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## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x \cdot$
$$
x^{2}+(y-2)^{2}=1
$$ | ## Solution
$$
\begin{aligned}
& x^{2}+(y-2)^{2}=1 \\
& (y-2)^{2}=1-x^{2} \\
& y-2= \pm \sqrt{1-x^{2}} \\
& y= \pm \sqrt{1-x^{2}}+2
\end{aligned}
$$
Thus, we have two functions:
$$
y=\sqrt{1-x^{2}}+2
$$
and
$$
y=-\sqrt{1-x^{2}}+2=2-\sqrt{1-x^{2}}
$$
Since \( x^{2}+(y-2)^{2}=1 \) is the equation of a circle in the... | 4\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,587 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
y=x^{2}, y=1, x=2
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration: From the problem statement, we already have: $x_{2}=2$ Now let's find the lower limit:
$$
y=x^{2}=1 \Rightarrow x_{1}=1
$$
We will find the... | \frac{26\pi}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,589 |
## Problem Statement
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.
$$
y=x^{3}, y=\sqrt{x}
$$ | ## Solution
Since the $O x$ axis is the axis of rotation, the volume is found using the formula from Integrals $21-15$:
$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration:
$$
V=\pi \int_{a}^{b} y^{2} d x
$$
Let's find the limits of integration:
$$
y=x^{2} \sin \lef... | 0.3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,591 |
## Problem Statement
Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O y$.
$$
y=\arccos \frac{x}{3}, y=\arccos x, y=0
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume $\quad$ Integrals 21-17 is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits ... | 6\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,593 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=x^{2}, x=2, y=0
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
$$
y=x^{2} \Rightarrow x=\sqrt{y}
$$
From the problem statement, we already have: $y_{1}=0$.
Now find the upper ... | 8\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,594 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=\sqrt{x-1}, y=0, y=1, x=0.5
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,597 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=(x-1)^{2}, y=1
$$ | ## Solution
Since the Oy axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express \( x \) in terms of \( y \) and find the li... | \frac{8\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,598 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y^{2}=x-2, y=0, y=x^{3}, y=1
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume is found using the formula:
Integrals $21-24$
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
$$
\begin{aligned}
& y^{2}=x-2 \Rightarrow x=y^{2}+2 \\
& y=x^{3} \Rightarrow x=\sqrt[3]{y}
\en... | \frac{24}{5}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,599 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=x^{3}, y=x^{2}
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume
Integrals $21-25$ are found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
$$
\begin{aligned}
& y=x^{2} \Rightarrow x=\sqrt{y} \\
& y=x^{3} \Rightarrow x=\sqrt[3]{y}
\en... | \frac{\pi}{10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,600 |
## Problem Statement
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=\arccos \frac{x}{5}, y=\arccos \frac{x}{3}, y=0
$$ | ## Solution
Since
the $O y$
axis
is
the axis
of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Let's find the limits of integration:
$$
\... | 4\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,601 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=\arcsin x, y=\arccos x, y=0
$$ | ## Solution
$$
V_{y}=\pi \int_{0}^{\frac{\pi}{4}}\left(\cos ^{2} y-\sin ^{2} y\right) d y=\frac{1}{2} \pi \int_{0}^{\frac{\pi}{4}} \cos 2 y d(2 y)=\left.\frac{1}{2} \pi \sin 2 y\right|_{0}... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,602 |
## Task Condition
Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O y$.
$$
y=x^{2}-2 x+1, x=2, y=0
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
^{2}, x=0, x=2, y=0
$$ | ## Solution
Since the $O y$ axis is the axis of rotation, the volume is found using the formula:
$$
V=\pi \cdot \int_{a}^{b} x^{2} d y
$$
Express $x$ in terms of $y$ and find the limits of integration:
=-2 c_{2}$ for any $a$ and $b$.
