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4.28. One of the sides of a pentagon has a length of 30 cm. The lengths of the other sides are expressed as integers and form an arithmetic progression with a difference of 2 cm, and the length of the smallest side does not exceed 7 cm. Find the lengths of the sides of all pentagons that meet these conditions. | 4.28. Let the side $A B$ of the pentagon $A B C D E$ be 30 cm (Fig. 4.25). Further, let $B C=x$; then $C D=x+2$, $D E=x+4$, $E A=x+6$, where $x \in Z, x \leq 7$. For the pentagon to be formed, the inequality $x+x+2+x+4+x+6>30$ must be satisfied, or $4 x>18$, from which $x>4.5$. Thus, from the condition $4.5<x \leq 7$, ... | 5,7,9,11,30;6,8,10,12,30;7,9,11,13,30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,290 |
4.29. A regular $n$-sided polygon is inscribed in a circle of radius $R$, the area of which is equal to $3 R^{2}$. Find $n$. | 4.29. We have $\angle A O B=\frac{360^{\circ}}{n}$ and, therefore, $S_{\triangle A O B}=\frac{R^{2}}{2} \sin \frac{360^{\circ}}{n}$ (Fig. 4.26). Thus, we obtain the equation $\frac{R^{2}}{2} \sin \frac{360^{\circ}}{n}=\frac{3 R^{2}}{n}$, which holds
 inside a regular polygon (Fig. 4.27). Connect it with the vertices of the polygon and drop perpendiculars from \( M \) to the sides of the polygon. Then, for the area of the polygon, we get the expression
. In it, the angles at the vertices are $\frac{(5-2) 180^{\circ}}{5}=108^{\circ}$. In $\triangle A B C$, draw $B F \perp A C$; since $\angle B A C=\angle B C A=\frac{1}{2}\left(180^{\circ}-108^{\circ}\right)=36^{\circ}$, then $\angle A C E=108^{\circ}-2 \cdot 36^{\... | 0.81:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,293 |
4.32. Let $n$ be the number of sides of a convex polygon, and $d$ be the number of its diagonals. Specify all values of $n$ for which $n>d$.
| 4.32. The number of diagonals of an $n$-sided polygon is determined by the formula $d=\frac{(n-3) n}{2}$. We need to select only those values of $d$ that satisfy the condition $d \leq n$. Substituting the formula for $d$, we get $\frac{(n-3) n}{2} \leq n$. Simplifying, we have $(n-3) n \leq 2n \Rightarrow n^2 - 3n \leq... | n=3n=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,294 |
4.33. Given two skew lines. Is it possible to draw two intersecting lines such that each of them intersects both given lines? | 4.33. Let $p$ and $q$ be two skew lines (Fig. 4.29). Take any point $A$ on line $p$ and connect this point with any two points $B_{1}$ and $B_{2}$ on line $q$. Thus, $A B_{1}$ and $A B_{2}$ are two intersecting lines that intersect both given skew lines.
Answer: Yes, it is possible.
 of a triangular pyramid are equal. Find the ratio of the radius of the sphere inscribed in the pyramid to its height. | 4.34. Let $a$ be the edge of the pyramid $SABC$, $OM=r$ be the radius of the inscribed sphere, and $SH=\boldsymbol{h}$ be the height of the pyramid (Fig. 4.30). In $\triangle SAF$, we have $SF=\frac{a \sqrt{3}}{2}$, $HF=\frac{1}{3} AF=\frac{a \sqrt{3}}{6}$; therefore,
$$
h=SH=\sqrt{SF^{2}-HF^{2}}=\sqrt{\frac{3 a^{2}}{... | 1:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,296 |
4.35. The generatrix of a truncated cone forms an angle $\alpha$ with the base plane. Inside the cone, there are two spheres touching each other and the lateral surface of the cone, with the first sphere touching the lower base of the cone and the second sphere touching the upper base. The distance between the centers ... | 4.35. Let $O A=R$ and $O_{1} B=r$ be the radii of the bases of a truncated cone, and $M O=x$ and $N O_{1}=y(y<x)$ be the radii of the inscribed spheres (Fig. 4.31). By the condition, $x+y=l, \angle A=\alpha$. In $\triangle A O M$ and $\triangle B O_{1} N$ we have
$$
R=x \operatorname{ctg} \frac{\alpha}{2}, r=y \operat... | \cos^{2}\frac{\alpha}{2}\operatorname{ctg}\frac{\alpha}{2},\sin^{2}\frac{\alpha}{2}\operatorname{tg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,297 |
5.1. In parallelogram $O A B C$, the vertices $O(0 ; 0 ;)$, $A(3 ; 6)$, and $B(8 ; 6)$ are given. Find the ratio of the lengths of the diagonals $O B$ and $A C$, and also write the equations of the sides of the parallelogram and the diagonal $A C$. | 5.1. Since the ordinates of vertices $A$ and $B$ are equal, $A B \| O x$ (Fig. 5.1). Of the three segments $O A, A B$, and $O B$, only $O A$ and $A B$ can be the sides of the parallelogram, as $O B$ is the diagonal by condition; therefore, $B C \| O A$ and $C(5 ; 0)$. Using formula (5.1), we find
$O B=\sqrt{8^{2}+6^{2... | OB:AC=\sqrt{2.5};0(OC),6(AB),2x(OA),2x-10(BC),-3x+15(AC) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,298 |
5.2. Given three points $A(2 ; 1), B(3 ;-1), C(-4 ; 0)$, which are vertices of an isosceles trapezoid $A B D C$. Find the coordinates of point $D$, if $\overline{A B}=k \overline{C D}$. | 5.2. According to the condition,
$\overline{A B}=k \overline{C D} \Rightarrow \overline{A B} \| \overline{C D}$.
We find the coordinates of vectors ${ }^{-} \overline{A B}$ and $\overline{C D}$ (Fig. 5.2): $\overline{A B}(1 ;-2)$, $\overline{C D}(x+4 ; y)$, where $x$ and $y$ are the coordinates of point $D$. From (1)... | (-1.4;-5.2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,299 |
5.3. The lengths of the diagonals $A C$ and $B D$ of a rhombus are 15 and 8 cm. The first diagonal is directed along the $O x$ axis, and the second along the $O y$ axis. Form the equations of the sides of the rhombus and find the distance from the origin to a side of the rhombus. | 5.3. According to the condition, $A C=15 \text{~cm}, B D=8 \text{~cm}$, from which we find the coordinates of the vertices of the rhombus: $A(-7.5 ; 0), B(0 ; 4), C(7.5 ; 0), D(0 ;-4)$ (Fig. 5.3) The slope of the line $B C$ is $k_{1}=\operatorname{tg} \angle B C N=$
$ moves to point $A_{1}(8 ; 6)$. Find the cosine of the rotation angle. | 5.4. Using formula (5.21), we find
$$
\cos \angle A O A_{1}=\frac{\overline{O A} \cdot \overline{O A_{1}}}{|\overline{O A}| \overline{O A_{1}}|}=\frac{6 \cdot 8+8 \cdot 6}{\sqrt{36+64} \cdot \sqrt{64+36}}=\frac{24}{25}
$$ | \frac{24}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,301 |
5.5. The vector $\overline{O A}$ forms angles with the axes $O x, O y, O z$ that are respectively equal to $\alpha=\frac{\pi}{3}, \beta=\frac{\pi}{3}, \gamma=\frac{\pi}{4}$; the point $B$ has coordinates $(-2; -2; -2 \sqrt{2})$. Find the angle between the vectors $\overline{O A}$ and $\overline{O B}$. | 5.5. Let $\bar{n}$ be a unit vector in the direction of $\overline{O A};$ then $\quad \bar{n}\left(\cos \frac{\pi}{3} ; \cos \frac{\pi}{3} ; \cos \frac{\pi}{4}\right)$, or $\quad \bar{n}\left(\frac{1}{2} ; \frac{1}{2} ; \frac{\sqrt{2}}{2}\right)$, and $\overline{O B}(-2 ;-2 ;-2 \sqrt{2})$. Denoting the angle between $\... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,302 |
5.6. Find the unit vector collinear with the vector directed along the bisector of angle $B A C$ of triangle $A B C$, if its vertices are given: $A(1 ; 1 ; 1), B(3 ; 0 ; 1), C(0 ; 3 ; 1)$. | 5.6. Let's find the coordinates and magnitudes of vectors $\overline{A B}$ and $\overline{A C}$; we have $\overline{A B}(2 ; -1 ; 0), \overline{A C}(-1 ; 2 ; 0)$,
$$
|\overrightarrow{A B}|=\sqrt{2^{2}+(-1)^{2}+0^{2}}=\sqrt{5},|\overrightarrow{A C}|=\sqrt{(-1)^{2}+2^{2}+0^{2}}=\sqrt{5}
$$
Since $|\overline{A B}|=|\ove... | (\frac{1}{\sqrt{2}};\frac{1}{\sqrt{2}};0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,303 |
5.7. Given two non-zero vectors $\bar{a}$ and $\bar{b}$ such that $|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|$. Prove that $\bar{a} \perp \bar{b}$. | 5.7. I m e t h o d. If vectors $\bar{a}$ and $\bar{b}$ are used as sides to construct a parallelogram, then the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$ will coincide with its diagonals, the lengths of which are $|\bar{a}+\bar{b}|$ and $|\bar{a}-\bar{b}|$. Since the lengths of the diagonals are equal by conditio... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,304 |
