
& =\frac{1}{\sqrt{x-2 \sqrt{2}}}-\frac{1}{\sqrt{x+2 \sqrt{2}}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.240. $\frac{\sqrt{\left(\frac{9-2 \sqrt{3}}{\sqrt{3}-\sqrt[3]{2}}+3 \sqrt[3]{2}\right) \cdot \sqrt{3}}}{3+\sqrt[6]{108}}$. | Solution.
$\frac{\sqrt{\left(\frac{9-2 \sqrt{3}}{\sqrt{3}-\sqrt[3]{2}}+3 \sqrt[3]{2}\right) \cdot \sqrt{3}}}{3+\sqrt[6]{108}}=\frac{\sqrt{\left(\frac{3^{2}-\sqrt{2^{2} \cdot 3}}{\sqrt{3}-\sqrt[3]{2}}+\sqrt[3]{3^{3} \cdot 2}\right) \cdot \sqrt{3}}}{3+\sqrt[6]{27 \cdot 4}}=$
$=\frac{\sqrt{\left(\frac{\sqrt[6]{3^{12}}-\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.254. $\left(\frac{x+2 y}{8 y^{3}\left(x^{2}+2 x y+2 y^{2}\right)}-\frac{(x-2 y): 8 y^{2}}{x^{2}-2 x y+2 y^{2}}\right)+\left(\frac{y^{-2}}{4 x^{2}-8 y^{2}}-\frac{1}{4 x^{2} y^{2}+8 y^{4}}\right)$ $x=\sqrt[4]{6}, \quad y=\sqrt[8]{2}$. | Solution.
$$
\begin{aligned}
& \left(\frac{x+2 y}{8 y^{3}\left(x^{2}+2 x y+2 y^{2}\right)}-\frac{(x-2 y): 8 y^{2}}{x^{2}-2 x y+2 y^{2}}\right)+\left(\frac{y^{-2}}{4 x^{2}-8 y^{2}}-\frac{1}{4 x^{2} y^{2}+8 y^{4}}\right)= \\
& =\left(\frac{x+2 y}{8 y^{3}\left(x^{2}+2 x y+2 y^{2}\right)}-\frac{x-2 y}{8 y^{3}\left(x^{2}-2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.259. $\left(\sqrt[3]{\frac{8 z^{3}+24 z^{2}+18 z}{2 z-3}}-\sqrt[3]{\frac{8 z^{2}-24 z^{2}+18 z}{2 z+3}}\right)-\left(\frac{1}{2} \sqrt[3]{\frac{2 z}{27}-\frac{1}{6 z}}\right)^{-1}$. | Solution.
Domain of definition: $z \neq \pm \frac{3}{2}, z \neq 0$.
$$
\begin{aligned}
& \left(\sqrt[3]{\frac{8 z^{3}+24 z^{2}+18 z}{2 z-3}}-\sqrt[3]{\frac{8 z^{2}-24 z^{2}+18 z}{2 z+3}}\right)-\left(\frac{1}{2} \sqrt[3]{\frac{2 z}{27}-\frac{1}{6 z}}\right)^{-1}= \\
& =\sqrt[3]{\frac{2 z\left(4 z^{2}+12 z+9\right)}{2... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.274. $\frac{8-m}{\sqrt[3]{m}+2}:\left(2+\frac{\sqrt[3]{m^{2}}}{\sqrt[3]{m}+2}\right)+\left(\sqrt[3]{m}+\frac{2 \sqrt[3]{m}}{\sqrt[3]{m}-2}\right) \cdot \frac{\sqrt[3]{m^{2}}-4}{\sqrt[3]{m^{2}}+2 \sqrt[3]{m}}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}m \neq 0, \\ m \neq \pm 8 .\end{array}\right.$
$\frac{8-m}{\sqrt[3]{m}+2}:\left(2+\frac{\sqrt[3]{m^{2}}}{\sqrt[3]{m}+2}\right)+\left(\sqrt[3]{m}+\frac{2 \sqrt[3]{m}}{\sqrt[3]{m}-2}\right) \cdot \frac{\sqrt[3]{m^{2}}-4}{\sqrt[3]{m^{2}}+2 \sqrt[3]{m}}=$
$=\fra... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.282. $\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^{2}}-1+x}\right) \cdot\left(\sqrt{\frac{1}{x^{2}}-1}-\frac{1}{x}\right)=0<x<1$. | ## Solution.