That is, $c_{1}=-2 \cdot c_{2}$, which means the vectors $c_{1}... | c_{1}=-2\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,608 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 0 ; 4), B(-3 ;-6 ; 1), C(-5 ;-10 ;-1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-3-0 ;-6-0 ; 1-4)=(-3 ;-6 ;-3)$
$\overrightarrow{A C}=(-5-0 ;-10-0 ;-1-4)=(-5 ;-10 ;-5)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overri... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,609 |
## problem statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p-q$
$b=p+2 q$
$|p|=3$
$|q|=4$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p-q) \times(p+2 q)=3 \cdot p \times p+3 \cdot 2 \cdot p \times q-q \times p-2 ... | 42\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,610 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{2 ; 3 ; 2\}$
$b=\{4 ; 7 ; 5\}$
$c=\{2 ; 0 ;-1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
2 & 3 & 2 \\
4 & 7 & 5 \\
2 & 0 & -1
\end{array}\right|= \\
& =2 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,611 |
## 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 | 47,612 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ;-4 ;-3)$
$M_{2}(5 ;-6 ; 0)$
$M_{3}(-1 ; 3 ;-3)$
$M_{0}(2 ;-10 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-2 & y-(-4) & z-(-3) \\ 5-2 & -6-(-4) & 0-(-3) \\ -1-2 & 3-(-4) & -3-(-3)\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y+4 & z+3 \\
3 & -2 & 3... | \frac{73}{\sqrt{83}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,613 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(0; -2; 8)$
$B(4; 3; 2)$
$C(1; 4; 3)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$\overrightarrow{B C}=\{1-4 ; 4-3 ; 3-2\}=\{-3 ; 1 ; 1\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -3 \cdot(x-0)+(y... | -3x+y+z-6=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,614 |
## Problem Statement
Find the angle between the planes:
$$
\begin{aligned}
& 2 x+2 y+z+9=0 \\
& x-y+3 z-1=0
\end{aligned}
$$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{2 ; 2 ; 1\}$
$\overrightarrow{n_{2}}=\{1 ;-1 ; 3\}$
The angle $\phi$ between the planes is determined by the formula:
$\cos \phi=\frac{\left(\overr... | \arccos\frac{1}{\sqrt{11}}\approx7227^{\}6^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,615 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(-2 ; 8 ; 10)$
$C(6 ; 11 ;-2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(8-y)^{2}+(10-0)^{2}}=\sqrt{4+64-16 y+y^{2}+100}=\sqrt{y^{2}-16 y+168} \\
& A C=\sqrt{(6-0)^{2}+(11-y)^{2}+(-2-0)^{2}}=\sqrt{36+121-22 y+y^{2}+4}=\sqrt{y^{2}-22 y+161}
\end{aligned}
$$
Since by the condition of the proble... | A(0;-\frac{7}{6};0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,616 |
## 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 | 47,617 |
## Task Condition
Write the canonical equations of the line.
$8 x-y-3 z-1=0$
$x+y+z+10=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+1}{2}=\frac{y+9}{-11}=\frac{z}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,618 |
Condition of the problem
Find the point of intersection of the line and the plane.
$\frac{x+3}{1}=\frac{y-2}{-5}=\frac{z+2}{3}$
$5 x-y+4 z+3=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+3}{1}=\frac{y-2}{-5}=\frac{z+2}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3+t \\
y=2-5 t \\
z=-2+3 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$5(-3+t)-(2-5 t)+4(-2+3 t)+3=0$
$-15+... | (-2,-3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,619 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(-1 ; 0 ; 1)$
$\frac{x+0.5}{0}=\frac{y-1}{0}=\frac{z-4}{2}$ | ## Solution
Find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as the normal vector of the plane:
$\vec{n}=\vec{s}=\{0 ; 0 ; 2\}$
Then the equation of the desired plane... | M^{\}(0;2;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,620 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-2 ; 4 ; 7\}$
$p=\{0 ; 1 ; 2\}$
$q=\{1 ; 0 ; 1\}$
$r=\{-1 ; 2 ; 4\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p-q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,621 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ;-2 ; 3\}$
$b=\{3 ; 0 ;-1\}$
$c_{1}=2 a+4 b$
$c_{2}=3 b-a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$c_{1}=2 a+4 b=\{2 \cdot 1+4 \cdot 3 ; 2 \cdot(-2)+4 \cdot 0 ; 2 \cdot 3+4 \cdot(-1)\}=\{14 ;-4 ; 2\}$
$c_{2}=3 b-a=\{3 \cdot 3-1 ;... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,622 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(1, -2, 3), B(0, -1, 2), C(3, -4, 5)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-1 ;-1-(-2) ; 2-3)=(-1 ; 1 ;-1)$
$\overrightarrow{A C}=(3-1 ;-4-(-2) ; 5-3)=(2 ;-2 ; 2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{alig... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,623 |
## 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:
$$
\begin{aligned}
& a \times b=(p+2 q) \times(3 p-q)=3 \cdot p \times p-p \times q+2 \... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,624 |
## Condition of the problem
Are the vectors $a, b$ and $c$ coplanar?