5.8. Find the modulus of the projection of vector $\bar{a}(7; -4)$ onto the axis parallel to vector $\bar{b}(-8; 6)$.
untranslated text:
5.8. Найти модуль проекции вектора $\bar{a}(7 ;-4)$ на ось, параллельную вектору $\bar{b}(-8 ; 6)$. | 5.8. We have $O K=$ proj $_{\bar{b}} \bar{a}=|\bar{a}| \cos \varphi$ (Fig. 5.5). From here, using
Fig. 5.5 the equality $\quad \bar{a} \bar{b}=|\bar{a}||\vec{b}| \cos \varphi, \quad$ we fin... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,305 |
5.9. For what values of $x$ are the vectors $(x^{3}-1) \bar{a}$ and $2 x \bar{a}$ collinear, if $\bar{a} \neq \overline{0}$? | 5.9. The vectors are collinear if \(x^{3}-1\) and \(2 x\) have the same sign. To determine the desired values of \(x\), we solve the inequality \(2 x\left(x^{3}-1\right)>0\) and obtain \(x \in(-\infty, 0) \bigcup(1, \infty)\).
Answer: for \(-\infty<x<0\) and for \(1<x<\infty\). | x\in(-\infty,0)\bigcup(1,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,306 |
5.10. Given vectors $\bar{a}(6 ;-8 ; 5 \sqrt{2})$ and $\bar{b}(2 ;-4 ; \sqrt{2})$. Find the angle formed by the vector $\bar{a}-\bar{b}$ with the $O z$ axis. | 5.10. The vector $\bar{a}-\bar{b}$ has coordinates $(4; -4; 4 \sqrt{2})$. Let $\bar{k}(0; 0; 1)$ be the unit vector directed along the $O z$ axis. Assuming $(\widehat{a}-\bar{b}; \bar{k})=\gamma$, we find
$$
\cos \gamma=\frac{(\bar{a}-\bar{b}) \bar{k}}{|\bar{a}-\bar{b}||\bar{k}|}=\frac{4 \sqrt{2}}{\sqrt{16+16+32 \cdot... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,307 |
5.11. Given a triangle $A B C ; B D$ is a median, $\angle D B C=90^{\circ}$, $B D=\frac{\sqrt{3}}{4} A B$. Find $\angle A B D$. | 5.11. According to the condition, $\angle D B C=90^{\circ}, B D=\frac{\sqrt{3}}{4} A B, A D=D C$ (Fig. 5.6); we need to find $\angle A B D=$ $=\alpha$. To find the angle $\alpha$, we will use the formula $\cos \alpha=\frac{\overline{B A} \cdot \overline{B D}}{|\overrightarrow{B A}||\overline{B D}|}$. Since $B D$ is the... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,308 |
5.12. Let $O$ be the point of intersection of the medians of triangle $ABC$ and $\overline{A O}=\bar{a}, \overline{A C}=\bar{b}$. Decompose $\overline{A B}$ and $\overline{B C}$ in terms of the vectors $\bar{a}$ and $\bar{b}$. | 5.12. Since $O$ is the point of intersection of the medians of $\triangle ABC$ (Fig. 5.7), then $\overline{AO}=\frac{2}{3} \overline{AM}_{1}$; thus $\overline{AM}_{1}=\frac{3}{2} \bar{a}, \overline{M_{1}C}=\overline{AC}-\overline{AM}_{1}=\bar{b}-\frac{3}{2} \bar{a}$, $BC=2 \overline{M_{1}C}=2 \bar{b}-3 \bar{a}, \overli... | \overline{AB}=3\bar{}-\bar{b},\overline{BC}=2\bar{b}-3\bar{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,309 |
5.14. In the pyramid $S A B C$, all faces are equilateral triangles; point $M$ is the center of triangle $A B C$, and point $P$ divides the edge $S C$ in half. Find the decomposition of the vector $\overline{M P}$ in terms of the vectors $\overline{A B}, \overline{A C}$, and $\overline{A S}$. | 5.14. We have (Fig. 5.8) $\overline{M P}=\overline{M C}-\overline{P C}$, where $\overline{P C}=\frac{1}{2} \overline{S C}=$
Fig. 5.8
$$
=\frac{1}{2}(\overline{A C}-\overline{A S})
$$
$\ov... | \overline{MP}=\frac{1}{6}\overline{AC}-\frac{1}{3}\overline{AB}+\frac{1}{2}\overline{AS} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,311 |
5.15. Given the vector $\bar{a}(1; -2; 5)$. Find the coordinates of the vector $\bar{b}$, lying in the $x Q y$ plane and perpendicular to the vector $\bar{a}$, if $|\vec{b}|=2 \sqrt{5}$ | 5.15. Since the vector $\bar{b} \epsilon=x O y$, it has coordinates $(x ; y ; 0)$. Using the conditions $|\vec{b}|=2 \sqrt{5}, \bar{b} \perp \bar{a}$, we can form the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=20 \\
x-2 y=0
\end{array}\right.
$$
Solving it, we get two vectors $(4 ; 2 ; 0)$ and $(-4 ;-... | (4;2;0)or(-4;-2;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,312 |
5.16. Given: cube $A B C D A_{1} B_{1} C_{1} D_{1}$ (vertices of the base $A B C D$ are arranged clockwise); $K$ is the midpoint of edge $A A_{1}$; $H$ is the midpoint of edge $A D$; $M$ is the center of face $C C_{1} D_{1} D$. Prove that line $K M$ is perpendicular to line $B_{1} H$. | 5.16. Let $\overline{A D}=\bar{a}, \overline{A B}=\bar{b}, \overline{A A_{1}}=\bar{c}$ (Fig. 5.9). We will find the decompositions of $\overline{B_{1} H}$ and $\overline{K M}$ in terms of the vectors $\bar{a}, \bar{b}$, and $\bar{c}$. We have
$$
\overline{B_{1} H}=\overline{B_{1} A_{1}}+\overline{A_{1} A}+\overline{A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,313 |
5.17. In triangle $A B C$, point $N$ lies on side $A B$ and $A N=$ $=3 N B$; median $A M$ intersects $C N$ at point $O$. Find $A B$, if $A M=C N=7$ cm and $\angle N O M=60^{\circ}$. | 5.17. By the condition, $A M$ is the median of $\triangle A B C$ (Fig. 5.10) and, therefore,
$$
\overline{A M}=\frac{1}{2}(\overline{A B}+\overline{A C})
$$
Further,
$$
\overline{A C}=\overline{A N}+\overline{N C}=\frac{3}{4} \overline{A B}-\overline{C N}
$$
Substituting (2) into (1), we get $\overline{A M}=\frac{7... | 4\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,314 |
5.18. In parallelogram $A B C D$, point $K$ is the midpoint of side $B C$, and point $M$ is the midpoint of side $C D$. Find $A D$, if $A K=6$ cm, $A M=3$ cm, and $\angle K A M=60^{\circ}$. | 5.18. I n d i c a t i o n. Decompose the vector $\overline{A D}$ into vectors $\overline{A K}$ and $\overline{A M}$.
Answer: 4 cm. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,315 |
5.19. Given three non-zero vectors $\bar{a}, \bar{b}, \bar{c}$, each pair of which are non-collinear. Find their sum, if $(\bar{a}+\bar{b}) \| \bar{c}$ and $(\bar{b}+\bar{c}) \| \bar{a}$. | 5.19. We have
$$
\begin{aligned}
& (\bar{a}+\bar{b}) \| \bar{c} \Rightarrow \bar{c}=m(\bar{a}+\bar{b})=m \bar{a}+m \bar{b} \\
& (\bar{b}+\bar{c}) \| \bar{a} \Rightarrow \bar{b}+\bar{c}=n \bar{a} \Rightarrow \bar{c}=n \bar{a}-\bar{b}
\end{aligned}
$$
Due to the uniqueness of the decomposition of $\bar{c}$ in terms of ... | \overline{0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,316 |
5.20. Unit vectors $\bar{e}_{1}, \bar{e}_{2}, \bar{e}_{3}$ satisfy the condition $\bar{e}_{1}+\bar{e}_{2}+\bar{e}_{3}=\overline{0}$. Find $\bar{e}_{1} \bar{e}_{2}+\bar{e}_{2} \bar{e}_{3}+\bar{e}_{3} \bar{e}_{1}$.