$$
\begin{aligned}
& \left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^{2}}-1+x}\right) \cdot\left(\sqrt{\frac{1}{x^{2}}-1}-\frac{1}{x}\right)= \\
& =\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{(1-x)^{2}}}{\sqrt{(1-x)(1+x)}-\sqrt{(1-x)^{2}}}\right) \cdot\left(\sqrt{\frac{1... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.283. $\frac{\left(p q^{-1}+1\right)^{2}}{p q^{-1}-p^{-1} q} \cdot \frac{p^{3} q^{-3}-1}{p^{2} q^{-2}+p q^{-1}+1}: \frac{p^{3} q^{-3}+1}{p q^{-1}+p^{-1} q-1}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}p \neq 0, \\ q \neq 0, \\ p \neq \pm q .\end{array}\right.$
$$
\begin{aligned}
& \frac{\left(p q^{-1}+1\right)^{2}}{p q^{-1}-p^{-1} q} \cdot \frac{p^{3} q^{-3}-1}{p^{2} q^{-2}+p q^{-1}+1}: \frac{p^{3} q^{-3}+1}{p q^{-1}+p^{-1} q-1}= \\
& =\frac{\left(\frac{p}{q}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.288. Check that the number $x=\sqrt[3]{4+\sqrt{80}}-\sqrt[3]{\sqrt{80}-4}$ is a root of the equation $x^{3}+12 x-8=0$. | ## Solution.
Let $x=\sqrt[3]{4+\sqrt{80}}-\sqrt[3]{\sqrt{80}-4}=\sqrt[3]{4+\sqrt{80}}+\sqrt[3]{4-\sqrt{80}}$.
Substituting this value of $x$ into the equation, we get
$$
(\sqrt[3]{4+\sqrt{80}}+\sqrt[3]{4-\sqrt{80}})^{3}+12(\sqrt[3]{4+\sqrt{80}}+\sqrt[3]{4-\sqrt{80}})-8=0
$$
$$
\begin{aligned}
& (\sqrt[3]{4+\sqrt{80... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
2.297. $\sqrt[3]{26+15 \sqrt{3}} \cdot(2-\sqrt{3})=1$. | Solution.
$\sqrt[3]{26+15 \sqrt{3}} \cdot(2-\sqrt{3})=\sqrt[3]{26+15 \sqrt{3}} \cdot \sqrt[3]{(2-\sqrt{3})^{3}}=$ $=\sqrt[3]{26+15 \sqrt{3}} \cdot \sqrt[3]{8-12 \sqrt{3}+18-3 \sqrt{3}}=\sqrt[3]{26+15 \sqrt{3}} \cdot \sqrt[3]{26-15 \sqrt{3}}=$ $=\sqrt[3]{(26+15 \sqrt{3})(26-15 \sqrt{3})}=\sqrt[3]{26^{2}-(15 \sqrt{3})^{... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
2.301. $\frac{\sqrt{5-2 \sqrt{6}} \cdot(5+2 \sqrt{6})(49-20 \sqrt{6})}{\sqrt{27}-3 \sqrt{18}+3 \sqrt{12}-\sqrt{8}}=1$. | Solution.