$$
\begin{aligned}
& a=\{2 ; 3 ; 1\} \\
& b=\{-1 ; 0 ;-1\} \\
& c=\{2 ; 2 ; 2\}
\end{aligned}
$$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
2 & 3 & 1 \\
-1 & 0 & -1 \\
2 & 2 & 2
\end{array}\right|= \\
& =2 \cdot\left|\b... | 2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,625 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 3 ; 6) \)
\( A_{2}(2 ; 2 ; 1) \)
\( A_{3}(-1 ; 0 ; 1) \)
\( A_{4}(-4 ; 6 ;-3) \) | ## Solution
From vertex $A_{1 \text { we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{2-1 ; 2-3 ; 1-6\}=\{1 ;-1 ;-5\}$
$\overrightarrow{A_{1} A_{3}}=\{-1-1 ; 0-3 ; 1-6\}=\{-2 ;-3 ;-5\}$
$\overrightarrow{A_{1} A_{4}}=\{-4-1 ; 6-3 ;-3-6\}=\{-5 ; 3 ;-9\}$
According to the geometric meaning of the scalar triple pr... | 23\frac{1}{3},2\sqrt{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,626 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-3 ; 4 ;-7)$
$M_{2}(1 ; 5 ;-4)$
$M_{3}(-5 ;-2 ; 0)$
$M_{0}(-12 ; 7 ;-1)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-(-3) & y-4 & z-(-7) \\ 1-(-3) & 5-4 & -4-(-7) \\ -5-(-3) & -2-4 & 0-(-7)\end{array}\right|=0$
Perform transformations:
$\left|\begin{array}{ccc}x+3 & y-4 & z+7 \\ 4 & 1 & 3 \\ -2 & -6 & 7\end{ar... | \frac{459}{\sqrt{2265}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,627 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(1 ; 0 ; -2)$
$B(2 ; -1 ; 3)$
$C(0 ; -3 ; 2)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{0-2; -3-(-1); 2-3\}=\{-2; -2; -1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-2 \cdot (x-1) + (-2) \cdot (y-0) - (... | 2x+2y+z=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,628 |
## Task Condition
Find the angle between the planes:
$x-3 y+5=0$
$2 x-y+5 z-16=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-3 ; 0\}$
$\overrightarrow{n_{2}}=\{2 ;-1 ; 5\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | \arccos\frac{\sqrt{3}}{6}\approx73^{0}13^{\}17^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,629 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(5 ; 1 ; 0)$
$C(0 ; 2 ; 3)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(5-0)^{2}+(1-0)^{2}+(0-z)^{2}}=\sqrt{25+1+z^{2}}=\sqrt{z^{2}+26}$
$A C=\sqrt{(0-0)^{2}+(2-0)^{2}+(3-z)^{2}}=\sqrt{0+4+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+13}$
Since by the condition of the problem $A B=A C$, then
$\sqrt{z^{2}+26}=\sqrt{z^{2}-6 z+13}$
$z^{2... | A(0;0;-2\frac{1}{6}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,630 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(1 ; 2 ;-1)$
$a: 2x + 3y + z - 1 = 0$
$k=2$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a:$
$a^{\prime}: 2 x+3 y+z-2=0$
Substitute the coordinates of point $A$ into the equat... | 5\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,631 |
## Task Condition
Write the canonical equations of the line.
$2 x+y+z-2=0$
$2 x-y-3 z+6=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text { - coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vec... | \frac{x+1}{-2}=\frac{y-4}{8}=\frac{z}{-4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,632 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-2}{-1}=\frac{y-3}{-1}=\frac{z+1}{4} \\
& x+2 y+3 z-14=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\frac{x-2}{-1}=\frac{y-3}{-1}=\frac{z+1}{4}=t \Rightarrow
$$
$$
\left\{\begin{array}{l}
x=2-t \\
y=3-t \\
z=-1+4 t
\end{array}\right.
$$
Substitute into the equation of the plane:
$$
\begin{aligned}
& (2-t)+2(3-t)+3(-1+4 t)-14=0 \\
& 2-t+6-2 t-3+12 t... | (1;2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,633 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(0 ;-3 ;-2) \\
& \frac{x-1}{1}=\frac{y+1.5}{-1}=\frac{z}{1}
\end{aligned}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{1 ;-1 ; 1\}$
Then the equation of the desired plane is:
$$
... | M^{\}(1;1;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,634 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+\sqrt{x}}}{x \sqrt[4]{x^{3}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+\sqrt{x}}}{x \sqrt[4]{x^{3}}} d x=\int x^{-\frac{7}{4}}\left(1+x^{\frac{1}{2}}\right)^{\frac{1}{2}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{7}{4} ; n=\frac{1}{2} ; p=\frac{1}{2}
$$
Since $\frac{m+1}{n}+p=\frac{... | -\frac{4\sqrt{(1+\sqrt{x})^{3}}}{3\sqrt[4]{x}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,635 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt{x}}}{x \sqrt[3]{x^{2}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{1+\sqrt{x}}}{x \sqrt[3]{x^{2}}} d x=\int x^{-\frac{5}{3}}\left(1+x^{\frac{1}{2}}\right)^{\frac{1}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{5}{3} ; n=\frac{1}{2} ; p=\frac{1}{3}
$$
Since $\frac{m+1}{n}+p=\fr... | -\frac{3\sqrt[3]{(1+\sqrt{x})^{4}}}{2\sqrt[3]{x^{2}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,636 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+\sqrt[3]{x}}}{x \sqrt{x}} d x=\int x^{-\frac{3}{2}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{2}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{3}{2} ; n=\frac{1}{3} ; p=\frac{1}{2}
$$