## Group b | 5.20. Indication. From the condition $\bar{e}_{1}+\bar{e}_{2}+\bar{e}_{3}=\overline{0}$ it follows that these vectors form an equilateral triangle and ( $=\left(\widehat{e_{3} ; \vec{e}_{1}}\right)=180^{\circ}-60^{\circ}=120^{\circ}$
Answer: $-1.5$. | -1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,317 |
5.21. Form the equation of the circle circumscribed around the triangle formed by the lines $y=0.2 x-0.4, y=x+2, y=8-x$. | 5.21. The angular coefficients of the lines $y=x+2$ and $y=8-x$ are $k_{1}=1$ and $k_{2}=-1$, respectively. Since $k_{1} k_{2} = -1$, the condition (5.9) for perpendicularity of the lines is satisfied; hence, $\triangle A B C$ is a right triangle (Fig. 5.11) and the center of the circle is the midpoint of its hypotenus... | (x-2)^{2}+y^{2}=26 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,318 |
5.22. In trapezoid $A B C D$ given: vertex $A(3 ; 0)$, midpoint of base $A B$ - point $E(6 ;-1)$, midpoint of base $C D$ - point $F(7 ; 2)$. Side $B C$ is parallel to the $O y$ axis. Prove that the trapezoid is isosceles, and find the angle at its base. | 5.22. Let $B\left(x_{1} ; y_{1}\right), C\left(x_{2} ; y_{2}\right), D\left(x_{3} ; y_{3}\right)$ be the unknown vertices of the trapezoid (Fig. 5.12). Since point $E(6 ;-1)$ is the midpoint of base $A B$, we have the system $\frac{3+x_{1}}{2}=6, \frac{0+y_{1}}{2}=-1$, from which $x_{1}=9$,
, B(4 ; 0)$. | ### 5.23. According to the condition, $A(2 ; 1)$,
$B(4 ; 0)$ are known, while $C\left(x_{1} ; y_{1}\right)$, $D\left(x_{2} ; y_{2}\right)$ are unknown vertices of the square (Fig. 5.13). Since the length of the side of the square $A B=\sqrt{2^{2}+1^{2}}=\sqrt{5}$ and $B C = A B$, we have
$$
B C^{2}=\left(x_{1}-4\righ... | C(5;2),D(3;3)orC(3;-2),D(1;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,320 |
5.24. In parallelogram $A B C D$ it is given: $M \in B C$ and $B M: M C=1: 2$; $N \in D C, D N: N C=1: 2 ; \overline{A M}=\bar{a} ; \overline{A N}=\bar{b}$. Express the vectors $\overline{A B}, \overline{A D}, \overline{M N}$ and $\overline{B D}$ in terms of $\bar{a}$ and $\bar{b}$. | 5.24. Given that $\overline{A M}=\bar{a}, \overline{A N}=\bar{b}$ (Fig. 5.14) and, therefore, $\overline{M N}=\overline{A N}-\overline{A M}=\bar{b}-\bar{a}. \quad$ Since
Fig. 5.14 $\frac{B ... | \overline{AB}=\frac{9}{8}\bar{}-\frac{3}{8}\bar{b},\overline{AD}=\frac{9}{8}\bar{b}-\frac{3}{8}\bar{},\overline{MN}=\bar{b}-\bar{},\overline{BD}=\frac{3}{2}\bar{b}- | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,321 |
5.25. Given two segments $A B$ and $C D$. Prove that if $A C^{2}+B D^{2}=$ $=A D^{2}+B C^{2}$, then $A B \perp C D$. Is the converse statement true? | 5.25. Consider the vectors $\overline{A B}, \overline{C D}, \overline{A C}, \overline{A D}, \overline{B D}, \overline{B C}$. Depending on their relative positions, a planar or spatial figure can be formed (Fig. 5.15). Given that
. Let $\overline{A D}=\bar{m}, \overline{A B}=\bar{n}, \overline{A S}=\bar{p}$ and $(D \widehat{C ; A M})=\varphi$. The angle $\varphi$ we are looking for will be found using the formula
$$
\cos ... | \arccos\frac{5\sqrt{13}}{26} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,323 |
5.27. Points $M$ and $N$ are the midpoints of segments $AB$ and $CD$. Prove that $MN \leq \frac{1}{2}(AC + BD), MN \leq \frac{1}{2}(BC + AD)$. | 5.27. Draw $A K \| B D$ and $D K|| A B$ (Fig. 5.17). Then $\overline{A K}=\overline{B D}, \overline{D K}=\overline{B A}$. Using the obvious equalities $\overline{A B}+\overline{B D}=\overline{A D}$, $\overline{A C}-\overline{A B}=\overline{B C}$; adding them term by term, we get
$$
\overline{A C}+\overline{B D}=\overl... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 50,324 |
5.28. The sides of triangle $ABC$ are related by the equation $a^{2}+$ $+b^{2}=5 c^{2}$. Prove that two medians of the triangle are perpendicular. Is the converse statement true? | 5.28. Let's introduce the following notations: $\overline{B C}=\bar{a}, \overline{C A}=\bar{b}, A B=\bar{c}$, $\overline{A M_{1}}=\overline{m_{1}}, \overline{B M_{2}}=\overline{m_{2}}$, where $A M_{1}$ and $B M_{2}$ are the medians of $\triangle A B C$ (Fig. 5.18). Since $\overline{m_{1}}=\bar{c}+\frac{1}{2} \bar{a}, \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,325 |
5.29. In triangle $ABC$, given: $AB = BC$; $D$ is the midpoint of side $AC$; $DK$ is perpendicular to $BC$; point $M$ is the midpoint of segment $DK$. Prove that lines $AK$ and $BM$ are perpendicular. | 5.29. Since in $\triangle ABC$ (Fig. 5.19) $AB=BC$, $D$ is the midpoint of $AC$, then $BD \perp AC$. Let's write the decompositions $\overline{BM}=\overline{BD}+\overline{DM}, \overline{AK}=\overline{AD}+$ $+\overline{DK}=\overline{AD}+2 \overline{DM}$. Therefore,
$$
\begin{gathered}
\overline{BM} \cdot \overline{AK}=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,326 |
5.30. A triangle \(ABC\) is inscribed in a circle. The line containing the median \(CC_1\) of the triangle intersects the circle again at point \(D\). Prove that \(CA^2 + CB^2 = 2 CC_1 \cdot CD\). | 5.30. We have $\overline{C A}=\overline{C C_{1}}+\overline{C_{1} A}, \overline{C B}=\overline{C C_{1}}+\overline{C_{1} B}$ (Fig. 5.20). Then $C A^{2}=C C_{1}^{2}+2 \overline{C C_{1}} \cdot \overline{C_{1} A}+C_{1} A^{2}, C B^{2}=C C_{1}^{2}+2 \overline{C C_{1}} \cdot \overline{C_{1} B}+C_{1} B^{2}$. Adding these equati... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,327 |
5.31. Prove that if the bisectors of two dihedral angles of a trihedral angle are perpendicular, then the bisector of the third dihedral angle is perpendicular to each of them. | 5.31. Let's choose unit vectors $\bar{a}_{0}, \bar{b}_{0}$, and $\bar{c}_{0}$ directed along the edges of a trihedral angle. Then the direction vectors $\bar{m}, \bar{n}$, and $\bar{p}$ of the bisectors of the three planar angles of the trihedral angle can be written as $\bar{m}=\bar{a}_{0}+\bar{b}_{0}, \bar{n}=\bar{c}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,328 |
5.32. Find the volume of a triangular pyramid constructed on vectors $\overline{O A}, \overline{O B}$, and $\overline{O C}$, if $|\overrightarrow{O A}|=5,|\overrightarrow{O B}|=2,|\overrightarrow{O C}|=6, \overline{O A} \cdot \overrightarrow{O B}=0$, $\overline{O A} \cdot \overline{O C}=0, \overline{O B} \cdot \overlin... | 5.32. Since $\overline{O A} \cdot \overline{O B}=0$ and $\overline{O A} \cdot \overline{O C}=0$, then $\overline{O A} \perp \overline{O B}$ and $\overline{O A} \perp \overline{O C}$, which means $\overline{O A} \perp(O B C)$, i.e., $O A$ is the height of the pyramid $O A B C$ (Fig. 5.21). Let $\angle B O C=\varphi$; th... | \frac{10\sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,329 |
5.33. Prove that for any four points $A, B, C, D$ the equality $\overline{A B} \cdot \overline{C D}+\overline{A C} \cdot \overline{D B}+\overline{A D} \cdot \overline{B C}=0$ holds. | 5.33. From an arbitrary point $O$, we draw the radius vectors $\overline{O A}=\bar{r}_{1}, \overline{O B}=\bar{r}_{2}, \overline{O C}=\bar{r}_{3}$, and $\overline{O D}=\bar{r}_{4}$ (Fig. 5.22). Now we express the required vectors through them: $\overline{A B}=\bar{r}_{2}-\bar{r}_{1}, \overline{C D}=\bar{r}_{4}-\bar{r}_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,330 |
5.34. Prove that if the sums of the squares of opposite edges of a tetrahedron are equal, then these edges are pairwise perpendicular. | 5.34. In the tetrahedron $SABC$, we draw $SO \perp (ABC)$ (Fig. 5.23) and express the vectors directed along the edges of the pyramid in terms of the radius vectors $\overline{OA}=\bar{r}_{1}, \overline{OB}=\bar{r}_{2}, \overline{OC}=\bar{r}_{3}, \overline{OS}=\bar{r}_{4}$. We have $\overline{SA}=\bar{r}_{1}-\bar{r}_{4... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,331 |
5.35. Prove that for any triangle $ABC$ the inequality $\cos A+\cos B+\cos C \leq \frac{3}{2}$ holds. | 5.35. On the sides of a triangle, construct unit vectors $\bar{e}_{1}, \bar{e}_{2}$, and $\bar{e}_{3}$ (Fig. 5.24). The sum of these vectors is some vector $\bar{d}$, i.e., $\bar{e}_{1}+\bar{e}_{2}+\bar{e}_{3}=\bar{d}$. Squaring both sides of the equation, we get:
$$
\bar{e}_{1}^{2}+\bar{e}_{2}^{2}+\bar{e}_{3}^{2}+2 \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 50,332 |
1.001. $\frac{(7-6.35): 6.5+9.9}{\left(1.2: 36+1.2: 0.25-1 \frac{5}{16}\right): \frac{169}{24}}$. | ## Solution.