$$
\begin{aligned}
& \frac{\sqrt{5-2 \sqrt{6}} \cdot(5+2 \sqrt{6})(49-20 \sqrt{6})}{\sqrt{27}-3 \sqrt{18}+3 \sqrt{12}-\sqrt{8}}= \\
& =\frac{\sqrt{3-2 \sqrt{3 \cdot 2}+2} \cdot(3+2 \sqrt{3 \cdot 2}+2)(49-20 \sqrt{6})}{\sqrt{9 \cdot 3}-3 \sqrt{9 \cdot 2}+3 \sqrt{4 \cdot 3}-\sqrt{4 \cdot 2}}=
\end{aligned}
$$
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.235. $\frac{1-2 \cos ^{2} \alpha}{2 \operatorname{tg}\left(2 \alpha-\frac{\pi}{4}\right) \sin ^{2}\left(\frac{\pi}{4}+2 \alpha\right)}=1$. | ## Solution.
$$
\begin{aligned}
& \frac{1-2 \cos ^{2} 2 \alpha}{2 \operatorname{tg}\left(2 \alpha-\frac{\pi}{4}\right) \sin ^{2}\left(\frac{\pi}{4}+2 \alpha\right)}=\frac{1-1-\cos 4 \alpha}{\frac{\sin \left(4 \alpha-\frac{\pi}{2}\right)}{1+\cos \left(4 \alpha-\frac{\pi}{2}\right)} \cdot\left(1-\cos \left(\frac{\pi}{4}... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.243. $(\cos 8 \alpha \tan 4 \alpha-\sin 8 \alpha)(\cos 8 \alpha \cot 4 \alpha+\sin 8 \alpha)$. | ## Solution.
$$
\begin{aligned}
& (\cos 8 \alpha \tan 4 \alpha - \sin 8 \alpha)(\cos 8 \alpha \cot 4 \alpha + \sin 8 \alpha)= \\
& =\left(\frac{\cos 8 \alpha \sin 4 \alpha}{\cos 4 \alpha} - \sin 8 \alpha\right)\left(\frac{\cos 8 \alpha \cos 4 \alpha}{\sin 4 \alpha} + \sin 8 \alpha\right)= \\
& =\frac{\sin 4 \alpha \co... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.244. $\sin ^{2} 2 \alpha+\sin ^{2} \beta+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)$. | ## Solution.
$$
\begin{aligned}
& \sin ^{2} 2 \alpha+\sin ^{2} \beta+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =\frac{1-\cos 4 \alpha}{2}+\frac{1-\cos 2 \beta}{2}+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =\frac{1}{2}(1-\cos 4 \alpha+1-\cos 2 \beta)+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =-\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.247. $\left(1-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi-2 \alpha\right)\right) \sin ^{2}\left(\frac{\pi}{2}+2 \alpha\right) \operatorname{tg}\left(\frac{5}{4} \pi-2 \alpha\right)+\cos \left(4 \alpha-\frac{\pi}{2}\right)$. | ## Solution.
$$
\begin{aligned}
& \left(1-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi-2 \alpha\right)\right) \sin ^{2}\left(\frac{\pi}{2}+2 \alpha\right) \operatorname{tg}\left(\frac{5}{4} \pi-2 \alpha\right)+\cos \left(4 \alpha-\frac{\pi}{2}\right)= \\
& =\left(1-\left(\operatorname{ctg}\left(\frac{3}{2} \pi-2 \alpha... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.282. $\frac{\sin 8 \alpha+\sin 9 \alpha+\sin 10 \alpha+\sin 11 \alpha}{\cos 8 \alpha+\cos 9 \alpha+\cos 10 \alpha+\cos 11 \alpha} \times$
$\times \frac{\cos 8 \alpha-\cos 9 \alpha-\cos 10 \alpha+\cos 11 \alpha}{\sin 8 \alpha-\sin 9 \alpha-\sin 10 \alpha+\sin 11 \alpha}$. | Solution.