Since $\frac{m+1}{n}+p=\frac{-\fr... | -2\sqrt{\frac{(1+\sqrt[3]{x})^{3}}{x}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,637 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x^{2}}}}{x \sqrt[9]{x^{8}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x^{2}}}}{x \sqrt[9]{x^{8}}} d x=\int x^{-\frac{17}{9}}\left(1+x^{\frac{2}{3}}\right)^{\frac{1}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{17}{9} ; n=\frac{2}{3} ; p=\frac{1}{3}
$$
Since $\frac{m+1}... | -\frac{9}{8}(\sqrt[3]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}})^{4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,639 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{(1+\sqrt[3]{x})^{2}}}{x \sqrt[9]{x^{5}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{(1+\sqrt[3]{x})^{2}}}{x \sqrt[9]{x^{5}}} d x=\int x^{-\frac{14}{9}}\left(1+x^{\frac{1}{3}}\right)^{\frac{2}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{14}{9} ; n=\frac{1}{3} ; p=\frac{2}{3}
$$
Since $\frac{m+... | -\frac{9}{5}(\sqrt[3]{\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}})^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,640 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[3]{x^{2}}\right)^{2}}}{x^{2} \cdot \sqrt[9]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[3]{x^{2}}\right)^{2}}}{x^{2} \cdot \sqrt[9]{x}} d x=\int x^{-\frac{19}{9}}\left(1+x^{\frac{2}{3}}\right)^{\frac{2}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{19}{9} ; n=\frac{2}{3} ; p=\frac{2}{3... | -\frac{9}{10}(\sqrt[3]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}})^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,641 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{(1+\sqrt{x})^{2}}}{x \sqrt[6]{x^{5}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{(1+\sqrt{x})^{2}}}{x \sqrt[6]{x^{5}}} d x=\int x^{-\frac{11}{6}}\left(1+x^{\frac{1}{2}}\right)^{\frac{2}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{11}{6} ; n=\frac{1}{2} ; p=\frac{2}{3}
$$
Since $\frac{m+1}{... | -\frac{6}{5}(\sqrt[3]{\frac{1+\sqrt{x}}{\sqrt{x}}})^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,642 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+\sqrt[3]{x^{2}}}}{x^{2}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+\sqrt[3]{x^{2}}}}{x^{2}} d x=\int x^{-2}\left(1+x^{\frac{2}{3}}\right)^{\frac{1}{2}} d x=
$$
Under the integral is a differential binomial $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-2 ; n=\frac{2}{3} ; p=\frac{1}{2}
$$
Since $\frac{m+1}{n}+p=\frac{-2+1}{\frac{2}{3}}+\frac{... | \frac{\sqrt{(1+\sqrt[3]{x^{2}})^{3}}}{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,643 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+x}}{x^{2} \cdot \sqrt{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+x}}{x^{2} \cdot \sqrt{x}} d x=\int x^{-\frac{5}{2}}(1+x)^{\frac{1}{2}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{5}{2} ; n=1 ; p=\frac{1}{2}
$$
Since $\frac{m+1}{n}+p=\frac{-\frac{5}{2}+1}{1}+\frac{1}{2}=-\frac{3... | -\frac{2}{3}(\sqrt{\frac{1+x}{x}})^{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,644 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[4]{(1+\sqrt{x})^{3}}}{x \sqrt[8]{x^{7}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[4]{(1+\sqrt{x})^{3}}}{x \sqrt[8]{x^{7}}} d x=\int x^{-\frac{15}{8}}\left(1+x^{\frac{1}{2}}\right)^{\frac{3}{4}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{15}{8} ; n=\frac{1}{2} ; p=\frac{3}{4}
$$
Since $\frac{m+1}{... | -\frac{8}{7}(\sqrt[4]{\frac{1+\sqrt{x}}{\sqrt{x}}})^{7}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,645 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[4]{(1+\sqrt[3]{x})^{3}}}{x \sqrt[12]{x^{7}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[4]{(1+\sqrt[3]{x})^{3}}}{x \sqrt[12]{x^{7}}} d x=\int x^{-\frac{19}{12}}\left(1+x^{\frac{1}{3}}\right)^{\frac{3}{4}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{19}{12} ; n=\frac{1}{3} ; p=\frac{3}{4}
$$
Since $\frac... | -\frac{12}{7}\cdot\sqrt[4]{(\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}})^{7}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,646 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{3}}}{x^{2} \cdot \sqrt[6]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[4]{\left(1+\sqrt[3]{x^{2}}\right)^{3}}}{x^{2} \cdot \sqrt[6]{x}} d x=\int x^{-\frac{13}{6}}\left(1+x^{\frac{2}{3}}\right)^{\frac{3}{4}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{13}{6} ; n=\frac{2}{3} ; p=\frac{3}{4... | -\frac{6\cdot\sqrt[4]{(1+\sqrt[3]{x^{2}})^{7}}}{7\sqrt[6]{x^{7}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,647 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+\sqrt[4]{x^{3}}}}{x^{2} \cdot \sqrt[8]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+\sqrt[4]{x^{3}}}}{x^{2} \cdot \sqrt[8]{x}} d x=\int x^{-\frac{17}{8}}\left(1+x^{\frac{3}{4}}\right)^{\frac{1}{2}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{17}{8} ; n=\frac{3}{4} ; p=\frac{1}{2}
$$