$$
\begin{aligned}
& \frac{(7-6.35): 6.5+9.9}{\left(1.2: 36+1.2: 0.25-1 \frac{5}{16}\right): \frac{169}{24}}=\frac{0.65: 6.5+9.9}{\left(\frac{1}{30}+\frac{24}{5}-\frac{21}{16}\right) \cdot \frac{24}{169}}=\frac{0.1+9.9}{\frac{169}{48} \cdot \frac{24}{169}}= \\
& =\frac{10}{\frac{1}{2}}=20 .
\end{aligned}
... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,333 |
1.002. $\left(\left(\frac{7}{9}-\frac{47}{72}\right): 1.25+\left(\frac{6}{7}-\frac{17}{28}\right):(0.358-0.108)\right) \cdot 1.6-\frac{19}{25}$. | ## Solution.
$$
\begin{aligned}
& \left(\left(\frac{7}{9}-\frac{47}{72}\right): 1.25+\left(\frac{6}{7}-\frac{17}{28}\right):(0.358-0.108)\right) \cdot 1.6-\frac{19}{25}= \\
& =\left(\frac{56-47}{72} \cdot \frac{4}{5}+\frac{24-17}{28}: 0.25\right) \cdot 1.6-\frac{19}{25}=(0.1+1) \cdot 1.6-\frac{19}{25}=1.76-0.76=1 .
\e... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,334 |
1.005. $\frac{2 \frac{3}{4}: 1.1+3 \frac{1}{3}}{2.5-0.4 \cdot 3 \frac{1}{3}}: \frac{5}{7}-\frac{\left(2 \frac{1}{6}+4.5\right) \cdot 0.375}{2.75-1 \frac{1}{2}}$. | Solution.
$$
\frac{2 \frac{3}{4}: 1.1+3 \frac{1}{3}}{2.5-0.4 \cdot 3 \frac{1}{3}}: \frac{5}{7}-\frac{\left(2 \frac{1}{6}+4.5\right) \cdot 0.375}{2.75-1 \frac{1}{2}}=\frac{\frac{5}{2}+\frac{10}{3}}{\frac{5}{2}-\frac{4}{3}} \cdot \frac{7}{5}-\frac{\frac{20}{3} \cdot \frac{3}{8}}{1.25}=7-2=5
$$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,335 |
1.006. $\frac{\left(13.75+9 \frac{1}{6}\right) \cdot 1.2}{\left(10.3-8 \frac{1}{2}\right) \cdot \frac{5}{9}}+\frac{\left(6.8-3 \frac{3}{5}\right) \cdot 5 \frac{5}{6}}{\left(3 \frac{2}{3}-3 \frac{1}{6}\right) \cdot 56}-27 \frac{1}{6}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(13.75+9 \frac{1}{6}\right) \cdot 1.2}{\left(10.3-8 \frac{1}{2}\right) \cdot \frac{5}{9}}+\frac{\left(6.8-3 \frac{3}{5}\right) \cdot 5 \frac{5}{6}}{\left(3 \frac{2}{3}-3 \frac{1}{6}\right) \cdot 56}-27 \frac{1}{6}=\frac{\left(\frac{55}{4}+\frac{55}{6}\right) \cdot \frac{6}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,336 |
1.008. $\left(\frac{3 \frac{1}{3}+2.5}{2.5-1 \frac{1}{3}} \cdot \frac{4.6-2 \frac{1}{3}}{4.6+2 \frac{1}{3}} \cdot 5.2\right):\left(\frac{0.05}{\frac{1}{7}-0.125}+5.7\right)$. | Solution.
$$
\begin{aligned}
& \left(\frac{3 \frac{1}{3}+2.5}{2.5-1 \frac{1}{3}} \cdot \frac{4.6-2 \frac{1}{3}}{4.6+2 \frac{1}{3}} \cdot 5.2\right):\left(\frac{0.05}{\frac{1}{7}-0.125}+5.7\right)= \\
& =\left(\frac{\frac{10}{3}+\frac{5}{2}}{\frac{5}{2}-\frac{4}{3}} \cdot \frac{\frac{23}{5}-\frac{7}{3}}{\frac{23}{5}+\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,337 |
1.009.
$$
\frac{0.4+8\left(5-0.8 \cdot \frac{5}{8}\right)-5: 2 \frac{1}{2}}{\left(1 \frac{7}{8} \cdot 8-\left(8.9-2.6: \frac{2}{3}\right)\right) \cdot 34 \frac{2}{5}} \cdot 90
$$ | Solution.
$$
\begin{aligned}
& \frac{0.4+8\left(5-0.8 \cdot \frac{5}{8}\right)-5: 2 \frac{1}{2}}{\left(1 \frac{7}{8} \cdot 8-\left(8.9-2.6: \frac{2}{3}\right)\right) \cdot 34 \frac{2}{5}} \cdot 90=\frac{\left(0.4+40-4-5 \cdot \frac{2}{5}\right) \cdot 90}{\left(\frac{15}{8} \cdot 8-\frac{89}{10}+\frac{13}{5} \cdot \fra... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,338 |
1.010. $\frac{\left(5 \frac{4}{45}-4 \frac{1}{6}\right): 5 \frac{8}{15}}{\left(4 \frac{2}{3}+0.75\right) \cdot 3 \frac{9}{13}} \cdot 34 \frac{2}{7}+\frac{0.3: 0.01}{70}+\frac{2}{7}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(5 \frac{4}{45}-4 \frac{1}{6}\right): 5 \frac{8}{15}}{\left(4 \frac{2}{3}+0.75\right) \cdot 3 \frac{9}{13}} \cdot 34 \frac{2}{7}+\frac{0.3: 0.01}{70}+\frac{2}{7}=\frac{\left(\frac{229}{45}-\frac{25}{6}\right): \frac{83}{15}}{\left(\frac{14}{3}+\frac{3}{4}\right) \cdot \frac... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,339 |
1.011. $\frac{\left(\frac{3}{5}+0.425-0.005\right): 0.1}{30.5+\frac{1}{6}+3 \frac{1}{3}}+\frac{6 \frac{3}{4}+5 \frac{1}{2}}{26: 3 \frac{5}{7}}-0.05$. | ## Solution.
$$
\frac{\left(\frac{3}{5}+0.425-0.005\right): 0.1}{30.5+\frac{1}{6}+3 \frac{1}{3}}+\frac{6 \frac{3}{4}+5 \frac{1}{2}}{26: 3 \frac{5}{7}}-0.05=
$$
$=\frac{(0.6+0.42) \cdot 10}{\frac{61}{2}+\frac{1}{6}+\frac{10}{3}}+\frac{12 \frac{1}{4} \cdot 26}{26 \cdot 7}-0.05=$
$=\frac{10.2}{34}+\frac{7}{4}-\frac{1}{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,340 |
1.012. $\frac{3 \frac{1}{3} \cdot 1.9+19.5: 4 \frac{1}{2}}{\frac{62}{75}-0.16}: \frac{3.5+4 \frac{2}{3}+2 \frac{2}{15}}{0.5\left(1 \frac{1}{20}+4.1\right)}$. | ## Solution.