$$
\begin{aligned}
& \frac{\sin 8 \alpha + \sin 9 \alpha + \sin 10 \alpha + \sin 11 \alpha}{\cos 8 \alpha + \cos 9 \alpha + \cos 10 \alpha + \cos 11 \alpha} \cdot \frac{\cos 8 \alpha - \cos 9 \alpha - \cos 10 \alpha + \cos 11 \alpha}{\sin 8 \alpha - \sin 9 \alpha - \sin 10 \alpha + \sin 11 \alpha} = \\
& = \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.332. $\frac{\sin 24^{\circ} \cos 6^{\circ}-\sin 6^{\circ} \sin 66^{\circ}}{\sin 21^{\circ} \cos 39^{\circ}-\sin 39^{\circ} \cos 21^{\circ}}=-1$. | Solution.
$$
\begin{aligned}
& \frac{\sin 24^{\circ} \cos 6^{\circ}-\sin 6^{\circ} \sin 66^{\circ}}{\sin 21^{\circ} \cos 39^{\circ}-\sin 39^{\circ} \cos 21^{\circ}}=\frac{\sin 24^{\circ} \cos 6^{\circ}-\sin 6^{\circ} \sin \left(90^{\circ}-24^{\circ}\right)}{\sin 21^{\circ} \cos 39^{\circ}-\sin 39^{\circ} \cos 21^{\cir... | -1 | Algebra | proof | Yes | Yes | olympiads | false |
3.333. $\frac{\sin 20^{\circ} \cos 10^{\circ}+\cos 160^{\circ} \cos 100^{\circ}}{\sin 21^{\circ} \cos 9^{\circ}+\cos 159^{\circ} \cos 99^{\circ}}=1$. | ## Solution.
$\frac{\sin 20^{\circ} \cos 10^{\circ}+\cos 160^{\circ} \cos 100^{\circ}}{\sin 21^{\circ} \cos 9^{\circ}+\cos 159^{\circ} \cos 99^{\circ}}=$
$$
\begin{aligned}
& =\frac{\sin 20^{\circ} \cos 10^{\circ}+\cos \left(180^{\circ}-20^{\circ}\right) \cos \left(90^{\circ}+10^{\circ}\right)}{\sin 21^{\circ} \cos 9... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.335.
$$
\frac{\cos 64^{\circ} \cos 4^{\circ}-\cos 86^{\circ} \cos 26^{\circ}}{\cos 71^{\circ} \cos 41^{\circ}-\cos 49^{\circ} \cos 19^{\circ}}=-1
$$ | Solution.
$$
\begin{aligned}
& \frac{\cos 64^{\circ} \cos 4^{\circ}-\cos 86^{\circ} \cos 26^{\circ}}{\cos 71^{\circ} \cos 41^{\circ}-\cos 49^{\circ} \cos 19^{\circ}}= \\
& =\frac{\cos \left(90^{\circ}-26^{\circ}\right) \cos 4^{\circ}-\cos \left(90^{\circ}-4^{\circ}\right) \cos 26^{\circ}}{\cos \left(90^{\circ}-19^{\ci... | -1 | Algebra | proof | Yes | Yes | olympiads | false |
3.336. $\frac{\cos 66^{\circ} \cos 6^{\circ}+\cos 84^{\circ} \cos 24^{\circ}}{\cos 65^{\circ} \cos 5^{\circ}+\cos 85^{\circ} \cos 25^{\circ}}=1$. | Solution.
$$
\begin{aligned}
& \frac{\cos 66^{\circ} \cos 6^{\circ}+\cos 84^{\circ} \cos 24^{\circ}}{\cos 65^{\circ} \cos 5^{\circ}+\cos 85^{\circ} \cos 25^{\circ}}= \\
& =\frac{\cos 66^{\circ} \cos 6^{\circ}+\cos \left(90^{\circ}-6^{\circ}\right) \cos \left(90^{\circ}-66^{\circ}\right)}{\cos 65^{\circ} \cos 5^{\circ}... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.341. $\frac{\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}}{\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}}=3$. | Solution.