Since $\frac{m... | -\frac{8}{9}\sqrt{(\frac{1+\sqrt[4]{x^{3}}}{\sqrt[4]{x^{3}}})^{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,648 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x^{3}}}}{x^{2}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x^{3}}}}{x^{2}} d x=\int x^{-2}\left(1+x^{\frac{3}{4}}\right)^{\frac{1}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-2 ; n=\frac{3}{4} ; p=\frac{1}{3}
$$
Since $\frac{m+1}{n}+p=\frac{-2+1}{\frac{3}{4}}+\fr... | -\frac{\sqrt[3]{(1+\sqrt[4]{x^{3}})^{4}}}{x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,649 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[4]{x^{3}}\right)^{2}}}{x^{2} \cdot \sqrt[4]{x}} d x
$$ | ## Solution
Let's represent the integral as a differential binomial:
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[4]{x^{3}}\right)^{2}}}{x^{2} \cdot \sqrt[4]{x}} d x=\int x^{-9 / 4}\left(1+x^{3 / 4}\right)^{2 / 3} d x
$$
Perform a change of variable:
$$
\begin{aligned}
& x=\left(t^{3 / 2}-1\right)^{-4 / 3} \Rightarrow t=\l... | -\frac{4}{5}(x^{-3/4}+1)^{5/3}+ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,650 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[5]{(1+\sqrt{x})^{4}}}{x \sqrt[10]{x^{9}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[8]{(1+\sqrt{x})^{4}}}{x \sqrt[10]{x^{9}}} d x=\int x^{-\frac{19}{10}}\left(1+x^{\frac{1}{2}}\right)^{\frac{4}{8}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{19}{10} ; n=\frac{1}{2} ; p=\frac{4}{5}
$$
Since $\frac{m+... | -\frac{10}{9}(\sqrt[5]{\frac{1+\sqrt{x}}{\sqrt{x}}})^{9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,651 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[5]{(1+\sqrt[3]{x})^{4}}}{x \sqrt[5]{x^{3}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[5]{(1+\sqrt[3]{x})^{4}}}{x \sqrt[5]{x^{3}}} d x=\int x^{-\frac{8}{5}}\left(1+x^{\frac{1}{3}}\right)^{\frac{4}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{8}{5} ; n=\frac{1}{3} ; p=\frac{4}{5}
$$
Since $\frac{m+1}... | -\frac{5}{3}(\sqrt[5]{\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}})^{9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,652 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[5]{\left(1+\sqrt[3]{x^{2}}\right)^{4}}}{x^{2} \cdot \sqrt[5]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[5]{\left(1+\sqrt[3]{x^{2}}\right)^{4}}}{x^{2} \cdot \sqrt[5]{x}} d x=\int x^{-\frac{11}{5}}\left(1+x^{\frac{2}{3}}\right)^{\frac{4}{5}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{11}{5} ; n=\frac{2}{3} ; p=\frac{4}{5... | -\frac{5}{6}(\sqrt[5]{\frac{1+\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}})^{9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,653 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[5]{\left(1+\sqrt[4]{x^{3}}\right)^{4}}}{x^{2} \cdot \sqrt[20]{x^{7}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[5]{\left(1+\sqrt[4]{x^{3}}\right)^{4}}}{x^{2} \cdot \sqrt[20]{x^{7}}} d x=\int x^{-\frac{47}{20}}\left(1+x^{\frac{3}{4}}\right)^{\frac{4}{5}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{47}{20} ; n=\frac{3}{4} ; p=\fr... | -\frac{20}{27}(\sqrt[5]{\frac{1+\sqrt[4]{x^{3}}}{\sqrt[4]{x^{3}}}})^{9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,654 |
## Condition of the problem
Find the indefinite integral:
$$
\int \frac{\sqrt[5]{\left(1+\sqrt[5]{x^{4}}\right)^{4}}}{x^{2} \cdot \sqrt[25]{x^{11}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[5]{x^{4}}\right)^{4}}}{x^{2} \cdot \sqrt[25]{x^{11}}} d x=\int x^{-\frac{61}{25}}\left(1+x^{\frac{4}{5}}\right)^{\frac{4}{5}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{61}{25} ; n=\frac{4}{5} ; p=\f... | -\frac{25}{36}(\sqrt[5]{\frac{1+\sqrt[5]{x^{4}}}{\sqrt[5]{x^{4}}}})^{9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,655 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[5]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[5]{x}} d x=\int x^{-\frac{11}{5}}\left(1+x^{\frac{4}{5}}\right)^{\frac{1}{2}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{11}{5} ; n=\frac{4}{5} ; p=\frac{1}{2}
$$
Since $\frac{m... | -\frac{5}{6}\cdot\frac{0\sqrt{(1+\sqrt[5]{x^{4}}.})^{3}}{x\sqrt[5]{x}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,656 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[15]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x^{4}}}}{x^{2} \cdot \sqrt[15]{x}} d x=\int x^{-\frac{31}{15}}\left(1+x^{\frac{4}{5}}\right)^{\frac{1}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{31}{15} ; n=\frac{4}{5} ; p=\frac{1}{3}
$$
Since $\... | -\frac{15(1+\sqrt[3]{x^{4}})^{\frac{4}{3}}}{16\sqrt[18]{x^{16}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,657 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[3]{x^{4}}\right)^{2}}}{x^{2} \cdot \sqrt[3]{x}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{\left(1+\sqrt[5]{x^{4}}\right)^{2}}}{x^{2} \cdot \sqrt[3]{x}} d x=\int x^{-\frac{7}{3}}\left(1+x^{\frac{4}{5}}\right)^{\frac{2}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{7}{3} ; n=\frac{4}{5} ; p=\frac{2}{3}
... | -\frac{3(1+\sqrt[5]{x^{4}})^{\frac{5}{3}}}{4\sqrt[3]{x^{4}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,658 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{x \sqrt[3]{x}} d x