$\frac{3 \frac{1}{3} \cdot 1.9 + 19.5 : 4 \frac{1}{2}}{\frac{62}{75} - 0.16} : \frac{3.5 + 4 \frac{2}{3} + 2 \frac{2}{15}}{0.5 \left(1 \frac{1}{20} + 4.1\right)} = \frac{\frac{10}{3} \cdot \frac{19}{10} + \frac{39}{2} \cdot \frac{2}{9}}{\frac{62}{75} - \frac{4}{25}} \cdot \frac{\frac{1}{2} \left(\frac{21}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,341 |
1.014. $\frac{\left(4.5 \cdot 1 \frac{2}{3}-6.75\right) \cdot \frac{2}{3}}{\left(3 \frac{1}{3} \cdot 0.3+5 \frac{1}{3} \cdot \frac{1}{8}\right): 2 \frac{2}{3}}+\frac{1 \frac{4}{11} \cdot 0.22: 0.3-0.96}{\left(0.2-\frac{3}{40}\right) \cdot 1.6}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(4.5 \cdot 1 \frac{2}{3}-6.75\right) \cdot \frac{2}{3}}{\left(3 \frac{1}{3} \cdot 0.3+5 \frac{1}{3} \cdot \frac{1}{8}\right): 2 \frac{2}{3}}+\frac{1 \frac{4}{11} \cdot 0.22: 0.3-0.96}{\left(0.2-\frac{3}{40}\right) \cdot 1.6}=\frac{\left(\frac{9}{2} \cdot \frac{5}{3}-\frac{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,342 |
1.016. $\left(16 \frac{1}{2}-13 \frac{7}{9}\right) \cdot \frac{18}{33}+2.2\left(\frac{8}{33}-\frac{1}{11}\right)+\frac{2}{11}$. | ## Solution.
$$
\begin{aligned}
& \left(16 \frac{1}{2}-13 \frac{7}{9}\right) \cdot \frac{18}{33}+2.2\left(\frac{8}{33}-\frac{1}{11}\right)+\frac{2}{11}=\left(\frac{33}{2}-\frac{124}{9}\right) \cdot \frac{6}{11}+ \\
& +\frac{22}{10}\left(\frac{8}{33}-\frac{3}{33}\right)+\frac{2}{11}=\frac{49}{18} \cdot \frac{6}{11}+\fr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,343 |
1.017. $\frac{0.128: 3.2+0.86}{\frac{5}{6} \cdot 1.2+0.8} \cdot \frac{\left(1 \frac{32}{63}-\frac{13}{21}\right) \cdot 3.6}{0.505 \cdot \frac{2}{5}-0.002}$. | ## Solution.
$$
\begin{aligned}
& \frac{0.128: 3.2+0.86}{\frac{5}{6} \cdot 1.2+0.8} \cdot \frac{\left(1 \frac{32}{63}-\frac{13}{21}\right) \cdot 3.6}{0.505 \cdot \frac{2}{5}-0.002}=\frac{0.04+0.86}{1+0.8} \cdot \frac{\left(\frac{95}{63}-\frac{39}{63}\right) \cdot \frac{18}{5}}{0.202-0.002}= \\
& =\frac{9}{18} \cdot \f... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,344 |
1.018. $\frac{3 \frac{1}{3}: 10+0.175: 0.35}{1.75-1 \frac{11}{17} \cdot \frac{51}{56}}-\frac{\left(\frac{11}{18}-\frac{1}{15}\right): 1.4}{\left(0.5-\frac{1}{9}\right) \cdot 3}$. | Solution.
$$
\begin{aligned}
& \frac{3 \frac{1}{3}: 10+0.175: 0.35}{1.75-1 \frac{11}{17} \cdot \frac{51}{56}}-\frac{\left(\frac{11}{18}-\frac{1}{15}\right): 1.4}{\left(0.5-\frac{1}{9}\right) \cdot 3}=\frac{\frac{1}{3}+\frac{1}{2}}{\frac{7}{4}-\frac{28}{17} \cdot \frac{51}{56}}-\frac{\frac{49}{90} \cdot \frac{5}{7}}{\f... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,345 |
1.019. $\frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3) \cdot 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5$. | ## Solution.
$$
\begin{aligned}
& \frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3): 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5=\frac{\frac{1}{2}+\frac{5}{8}}{\left(10-\frac{220}{23}\right) \cdot \frac{23}{50}+\frac{8}{5}}+\frac{17}{40}+\frac{19}{20}= \\
& =\frac{\frac{9}{8}}{\frac{1}{5}+\frac{8}{5}}+\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,346 |
1.020. $\left(\left(1 \frac{1}{7}-\frac{23}{49}\right): \frac{22}{147}-\left(0.6: 3 \frac{3}{4}\right) \cdot 2 \frac{1}{2}+3.75: 1 \frac{1}{2}\right): 2.2$. | ## Solution.
$$
\begin{aligned}
& \left(\left(1 \frac{1}{7}-\frac{23}{49}\right): \frac{22}{147}-\left(0.6: 3 \frac{3}{4}\right) \cdot 2 \frac{1}{2}+3.75: 1 \frac{1}{2}\right): 2.2= \\
& =\left(\left(\frac{8}{7}-\frac{23}{49}\right) \frac{147}{22}-0.16 \cdot 2.5+2.5\right): 2.2=\left(\frac{33}{49} \cdot \frac{147}{22}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,347 |
1.021. $\left(2: 3 \frac{1}{5}+\left(3 \frac{1}{4}: 13\right): \frac{2}{3}+\left(2 \frac{5}{18}-\frac{17}{36}\right) \cdot \frac{18}{65}\right) \cdot \frac{1}{3}$. | Solution.
$$
\begin{aligned}
& \left(2: 3 \frac{1}{5}+\left(3 \frac{1}{4}: 13\right): \frac{2}{3}+\left(2 \frac{5}{18}-\frac{17}{36}\right) \cdot \frac{18}{65}\right) \cdot \frac{1}{3}= \\
& =\left(2 \cdot \frac{5}{16}+\frac{1}{4} \cdot \frac{3}{2}+\frac{65}{36} \cdot \frac{18}{65}\right) \cdot \frac{1}{3}=\left(\frac... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,348 |
1.022. $\frac{0.5+\frac{1}{4}+\frac{1}{6}+0.125}{\frac{1}{3}+0.4+\frac{14}{15}}+\frac{(3.75-0.625) \frac{48}{125}}{12.8 \cdot 0.25}$. | Solution.
$\frac{0.5+\frac{1}{4}+\frac{1}{6}+0.125}{\frac{1}{3}+0.4+\frac{14}{15}}+\frac{(3.75-0.625) \frac{48}{125}}{12.8 \cdot 0.25}=\frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}}{\frac{1}{3}+\frac{2}{5}+\frac{14}{15}}+\frac{3.125 \cdot 48}{3.2 \cdot 125}=$
$=\frac{25}{24} \cdot \frac{3}{5}+\frac{1.2}{3.2}=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,349 |
1.024. $\frac{0.725+0.6+\frac{7}{40}+\frac{11}{20}}{0.128 \cdot 6 \frac{1}{4}-0.0345: \frac{3}{25}} \cdot 0.25$. | ## Solution.
$$
\begin{aligned}
& \frac{0.725+0.6+\frac{7}{40}+\frac{11}{20}}{0.128 \cdot 6 \frac{1}{4}-0.0345: \frac{3}{25}} \cdot 0.25=\frac{1.325+\frac{29}{40}}{0.128 \cdot 6.25-0.0345: 0.12} \cdot 0.25= \\
& =\frac{1.325+0.725}{0.8-0.2875} \cdot 0.25=\frac{2.05}{0.5125} \cdot 0.25=1
\end{aligned}
$$
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,350 |
1.025. $\left((520 \cdot 0.43): 0.26-217 \cdot 2 \frac{3}{7}\right)-\left(31.5: 12 \frac{3}{5}+114 \cdot 2 \frac{1}{3}+61 \frac{1}{2}\right)$. | Solution.
$$
\begin{aligned}
& \left((520 \cdot 0.43): 0.26-217 \cdot 2 \frac{3}{7}\right)-\left(31.5: 12 \frac{3}{5}+114 \cdot 2 \frac{1}{3}+61 \frac{1}{2}\right)= \\
& =\left(223.6: 0.26-217 \cdot \frac{17}{7}\right)-\left(\frac{63}{2} \cdot \frac{5}{63}+114 \cdot \frac{7}{3}+\frac{123}{2}\right)= \\
& =(860-527)-\l... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,351 |
1.026. $\frac{(3.4-1.275) \cdot \frac{16}{17}}{\frac{5}{18} \cdot\left(1 \frac{7}{85}+6 \frac{2}{17}\right)}+0.5\left(2+\frac{12.5}{5.75+\frac{1}{2}}\right)$. | Solution.
$$
\begin{aligned}
& \frac{(3.4-1.275) \cdot \frac{16}{17}}{\frac{5}{18} \cdot\left(1 \frac{7}{85}+6 \frac{2}{17}\right)}+0.5\left(2+\frac{12.5}{5.75+\frac{1}{2}}\right)=\frac{2.125 \cdot \frac{16}{17}}{\frac{5}{18}\left(\frac{92}{85}+\frac{104}{17}\right)}+\frac{1}{2}\left(2+\frac{12.5}{6.25}\right)= \\
& =... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,352 |
1.027. $\left(\frac{3.75+2 \frac{1}{2}}{2 \frac{1}{2}-1.875}-\frac{2 \frac{3}{4}+1.5}{2.75-1 \frac{1}{2}}\right) \cdot \frac{10}{11}$. | Solution.