$$
\begin{aligned}
& \frac{\left(\sin 20^{\circ} \sin 40^{\circ}\right)\left(\sin 60^{\circ} \sin 80^{\circ}\right)}{\left(\sin 10^{\circ} \sin 30^{\circ}\right)\left(\sin 50^{\circ} \sin 70^{\circ}\right)}=\left[\sin x \sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))\right]= \\
& =\frac{\left(\cos 20^{\circ}-\cos ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.347. $8 \cos \frac{4 \pi}{9} \cos \frac{2 \pi}{9} \cos \frac{\pi}{9}=1$. | ## Solution.
$$
8 \cos \frac{4 \pi}{9} \cos \frac{2 \pi}{9} \cos \frac{\pi}{9}=8 \cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ}=
$$
$=\frac{8 \cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} \sin 20^{\circ}}{\sin 20^{\circ}}=\frac{4 \cos 80^{\circ} \cos 40^{\circ}\left(2 \cos 20^{\circ} \sin 20^{\circ}\right)}{\sin 2... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.348. $\operatorname{tg} 9^{\circ}+\operatorname{tg} 15^{\circ}-\operatorname{tg} 27^{\circ}-\operatorname{ctg} 27^{\circ}+\operatorname{ctg} 9^{\circ}+\operatorname{ctg} 15^{\circ}=8$. | ## Solution.
$$
\begin{aligned}
& \operatorname{tg} 9^{\circ}+\operatorname{tg} 15^{\circ}-\operatorname{tg} 27^{\circ}-\operatorname{ctg} 27^{\circ}+\operatorname{ctg} 9^{\circ}+\operatorname{ctg} 15^{\circ}= \\
& =\left(\operatorname{tg} 9^{\circ}+\operatorname{ctg} 9^{\circ}\right)+\left(\operatorname{tg} 15^{\circ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.349. $\frac{\sin \left(\alpha-\frac{3}{2} \pi\right) \tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}{1+\cos \left(\alpha-\frac{5}{2} \pi\right)}=1$. | Solution.
$$
\begin{aligned}
& \frac{\sin \left(\alpha-\frac{3}{2} \pi\right) \tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}{1+\cos \left(\alpha-\frac{5}{2} \pi\right)}=\frac{-\sin \left(\frac{3}{2} \pi-\alpha\right) \tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)}{1+\cos \left(\frac{5}{2} \pi-\alpha\right)}= \\
& =... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.353. $\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=4$. | Solution.
$$
\begin{aligned}
& \frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}=\frac{2\left(\frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ}\right)}{\sin 10^{\circ} \cos 10^{\circ}}= \\
& =\frac{2 \cdot 2\left(\si... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.358. $\frac{\cos 68^{\circ} \cos 8^{\circ}-\cos 82^{\circ} \cos 22^{\circ}}{\cos 53^{\circ} \cos 23^{\circ}-\cos 67^{\circ} \cos 37^{\circ}}$. | ## Solution.
$$
\begin{aligned}
& \frac{\cos 68^{\circ} \cos 8^{\circ}-\cos 82^{\circ} \cos 22^{\circ}}{\cos 53^{\circ} \cos 23^{\circ}-\cos 67^{\circ} \cos 37^{\circ}}= \\
& =\frac{\cos 68^{\circ} \cos 8^{\circ}-\cos \left(90^{\circ}-8^{\circ}\right) \cos \left(90^{\circ}-68^{\circ}\right)}{\cos 53^{\circ} \cos 23^{\... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.359. $\frac{\cos 70^{\circ} \cos 10^{\circ}+\cos 80^{\circ} \cos 20^{\circ}}{\cos 69^{\circ} \cos 9^{\circ}+\cos 81^{\circ} \cos 21^{\circ}}$. | ## Solution.