$$ | ## Solution
## Problem 13.
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{x \sqrt[3]{x}} d x
$$
Represent the integral in the form of a differential binomial:
$$
\int \frac{\sqrt[3]{1+\sqrt[4]{x}}}{x \sqrt[3]{x}} d x=\int x^{-4 / 3}\left(1+x^{1 / 4}\right)^{1 / 3} d x
$$
Perform the substitu... | -3(x^{-1/4}+1)^{4/3}+ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,660 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x=\int x^{-\frac{17}{12}}\left(1+x^{\frac{1}{4}}\right)^{\frac{2}{3}} d x=
$$
By Chebyshev's substitutions, this is the 3rd case. Therefore, we get:
$$
\begin{aligned}
& \int \frac{\sqrt[3]{(1+\sqrt[4]{x})^{2}}}{x \sqrt[12]{x^{5}}} d x=\l... | -\frac{12}{5}\sqrt[3]{(\frac{1+\sqrt[4]{x}}{\sqrt[4]{x}})^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,661 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x}}}{x \sqrt[12]{x^{5}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x}}}{x \sqrt[12]{x^{5}}} d x=\int x^{-\frac{17}{12}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{4}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{17}{12} ; n=\frac{1}{3} ; p=\frac{1}{4}
$$
Since $\frac{m+1}{... | -\frac{12}{5}(\sqrt[4]{\frac{1+\sqrt[3]{x}}{\sqrt[3]{x}}})^{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,662 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x^{2}}}}{x \sqrt[6]{x^{5}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[4]{1+\sqrt[3]{x^{2}}}}{x \sqrt[6]{x^{5}}} d x=\int x^{-\frac{11}{6}}\left(1+x^{\frac{2}{3}}\right)^{\frac{1}{4}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{11}{6} ; n=\frac{2}{3} ; p=\frac{1}{4}
$$
Since $\frac{m+1}... | -\frac{6\cdot\sqrt[4]{(1+\sqrt[3]{x^{2}})^{5}}}{5\sqrt[6]{x^{5}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,663 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x}}}{x \sqrt[15]{x^{4}}} d x
$$ | ## Solution
$$
\int \frac{\sqrt[3]{1+\sqrt[5]{x}}}{x \sqrt[15]{x^{4}}} d x=\int x^{-\frac{19}{15}}\left(1+x^{\frac{1}{5}}\right)^{\frac{1}{3}} d x=
$$
Under the integral is a binomial differential $x^{m}\left(a+b x^{n}\right)^{p}$, from which
$$
m=-\frac{19}{15} ; n=\frac{1}{5} ; p=\frac{1}{3}
$$
Since $\frac{m+1}{... | -\frac{15(1+\sqrt[5]{x})^{4/3}}{4x^{4/15}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,664 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x
$$ | ## Solution
Transform the integrand:
$$
\int \frac{\sqrt[3]{1+\sqrt[3]{x}}}{x \sqrt[5]{x^{2}}} d x=\int \frac{\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{5}}}{x \cdot x^{\frac{2}{3}}} d x=\int x^{-\frac{7}{5}}\left(1+x^{\frac{1}{3}}\right)^{\frac{1}{3}} d x
$$
This integral represents a differential binomial.