$\left(\frac{3.75+2 \frac{1}{2}}{2 \frac{1}{2}-1.875}-\frac{2 \frac{3}{4}+1.5}{2.75-1 \frac{1}{2}}\right) \cdot \frac{10}{11}=\left(\frac{3.75+2.5}{2.5-1.875}-\frac{2.75+1.5}{2.75-1.5}\right) \cdot \frac{10}{11}=$
$=\left(\frac{6.25}{0.625}-\frac{4.25}{1.25}\right) \cdot \frac{10}{11}=\left(10-\frac{17}{5}\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,353 |
1.028. $((21.85: 43.7+8.5: 3.4): 4.5): 1 \frac{2}{5}+1 \frac{11}{21}$. | ## Solution.
$$
\begin{aligned}
& ((21.85: 43.7+8.5: 3.4): 4.5): 1 \frac{2}{5}+1 \frac{11}{21}=\left((0.5+2.5): 4 \frac{1}{2}\right): \frac{7}{5}+\frac{32}{21}= \\
& =\left(3 \cdot \frac{2}{9}\right) \cdot \frac{5}{7}+\frac{32}{21}=\frac{10}{21}+\frac{32}{21}=\frac{42}{21}=2
\end{aligned}
$$
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,354 |
1.029. $\left(1 \frac{2}{5}+3.5: 1 \frac{1}{4}\right): 2 \frac{2}{5}+3.4: 2 \frac{1}{8}-0.35$. | ## Solution.
$$
\begin{aligned}
& \left(1 \frac{2}{5}+3.5: 1 \frac{1}{4}\right): 2 \frac{2}{5}+3.4: 2 \frac{1}{8}-0.35= \\
& =(1.4+3.5: 1.25): 2.4+3.4: 2.125-0.35=(1.4+2.8): 2.4+ \\
& +1.6-0.35=4.2: 2.4+1.25=1.75+1.25=3
\end{aligned}
$$
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,355 |
1.030. $\frac{\left(0.3275-\left(2 \frac{15}{88}+\frac{4}{33}\right): 12 \frac{2}{9}\right): 0.07}{(13-0.416): 6.05+1.92}$. | Solution.
$$
\begin{aligned}
& \frac{\left(0.3275-\left(2 \frac{15}{88}+\frac{4}{33}\right): 12 \frac{2}{9}\right): 0.07}{(13-0.416): 6.05+1.92}=\frac{\left(0.3275-\left(\frac{191}{88}+\frac{4}{33}\right) \cdot \frac{9}{110}\right): 0.07}{12.584: 6.05+1.92}= \\
& =\frac{\left(\frac{131}{400}-\frac{605}{264} \cdot \fra... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,356 |
1.031. $\frac{\frac{5}{6}-\frac{21}{45}}{1 \frac{5}{6}} \cdot \frac{1,125+1 \frac{3}{4}-\frac{5}{12}}{0.59}$. | Solution.
$$
\begin{aligned}
& \frac{\frac{5}{6}-\frac{21}{45}}{1 \frac{5}{6}} \cdot \frac{1.125+1 \frac{3}{4}-\frac{5}{12}}{0.59}=\frac{\frac{5}{6}-\frac{7}{15}}{\frac{11}{6}} \cdot \frac{\frac{9}{8}+\frac{7}{4}-\frac{5}{12}}{\frac{59}{100}}=\frac{11}{30} \cdot \frac{6}{11} \cdot \frac{59}{24} \cdot \frac{100}{59}= \... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,357 |
1.032. $\frac{\left(3^{-1}-\sqrt{1 \frac{7}{9}}\right)^{-2}: 0.25}{\frac{37}{300}: 0.0925}+12.5 \cdot 0.64$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(3^{-1}-\sqrt{1 \frac{7}{9}}\right)^{-2}: 0.25}{\frac{37}{300}: 0.0925}+12.5 \cdot 0.64=\frac{\left(\frac{1}{3}-\sqrt{\frac{16}{9}}\right)^{-2} \cdot 4}{\frac{37}{300} \cdot \frac{400}{37}}+8= \\
& =\frac{\left(\frac{1}{3}-\frac{4}{3}\right)^{-2} \cdot 4}{\frac{4}{3}}+8=3(-... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,358 |
1.033. $\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5$. | ## Solution.
$\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5=\frac{\left(\frac{5}{8}+\frac{65}{24}\right) \cdot \frac{2}{5} \cdot \frac{1}{2}}{\left(\frac{13}{10}+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}}=\frac{\frac{10}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,359 |
1.034. $\frac{((7-6.35): 6.5+9.9) \cdot \frac{1}{12.8}}{\left(1.2: 36+1 \frac{1}{5}: 0.25-1 \frac{5}{6}\right) \cdot 1 \frac{1}{4}}: 0.125$. | ## Solution.
$$
\frac{((7-6.35): 6.5+9.9) \cdot \frac{1}{12.8}}{\left(1.2: 36+1 \frac{1}{5}: 0.25-1 \frac{5}{6}\right) \cdot 1 \frac{1}{4}}: 0.125=\frac{(0.65: 6.5+9.9) \cdot \frac{5}{64} \cdot 8}{\left(\frac{6}{5} \cdot \frac{1}{36}+\frac{6}{5} \cdot 4-\frac{11}{6}\right) \cdot \frac{5}{4}}=
$$
$$
=\frac{(0.1+9.9) \... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,360 |
1.037. $\frac{\left(\left(4,625-\frac{13}{18} \cdot \frac{9}{26}\right): \frac{9}{4}+2.5: 1.25: 6.75\right): 1 \frac{53}{68}}{\left(\frac{1}{2}-0.375\right): 0.125+\left(\frac{5}{6}-\frac{7}{12}\right):(0.358-1.4796: 13.7)}$. | Solution.
$$
\begin{aligned}
& \frac{\left(\left(4,625-\frac{13}{18} \cdot \frac{9}{26}\right): \frac{9}{4}+2.5: 1.25: 6.75\right): 1 \frac{53}{68}}{\left(\frac{1}{2}-0.375\right): 0.125+\left(\frac{5}{6}-\frac{7}{12}\right):(0.358-1.4796: 13.7)}= \\
& =\frac{\left(\left(\frac{37}{8}-\frac{1}{4}\right) \cdot \frac{4}{... | \frac{17}{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,361 |
1.039. $\left(\frac{(3.2-1.7): 0.003}{\left(\frac{29}{35}-\frac{3}{7}\right) \cdot 4: 0.2}-\frac{\left(1 \frac{13}{20}-1.5\right) \cdot 1.5}{\left(2.44+1 \frac{14}{25}\right) \cdot \frac{1}{8}}\right): 62 \frac{1}{20}+1.364: 0.124$. | Solution.
$$
\begin{aligned}
& \left(\frac{(3.2-1.7): 0.003}{\left(\frac{29}{35}-\frac{3}{7}\right) \cdot 4: 0.2}-\frac{\left(1 \frac{13}{20}-1.5\right) \cdot 1.5}{\left(2.44+1 \frac{14}{25}\right) \cdot \frac{1}{8}}\right): 62 \frac{1}{20}+1.364: 0.124= \\
& =\left(\frac{1.5: 0.003}{\frac{14}{35} \cdot 4 \cdot 5}-\fr... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,362 |
1.041. $\frac{\left(4-3.5 \cdot\left(2 \frac{1}{7}-1 \frac{1}{5}\right)\right): 0.16}{X}=\frac{3 \frac{2}{7}-\frac{3}{14}: \frac{1}{6}}{41 \frac{23}{84}-40 \frac{49}{60}}$. | ## Solution.
$$
\begin{aligned}
& X=\frac{\left(4-3.5 \cdot\left(2 \frac{1}{7}-1 \frac{1}{5}\right)\right): 0.16 \cdot\left(41 \frac{23}{84}-40 \frac{49}{60}\right)}{3 \frac{2}{7}-\frac{3}{14}: \frac{1}{6}}= \\
& =\frac{\left(4-3.5 \cdot\left(\frac{15}{7}-\frac{6}{5}\right)\right): 0.16 \cdot \frac{16}{35}}{\frac{23}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,364 |
1.042. $\frac{1.2: 0.375-0.2}{6 \frac{4}{25}: 15 \frac{2}{5}+0.8}=\frac{0.016: 0.12+0.7}{X}$. | Solution.
$$
\begin{aligned}
& X=\frac{(0.016: 0.12+0.7)\left(6 \frac{4}{25}: 15 \frac{2}{5}+0.8\right)}{1.2: 0.375-0.2}=\frac{\left(\frac{2}{125} \cdot \frac{3}{25}+\frac{7}{10}\right)\left(\frac{154}{25} \cdot \frac{77}{5}+\frac{4}{5}\right)}{3.2-0.2}= \\
& =\frac{\left(\frac{2}{15}+\frac{7}{10}\right)\left(\frac{2}... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,365 |
1.043. $\frac{0.125 X}{\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7}{0.675 \cdot 2.4-0.02}$. | Solution.