$$
\frac{\cos 70^{\circ} \cos 10^{\circ}+\cos 80^{\circ} \cos 20^{\circ}}{\cos 69^{\circ} \cos 9^{\circ}+\cos 81^{\circ} \cos 21^{\circ}}=
$$
$$
\begin{aligned}
& =\frac{\cos \left(90^{\circ}-20^{\circ}\right) \cos 10^{\circ}+\cos \left(90^{\circ}-10^{\circ}\right) \cos 20^{\circ}}{\cos \left(90^{\circ}-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.360. $\frac{\cos 67^{\circ} \cos 7^{\circ}-\cos 83^{\circ} \cos 23^{\circ}}{\cos 128^{\circ} \cos 68^{\circ}-\cos 38^{\circ} \cos 22^{\circ}}-\operatorname{tg} 164^{\circ}$.
| ## Решение.
$$
\frac{\cos 67^{\circ} \cos 7^{\circ}-\cos 83^{\circ} \cos 23^{\circ}}{\cos 128^{\circ} \cos 68^{\circ}-\cos 38^{\circ} \cos 22^{\circ}}-\operatorname{tg} 164^{\circ}=
$$
$$
=\frac{\cos 67^{\circ} \cos \left(90^{\circ}-83^{\circ}\right)-\cos 83^{\circ} \cos \left(90^{\circ}-67^{\circ}\right)}{\cos \left... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.361. $\frac{\sin 22^{\circ} \cos 8^{\circ}+\cos 158^{\circ} \cos 98^{\circ}}{\sin 23^{\circ} \cos 7^{\circ}+\cos 157^{\circ} \cos 97^{\circ}}$. | Solution.
$$
\begin{aligned}
& \frac{\sin 22^{\circ} \cos 8^{\circ}+\cos 158^{\circ} \cos 98^{\circ}}{\sin 23^{\circ} \cos 7^{\circ}+\cos 157^{\circ} \cos 97^{\circ}}= \\
& =\frac{\sin 22^{\circ} \cos 8^{\circ}+\cos \left(180^{\circ}-22^{\circ}\right) \cos \left(90^{\circ}+8^{\circ}\right)}{\sin 23^{\circ} \cos 7^{\ci... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.374. $\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{tg} \frac{C}{2} \operatorname{tg} \frac{A}{2}=1$.
3.374. $\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$. | Solution.
$$
\begin{aligned}
& \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = \tan \frac{A}{2} \tan \frac{B}{2} + \left( \tan \frac{B}{2} + \tan \frac{A}{2} \right) \tan \frac{C}{2} = \\
& = \tan \frac{A}{2} \tan \frac{B}{2} + \left( \tan \frac{B}{2} + \tan ... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
3.384. Prove that the expression $\frac{1-2 \sin ^{2}\left(\alpha-\frac{3}{2} \pi\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3}{2} \pi\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}$
does not depend on $\alpha$, where $\alpha \neq \frac{\pi n}{2}+\frac{\pi}{12}$. | ## Solution.
$$
\begin{aligned}
& \frac{1-2 \sin ^{2}\left(\alpha-\frac{3}{2} \pi\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3}{2} \pi\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}= \\
& =\frac{1-2\left(-\sin \left(\frac{3}{2} \pi-\alpha\right)\right)^{2}+\sqrt{3} \cos \left(\frac{3}{2} \pi+2 \alpha\right)}{\sin \l... | -2 | Algebra | proof | Yes | Yes | olympiads | false |
3.389. Prove that the expression $\cos ^{2} \alpha+\cos ^{2} \varphi+\cos ^{2}(\alpha+\varphi)-$ $2 \cos \alpha \cos \varphi \cos (\alpha+\varphi)$ does not depend on either $\alpha$ or $\varphi$. | ## Solution.
$\cos ^{2} \alpha+\cos ^{2} \varphi+\cos ^{2}(\alpha+\varphi)-2 \cos \alpha \cos \varphi \cos (\alpha+\varphi)=$ $=\cos ^{2} \alpha+\cos ^{2} \varphi+(\cos (\alpha+\varphi))^{2}-2 \cos \alpha \cos \varphi \times$
$x(\cos \alpha \cos \varphi-\sin \alpha \sin \varphi)=\cos ^{2} \alpha+\cos ^{2} \varphi+\cos... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
4.041. Find the positive integer $n$ from the equation
$$
(3+6+9+\ldots+3(n-1))+\left(4+5.5+7+\ldots+\frac{8+3 n}{2}\right)=137
$$ | ## Solution.