Since $\... | -\frac{5}{2}\sqrt[5]{\frac{1+\sqrt[3]{x}}{x^{2}}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,665 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-15 ; 5 ; 6\}$
$p=\{0 ; 5 ; 1\}$
$q=\{3 ; 2 ;-1\}$
$r=\{-1 ; 1 ; 0\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p-4q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,666 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{3 ; 7 ; 0\}$
$b=\{4 ; 6 ;-1\}$
$c_{1}=3 a+2 b$
$c_{2}=5 a-7 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=3 a+2 b=\{3 \cdot 3+2 \cdot 4 ; 3 \cdot 7+2 \cdot 6 ; 3 \cdot 0+2 \cdot(-1)\}=\{17 ; 33 ;-2\} \\
& c_{2}=... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,667 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(7 ; 0 ; 2), B(7 ; 1 ; 3), C(8 ;-1 ; 2)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(7-7 ; 1-0 ; 3-2)=(0 ; 1 ; 1)$
$\overrightarrow{A C}=(8-7 ;-1-0 ; 2-2)=(1 ;-1 ; 0)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarrow{A ... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,668 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=6 p-q$
$b=p+2 q$
$|p|=8$
$|q|=\frac{1}{2}$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(6 p-q) \times(p+2 q)=6 \cdot p \times p+6 \cdot 2 \cdot p \times q-q \times p-2 ... | 26\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,669 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{5 ; 3 ; 4\}$
$b=\{4 ; 3 ; 3\}$
$c=\{9 ; 5 ; 8\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
5 & 3 & 4 \\
4 & 3 & 3 \\
9 & 5 & 8
\end{array}\right|= \\
& =5 \cdot\left|\beg... | 2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,670 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 2 ; 0) \)
\( A_{2}(1 ;-1 ; 2) \)
\( A_{3}(0 ; 1 ;-1) \)
\( A_{4}(-3 ; 0 ; 1) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$\overrightarrow{A_{1} A_{2}}=\{1-1 ;-1-2 ; 2-0\}=\{0 ;-3 ; 2\}$
$\overrightarrow{A_{1} A_{3}}=\{0-1 ; 1-2 ;-1-0\}=\{-1 ;-1 ;-1\}$
$\overrightarrow{A_{1} A_{4}}=\{-3-1 ; 0-2 ; 1-0\}=\{-4 ;-2 ; 1\}$
According to the geometric meaning of the scalar triple product, we... | \frac{19}{6},\sqrt{\frac{19}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,671 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(0 ;-1 ;-1)$
$M_{2}(-2 ; 3 ; 5)$
$M_{3}(1 ;-5 ;-9)$
$M_{0}(-4 ;-13 ; 6)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-0 & y-(-1) & z-(-1) \\
-2-0 & 3-(-1) & 5-(-1) \\
1-0 & -5-(-1) & -9-(-1)
\end{array}\right|=0
$$
Perform transformations:
$\left|\begin{array}{ccc}x & y+1 & z+1 \\ -2 & 4 & 6 \\ 1 & -4 & -8\e... | 2\sqrt{45} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,672 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3 ; 1 ; 0)$
$B(6 ; 3 ; 3)$
$C(9 ; 4 ;-2)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{9-6 ; 4-3 ;-2-3\}=\{3 ; 1 ;-5\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$3 \cdot(x-(-3))+(y-1)-5 \cdot(z-0)=0$
$3... | 3x+y-5z+8=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,673 |
## problem statement
Find the angle between the planes:
$x+2 y-1=0$
$x+y+6=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{1 ; 2 ; 0\}$
$\overrightarrow{n_{2}}=\{1 ; 1 ; 0\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left(... | \arccos\frac{3}{\sqrt{10}}\approx1826^{\}6^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,674 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(0 ;-2 ; 4)$
$C(-4 ; 0 ; 4)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(-2-y)^{2}+(4-0)^{2}}=\sqrt{0+4+4 y+y^{2}+16}=\sqrt{y^{2}+4 y+20} \\
& A C=\sqrt{(-4-0)^{2}+(0-y)^{2}+(4-0)^{2}}=\sqrt{16+y^{2}+16}=\sqrt{y^{2}+32}
\end{aligned}
$$
Since according to the problem $A B=A C$, then
$\sqrt{y^... | A(0;3;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,675 |
## 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 | 47,676 |
## Problem Statement
Write the canonical equations of the line.
\[
\begin{aligned}
& x+y-2z-2=0 \\
& x-y+z+2=0
\end{aligned}
\] | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vect... | \frac{x}{-1}=\frac{y-2}{-3}=\frac{z}{-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,677 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z+5}{0}$
$x+7 y+3 z+11=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-3}{1}=\frac{y-1}{-1}=\frac{z+5}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+t \\
y=1-t \\
z=-5
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(3+t)+7(1-t)+3 \cdot(-5)+11=0$ $3+t+7-7 t-15... | (4;0;-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,678 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(2; -1; 1)$
$x - y + 2z - 2 = 0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{1 ;-1 ; 2\}$
Then the equation of the desired line is:
$\f... | M^{\}(1;0;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,679 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+1}{x^{2}-x} d x
$$ | ## Solution
$$
\int \frac{x^{3}+1}{x^{2}-x} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& \begin{array}{ll}
x^{3}+1 & \frac{\mid x^{2}-x}{x+1} \\