$X=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7 \cdot\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}{(0.675 \cdot 2.4-0.02) \cdot 0.125}=\frac{\left(\frac{91}{63}-\frac{17}{21}\right) \cdot \frac{7}{10} \cdot \frac{4}{15} \cdot \frac{135}{16}}{(1.62-0.02) \cdot 0.125}=$ $=\frac{\... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,366 |
1.046.
$$
\frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{(6.3+1.7)^{2}-4 \cdot 6.3 \cdot 1.7}}
$$ | Solution.
$$
\begin{aligned}
& \frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{(6.3+1.7)^{2}-4 \cdot 6.3 \cdot 1.7}}=\frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{6.3^{2}+2 \cdot 6.3 \cdot 1.7+1.7^{2}-4 \cdot 6.3 \cdo... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,369 |
1.047. $\left(\frac{\sqrt{561^{2}-459^{2}}}{4 \frac{2}{7} \cdot 0.15+4 \frac{2}{7}: \frac{20}{3}}+4 \sqrt{10}\right): \frac{1}{3} \sqrt{40}$. | Solution.
$$
\begin{aligned}
& \left(\frac{\sqrt{561^{2}-459^{2}}}{4 \frac{2}{7} \cdot 0.15+4 \frac{2}{7} : \frac{20}{3}}+4 \sqrt{10}\right): \frac{1}{3} \sqrt{40}=\left(\frac{\sqrt{(561+459)(561-459)}}{\frac{30}{7} \cdot \frac{3}{20}+\frac{30}{7} \cdot \frac{3}{20}}+4 \sqrt{10}\right) \times \\
& \times \frac{3}{2 \s... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,370 |
1.049. $\frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+\left(\frac{2}{3}\right)^{-2}}+4.75$. | Solution.
$$
\begin{aligned}
& \frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+\left(\frac{2}{3}\right)^{-2}}+4.75=\frac{\frac{1}{2^{2}}+1}{-\frac{1}{(0.5)^{2}}-\frac{5}{(-2)^{-2}}+\left(\frac{3}{2}\right)^{2}}+4.75= \\
& =\frac{\frac{1}{4}+1}{\frac{1}{0.25}-\frac{5}{4}+\frac{9}{4}}+4.75=\frac{-\frac{4}{4}}{4+1}+4.75=\frac{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,371 |
2.003
$$
\frac{\left(\sqrt{a^{2}+a \sqrt{a^{2}-b^{2}}}\right)-\left(\sqrt{a^{2}-a \sqrt{a^{2}-b^{2}}}\right)^{2}}{2 \sqrt{a^{3} b}} \cdot\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-2\right)(a>b>0)
$$ | Solution.
Let
$$
X=\frac{\left(\sqrt{a^{2}+a \sqrt{a^{2}-b^{2}}}\right)-\left(\sqrt{a^{2}-a \sqrt{a^{2}-b^{2}}}\right)^{2}}{2 \sqrt{a^{3} b}}=
$$
$$
\begin{aligned}
& =\frac{\left.\left(\sqrt{a\left(a+\sqrt{a^{2}-b^{2}}\right)}-\sqrt{a\left(a-\sqrt{a^{2}-b^{2}}\right.}\right)\right)^{2}}{2 a \sqrt{a b}}= \\
& =\frac... | \frac{(\sqrt{}+\sqrt{b})^{2}}{-b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,374 |
2.004. $\left(\frac{(a+b)^{-n / 4} \cdot c^{1 / 2}}{a^{2-n} b^{-3 / 4}}\right)^{4 / 3}:\left(\frac{b^{3} c^{4}}{(a+b)^{2 n} a^{16-8 n}}\right)^{1 / 6} ; b=0.04$. | ## Solution.
Domain of definition: $a \neq -b \neq -0.04$.
Let $X=\left(\frac{(a+b)^{-n / 4} \cdot c^{1 / 2}}{a^{2-n} b^{-3 / 4}}\right)^{4 / 3}=\frac{(a+b)^{-n / 3} \cdot c^{2 / 3}}{a^{(8-4 n) / 3} b^{-1}}=\frac{b \cdot c^{2 / 3}}{a^{(8-4 n) / 3} \cdot(a+b)^{n / 3}} ;$
$$
Y=\left(\frac{b^{3} c^{4}}{(a+b)^{2 n} a^{1... | 0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,375 |
2.008. $\left(\left(\frac{2^{3 / 2}+27 y^{3 / 5}}{\sqrt{2}+3 \sqrt[5]{y}}+3 \sqrt[10]{32 y^{2}}-2\right) \cdot 3^{-2}\right)^{5}$. | ## Solution.
Domain of definition: $\sqrt{2}+3 \sqrt[s]{y} \neq 0, \Leftrightarrow y \neq-\left(\frac{\sqrt{2}}{3}\right)^{5}$.
$\left(\left(\frac{2^{3 / 2}+27 y^{3 / 5}}{\sqrt{2}+3 \sqrt[5]{y}}+3 \sqrt[10]{32 y^{2}}-2\right) \cdot 3^{-2}\right)^{5}=$
$=\left(\left(\frac{(\sqrt{2})^{3}+(3 \sqrt[5]{y})^{3}}{\sqrt{2}+... | y^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,378 |
2.009. $\frac{2 \sqrt{1+\frac{1}{4}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)^{2}}}{\sqrt{1+\frac{1}{4}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)^{2}}-\frac{1}{2}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)}$. | Solution.
Domain of definition: $0<t \neq 1$.
$$
\begin{aligned}
& \frac{\sqrt[2]{1+\frac{1}{4}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)^{2}}}{\sqrt{1+\frac{1}{4}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)^{2}}-\frac{1}{2}\left(\sqrt{\frac{1}{t}}-\sqrt{t}\right)}=\frac{2 \sqrt{1+\frac{1}{4}\left(\frac{1}{t}-2+t\right)}}{... | \frac{1+}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,379 |
2.011. $\left(\frac{1+\sqrt{x}}{\sqrt{1+x}}-\frac{\sqrt{1+x}}{1+\sqrt{x}}\right)^{2}-\left(\frac{1-\sqrt{x}}{\sqrt{1+x}}-\frac{\sqrt{1+x}}{1-\sqrt{x}}\right)^{2}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}1+x \geq 0, \\ x \geq 0, \\ 1-\sqrt{x} \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ x \neq 1 .\end{array}\right.\right.$
$$
\begin{aligned}
& \left(\frac{1+\sqrt{x}}{\sqrt{1+x}}-\frac{\sqrt{1+x}}{1+\sqrt{x}}\right)^{2}-\left(\frac{1-\sqrt... | \frac{16x\sqrt{x}}{(1-x^{2})(x-1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,381 |
2.013. $\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)$. | Solution.
Domain of definition: $\left\{\begin{array}{l}a \geq 0, \\ \sqrt{a}-\sqrt{a-1} \neq 0, \Leftrightarrow\left\{\begin{array}{l}a \geq 0, \\ a>1\end{array}\right. \\ \frac{a+1}{a-1} \geq 0\end{array} \Leftrightarrow a>1\right.$.
Let $X$ be the expression in the first parentheses, $Y$ be the expression in the s... | \sqrt{-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,383 |
2.014. $\frac{x-y}{x^{3 / 4}+x^{1 / 2} y^{1 / 4}} \cdot \frac{x^{1 / 2} y^{1 / 4}+x^{1 / 4} y^{1 / 2}}{x^{1 / 2}+y^{1 / 2}} \cdot \frac{x^{1 / 4} y^{-1 / 4}}{x^{1 / 2}-2 x^{1 / 4} y^{1 / 4}+y^{1 / 2}}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0, \\ x \neq y .\end{array}\right.$
$$
\begin{aligned}
& \frac{x-y}{x^{3 / 4}+x^{1 / 2} y^{1 / 4}} \cdot \frac{x^{1 / 2} y^{1 / 4}+x^{1 / 4} y^{1 / 2}}{x^{1 / 2}+y^{1 / 2}} \cdot \frac{x^{1 / 4} y^{-1 / 4}}{x^{1 / 2}-2 x^{1 / 4} y^{1 / 4}+... | \frac{\sqrt[4]{x}+\sqrt[4]{y}}{\sqrt[4]{x}-\sqrt[4]{y}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,384 |
2.016. $\left(\frac{\left(z^{2 / p}+z^{2 / q}\right)^{2}-4 z^{2 / p+2 / q}}{\left(z^{1 / p}-z^{1 / q}\right)^{2}+4 z^{1 / p+1 / q}}\right)^{1 / 2}$. | ## Solution.