In the first parentheses, there is the sum of the terms of an arithmetic progression $S_{k}$ where $a_{1}=3, d=3, a_{k}=3(n-1), k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-3-3}{3}+1=n-1$; in the second parentheses, there is the sum of the terms of an arithmetic progression where $b_{1}=4, d=1.5, a_{m}=\frac{8+3 n... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.045. The sum of the first three terms of a geometric progression is 91. If 25, 27, and 1 are added to these terms respectively, the resulting three numbers form an arithmetic progression. Find the seventh term of the geometric progression. | ## Solution.
From the condition, we have:
$b_{1}, b_{1} q, b_{1} q^{2}$ - terms of a geometric progression,
$b_{1}+25, b_{1} q+27, b_{1} q^{2}$ - terms of an arithmetic progression, then we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 9 1 , } \\
{ 2 ( b _ { 1 } q + 2 7 ) = b ... | 5103 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.050. Find the sum of all three-digit numbers divisible by 7. | Solution.
We have:
$$
\left\{\begin{array}{l}
a_{1}=105 \\
a_{n}=994, \\
d=7
\end{array}\right.
$$
Answer: 70336. | 70336 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.056. Find the fifth term of an increasing geometric progression, given that its first term is equal to $7-3 \sqrt{5}$ and that each of its terms, starting from the second, is equal to the difference of the two adjacent terms. | Solution.
We have:
$$
\left\{\begin{array}{l}
b_{1}=7-3 \sqrt{5} \\
b_{2}=b_{3}-b_{1}, \\
|q|>1
\end{array}\right.
$$
$q_{1}=\frac{1-\sqrt{5}}{2}$ does not fit, as $\left|q_{1}\right|<1 ; q_{2}=\frac{1+\sqrt{5}}{2}$.
From this,
$$
b_{5}=b_{1} q^{4}=\frac{(7-3 \sqrt{5})(1+\sqrt{5})^{4}}{16}=\frac{(17-3 \sqrt{5}) 8(... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.061. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x}+\ldots+\frac{1}{x}=3
$$
where $x$ is a positive integer. | ## Solution.
Multiplying both sides of the equation by $x$, we have
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1, a_{n}=1$,
$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{1-x+1}{-1}+1=x-1 . \text { Since } S_{n}=\frac{a_{1}+a_{n... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.063. Given two geometric progressions consisting of the same number of terms. The first term and the common ratio of the first progression are 20 and $3 / 4$, respectively, while the first term and the common ratio of the second progression are 4 and $2 / 3$, respectively. If the terms of these progressions with the ... | ## Solution.
Let $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ be the terms of two geometric progressions, where $a_{1}=20, q_{1}=\frac{3}{4}; b_{1}=4, q_{2}=\frac{2}{3}$ and $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\ldots=$ $=158.75$. We find $a_{2}=a_{1} q_{1}=20 \cdot \frac{3}{4}=15, a_{3}=a_{1} q_{1}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
6.136. $\frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
(Note: The equation is the same in both languages, so the translation is identical to the original text.) | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq 2 .\end{array}\right.$
After bringing all terms of the equation to a common denominator, we have
$$
\begin{aligned}
& \frac{\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)+2(x+1)(x-2)}{(x+1)(x-2)}=0 \Leftrightarrow \\
& \Leftrightarro... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.145. $(x+1)^{2}(x+2)+(x-1)^{2}(x-2)=12$. | ## Solution.