\frac{x^{3}-x^{2}}{x^{2}+1} & x+1
\end{array} \\
& \frac{x^{2}-x}{x+1}
\end{aligned}
$$
We get:
$$
=\in... | \frac{x^{2}}{2}+x-\ln|x|+2\ln|x-1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,680 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{3}+1}{x^{2}-1} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+1}{x^{2}-1} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
\frac{3 x^{3}+1}{} \\
\frac{3 x^{3}-3 x}{3 x+1} & \frac{x^{2}-1}{3 x}
\end{array}
$$
We get:
$$
=\int\left(3 x+\frac{3 x+1}{x^{2}-1}\right) d x=\int 3 x d x+... | \frac{3x^{2}}{2}+2\ln|x-1|+\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,681 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-17}{x^{2}-4 x+3} d x
$$ | ## Solution
$$
\int \frac{x^{3}-17}{x^{2}-4 x+3} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{aligned}
& x^{3}-17 \\
& x^{2}-4 x+3 \\
& \frac{x^{3}-4 x^{2}+3 x}{4 x^{2}-3 x}-17 \\
& \frac{4 x^{2}-16 x+12}{13 x-29}
\end{aligned}
$$
We get:
$$
=\int\left(x+4+\f... | \frac{x^{2}}{2}+4x+8\ln|x-1|+5\ln|x-3|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,682 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+5}{x^{2}-x-2} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+5}{x^{2}-x-2} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
\frac{2 x^{3}+5}{2 x^{3}-2 x^{2}-4 x} & \frac{\mid x^{2}-x-2}{2 x+2} \\
2 x^{2}+4 x+5 & \\
\frac{2 x^{2}-2 x-4}{6 x+9} &
\end{array}
$$
We get:
$$
=\int\le... | x^{2}+2x+7\ln|x-2|-\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,683 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}-1}{x^{2}+x-6} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}-1}{x^{2}+x-6} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
2 x^{3}-1 & \frac{x^{2}+x-6}{2 x-2} \\
\frac{2 x^{3}+2 x^{2}-12 x}{-2 x^{2}+12 x-1}-1 & \\
\frac{-2 x^{2}-2 x+12}{14 x-13} &
\end{array}
$$
We get:
$$
=\in... | x^{2}-2x+11\cdot\ln|x+3|+3\cdot\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,684 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{3}+25}{x^{2}+3 x+2} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+25}{x^{2}+3 x+2} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
\frac{3 x^{3}+25}{3 x^{3}+9 x^{2}+6 x} & \left\lvert\, \frac{x^{2}+3 x+2}{3 x-9}\right. \\
-9 x^{2}-6 x+25 & \\
\frac{-9 x^{2}-27 x-18}{21 x+43} &
\end{ar... | \frac{3x^{2}}{2}-9x+22\cdot\ln|x+1|-\ln|x+2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,685 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+2 x^{2}+3}{(x-1)(x-2)(x-3)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+2 x^{2}+3}{(x-1)(x-2)(x-3)} d x=\int \frac{x^{3}+2 x^{2}+3}{x^{3}-6 x^{2}+11 x-6} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
x^{3}+2 x^{2}+3 & \mid x^{3}-6 x^{2}+11 x-6 \\
\frac{x^{3}-6 x^{2}+11 x-6}{8 x^{2}-11 x+9} ... | x+3\cdot\ln|x-1|-19\cdot\ln|x-2|+24\cdot\ln|x-3|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,686 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{3 x^{3}+2 x^{2}+1}{(x+2)(x-2)(x-1)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+2 x^{2}+1}{(x+2)(x-2)(x-1)} d x=\int \frac{3 x^{3}+2 x^{2}+1}{x^{3}-x^{2}-4 x+4} d x=
$$
Under the integral, there is an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
3 x^{3}+2 x^{2}+1 & \mid x^{3}-x^{2}-4 x+4 \\
\frac{3 x^{3}-3 x^{2}-12 x+12}{5 x^{2}+12 x... | 3x-\frac{5}{4}\cdot\ln|x+2|+\frac{33}{4}\cdot\ln|x-2|-2\cdot\ln|x-1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,687 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}}{(x-1)(x+1)(x+2)} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x^{3}}{(x-1)(x+1)(x+2)} d x=\int \frac{x^{3}}{x^{3}+2 x^{2}-x-2} d x= \\
& =\int\left(1+\frac{-2 x^{2}+x+2}{x^{3}+2 x^{2}-x-2}\right) d x=\int d x+\int \frac{-2 x^{2}+x+2}{(x-1)(x+1)(x+2)} d x=
\end{aligned}
$$
Decompose the proper rational fraction into partial fractions u... | x+\frac{1}{6}\cdot\ln|x-1|+\frac{1}{2}\cdot\ln|x+1|-\frac{8}{3}\cdot\ln|x+2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,688 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3)(x-2)} d x
$$ | ## Solution
$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3)(x-2)} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-9 x^{2}+26 x-24} d x=
$$
Under the integral, we have an improper fraction. Let's separate the integer part:
$$
\begin{array}{ll}
x^{3}-3 x^{2}-12 & \mid x^{3}-9 x^{2}+26 x-24 \\
\frac{x^{3}-9 x^{2}+26 x-24}{6 x^{2}-26... | x+2\cdot\ln|x-4|+12\cdot\ln|x-3|-8\cdot\ln|x-2|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,689 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3) x} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3) x} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-7 x^{2}+12 x} d x= \\
& =\int\left(1+\frac{4 x^{2}-12 x-12}{x^{3}-7 x^{2}+12 x}\right) d x=\int d x+\int \frac{4 x^{2}-12 x-12}{(x-4)(x-3) x} d x=
\end{aligned}
$$
We decompose the proper rational fracti... | x+\ln|x-4|+4\cdot\ln|x-3|-\ln|x|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,690 |
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