We have:
$$
\begin{aligned}
& \left(x^{2}+2 x+1\right)(x+2)+\left(x^{2}-2 x+1\right)(x-2)-12=0 \Leftrightarrow \\
& \Leftrightarrow x^{3}+2 x^{2}+2 x^{2}+4 x+x+2+x^{3}-2 x^{2}-2 x^{2}+4 x+x-2-12=0 \\
& 2 x^{3}+10 x-12=0, x^{3}+5 x-6=0
\end{aligned}
$$
The last equation can be rewritten as:
$x^{3}+5 x-5... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.146. $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-1, \\ x \neq-2, \\ x \neq-3, \\ x \neq-4 .\end{array}\right.$
$$
\begin{aligned}
& \text { From the condition } \frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}-1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(x-1)(x-4)(x-2)(x-3)-(x+1)(x+4)(x+2)(x+3)}{(x+... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.158. $\sqrt[3]{9-\sqrt{x+1}}+\sqrt[3]{7+\sqrt{x+1}}=4$. | ## Solution.
Domain of definition: $x+1 \geq 0$ or $x \geq-1$.
Let $\sqrt{x+1}=y \geq 0$. The equation in terms of $y$ becomes $\sqrt[3]{9-y}+\sqrt[3]{7+y}=4$. Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 9-y+\sqrt[3]{(9-y)^{2}(7+y)}+3 \sqrt[3]{(9-y)(7+y)^{2}}+7+y=64 \Leftrighta... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
6.159. $\sqrt{x+2}-\sqrt[3]{3 x+2}=0$. | ## Solution.
Domain of definition: $x+2 \geq 0, x \geq-2$.
$$
\begin{aligned}
& \text { From the condition } \sqrt{x+2}=\sqrt[3]{3 x+2} \Rightarrow(x+2)^{3}=(3 x+2)^{2} \\
& x^{3}+6 x^{2}+12 x+8=9 x^{2}+12 x+4 \Leftrightarrow x^{3}-3 x^{2}+4=0 \Leftrightarrow \\
& \Leftrightarrow(x+1)\left(x^{2}-x+1\right)-3(x-1)(x+1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.160. $\sqrt{\frac{20+x}{x}}+\sqrt{\frac{20-x}{x}}=\sqrt{6}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{20+x}{x} \geq 0, \\ \frac{20-x}{x} \geq 0,\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x+20) \geq 0, \\ x(x-20) \geq 0, \\ x \neq 0,0<x \leq 20 .\end{array}\right.\right.$
By squaring both sides of the equation, we have
$$
\begin{aligned}
& \frac{2... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.164. $\sqrt{x+8+2 \sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4$. | Solution.
Let $\sqrt{x+7}=y \geq 0, x+7=y^{2}, x=y^{2}-7$.
With respect to $y$, the equation takes the form
$$
\begin{aligned}
& \sqrt{y^{2}+2 y+1}+\sqrt{y^{2}-y-6}=4, \sqrt{(y+1)^{2}}+\sqrt{y^{2}-y-6}=4 \\
& |y+1|+\sqrt{y^{2}-y-6}=4
\end{aligned}
$$
Since $y \geq 0$, we have $y+1+\sqrt{y^{2}-y-6}=4, \sqrt{y^{2}-y-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$.
6.169. $\sqrt{x}+\frac{2 x+1}{x+2}=2$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition, we have $\sqrt{x}=2-\frac{2 x+1}{x+2} \Leftrightarrow \sqrt{x}=\frac{2 x+4-2 x-1}{x+2}, \sqrt{x}=\frac{3}{x+2}$.
Squaring both sides of the equation, we get $x=\frac{3}{x^{2}+4 x+4} \Leftrightarrow$ $\Leftrightarrow \frac{x^{3}+4 x^{2}+4 x-9}{x^{2}+4... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.170. $\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^{2}-16}-6$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+4 \geq 0, \\ x-4 \geq 0,\end{array} \Leftrightarrow x \geq 4\right.$.
By squaring both sides of the equation, we get
$$
\begin{aligned}
& \frac{x+4+2 \sqrt{(x+4)(x-4)}+x-4}{4}=\left(x+\sqrt{x^{2}-16}\right)^{2}-12\left(x+\sqrt{x^{2}-16}\right)+36 \\
& 2\le... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.