Domain of definition: $z \neq 0, p \neq 0, q \neq 0$.
$$
\left(\frac{\left(z^{2 / p}+z^{2 / q}\right)^{2}-4 z^{2 / p+2 / q}}{\left(z^{1 / p}-z^{1 / q}\right)^{2}+4 z^{1 / p+1 / q}}\right)^{1 / 2}=\left(\frac{\left(z^{2 / p}\right)^{2}+2 z^{2 / p+2 / q}+\left(z^{2 / q}\right)^{2}-4 z^{2 / p+2 / q}}{\left(... | |z^{1/p}-z^{1/q}| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,386 |
2.018. $\left(\frac{1+x+x^{2}}{2 x+x^{2}}+2-\frac{1-x+x^{2}}{2 x-x^{2}}\right)^{-1} \cdot\left(5-2 x^{2}\right) ; x=\sqrt{3.92 .}$ | ## Solution.
$$
\begin{aligned}
& \left(\frac{1+x+x^{2}}{2 x+x^{2}}+2-\frac{1-x+x^{2}}{2 x-x^{2}}\right)^{-1} \cdot\left(5-2 x^{2}\right)= \\
& =\left(\frac{1+x+x^{2}}{x(2+x)}+2-\frac{1-x+x^{2}}{x(2-x)}\right)^{-1} \cdot\left(5-2 x^{2}\right)= \\
& =\left(\frac{(2-x)\left(1+x+x^{2}\right)+2 x(2+x)(2-x)-(2+x)\left(1-x+... | 0.04 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,388 |
2.019. $\frac{\left(x^{2}-y^{2}\right)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}}}-\left(\sqrt[3]{x y}+\sqrt[3]{y^{2}}\right) x=64$. | Solution.
$$
\begin{aligned}
& \text { Domain: } z=\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}} \neq 0 \\
& \frac{\left(x^{2}-y^{2}\right)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}}}-\left(\sqrt[3]{x y}+\sqrt[3]{y^{2}}\right)= \\... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,389 |
2.021. $\frac{4 x\left(x+\sqrt{x^{2}-1}\right)^{2}}{\left(x+\sqrt{x^{2}-1}\right)^{4}-1}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\left(x+\sqrt{x^{2}-1}\right)^{4}-1 \neq 0, \\ x^{2}-1 \geq 0\end{array}\right.$
$\frac{4 x\left(x+\sqrt{x^{2}-1}\right)^{2}}{\left(x+\sqrt{x^{2}-1}\right)^{4}-1}=\frac{4 x\left(x+\sqrt{x^{2}-1}\right)^{2}}{\left(x^{2}+2 x \sqrt{x^{2}-1}+x^{2}-1\right)^{2}-1}=$
... | \frac{1}{\sqrt{x^{2}-1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,391 |
2.022. $\frac{\sqrt{(x+2)^{2}-8 x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}$.
2.022. $\frac{\sqrt{(x+2)^{2}-8x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}$.
| ## Solution.
Domain of definition: $0<x \neq 2$.
$$
\begin{aligned}
& \frac{\sqrt{(x+2)^{2}-8 x}}{\sqrt{x}-\frac{2}{\sqrt{x}}}=\frac{\sqrt{x^{2}+4 x+4-8 x}}{\frac{(\sqrt{x})^{2}-2}{\sqrt{x}}}=\frac{\sqrt{x} \sqrt{x^{2}-4 x-4}}{x-2}= \\
& =\frac{\sqrt{x} \sqrt{\left(x^{2}-2\right)^{2}}}{x-2}=\frac{\sqrt{x} \cdot|x-2|}... | -\sqrt{x}forx\in(0;2);\sqrt{x}forx\in(2;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,392 |
2.023. $\sqrt[4]{6 x(5+2 \sqrt{6})} \cdot \sqrt{3 \sqrt{2 x}-2 \sqrt{3 x}}$.
2.023. $\sqrt[4]{6 x(5+2 \sqrt{6})} \cdot \sqrt{3 \sqrt{2 x}-2 \sqrt{3 x}}$. | ## Solution.
Domain of definition: $x \geq 0$.
$$
\begin{aligned}
& \sqrt[4]{6 x(5+2 \sqrt{6})} \cdot \sqrt{3 \sqrt{2 x}-2 \sqrt{3 x}}=\sqrt[4]{6 x(5+2 \sqrt{6})} \cdot \sqrt{\sqrt{6 x}(\sqrt{3}-\sqrt{2})}= \\
& =\sqrt[4]{6 x(5+2 \sqrt{6})} \cdot \sqrt[4]{(\sqrt{6 x}(\sqrt{3}-\sqrt{2}))^{2}}=\sqrt[4]{6 x(5+2 \sqrt{6}... | \sqrt{6x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,393 |
2.025. $\frac{a^{3}-a-2 b-b^{2} / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^{3}+a^{2}+a b+a^{2} b}{a^{2}-b^{2}}+\frac{b}{a-b}\right) ;$
$a=23 ; b=22$. | Solution.
$$
\begin{aligned}
& \frac{a^{3}-a-2 b-b^{2} / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^{3}+a^{2}+a b+a^{2} b}{a^{2}-b^{2}}+\frac{b}{a-b}\right)= \\
& =\frac{\frac{a^{4}-a^{2}-2 a b-b^{2}}{a}}{\left(1-\sqrt{\frac{a+b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,395 |
2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.
2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.
(Note: The original text and the translation are identical as the expression... | Solution.
Domain of definition: $\left\{\begin{array}{l}a>0, \\ a \neq \sqrt{2} .\end{array}\right.$
$$
\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a^{2} \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}=\sqrt{\frac{a^{2}+2 a \sqrt{2}+(\sqrt{2})^{2}}{a \sqrt{2}}}-
$$
$$
\begin{aligned}
& -\frac{a^{2} ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,399 |
2.031. $\left(\frac{\sqrt[4]{a^{3}}-1}{\sqrt[4]{a}-1}+\sqrt[4]{a}\right)^{1 / 2} \cdot\left(\frac{\sqrt[4]{a^{3}}+1}{\sqrt[4]{a}+1}-\sqrt{a}\right) \cdot\left(a-\sqrt{a^{3}}\right)^{-1}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}a>0, \\ a \neq 1 .\end{array}\right.$
$$
\begin{aligned}
& \left(\frac{\sqrt[4]{a^{3}}-1}{\sqrt[4]{a}-1}+\sqrt[4]{a}\right)^{1 / 2} \cdot\left(\frac{\sqrt[4]{a^{3}}+1}{\sqrt[4]{a}+1}-\sqrt{a}\right) \cdot\left(a-\sqrt{a^{3}}\right)^{-1}= \\
& =\left(\frac{(\s... | \frac{1}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,400 |
2.033. $\frac{\sqrt{(2 p+1)^{3}}+\sqrt{(2 p-1)^{3}}}{\sqrt{4 p+2 \sqrt{4 p^{2}-1}}}$.
2.033. $\frac{\sqrt{(2 p+1)^{3}}+\sqrt{(2 p-1)^{3}}}{\sqrt{4 p+2 \sqrt{4 p^{2}-1}}}$. | ## Solution.
Domain of definition: $p \geq \frac{1}{2}$.
$$
\begin{aligned}
& \frac{\sqrt{(2 p+1)^{3}}+\sqrt{(2 p-1)^{3}}}{\sqrt{4 p+2 \sqrt{4 p^{2}-1}}}= \\
& =\frac{(\sqrt{2 p+1}+\sqrt{2 p-1})\left((\sqrt{2 p+1})^{2}-\sqrt{2 p+1} \cdot \sqrt{2 p-1}+(\sqrt{2 p-1})^{2}\right)}{\sqrt{2 p+1+2 \sqrt{4 p^{2}-1}+2 p-1}}= ... | 4p-\sqrt{4p^{2}-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,402 |
2.035. $\left(\frac{a+2}{\sqrt{2 a}}-\frac{a}{\sqrt{2 a}+2}+\frac{2}{a-\sqrt{2 a}}\right) \cdot \frac{\sqrt{a}-\sqrt{2}}{a+2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}a>0, \\ a \neq 2 .\end{array}\right.$
$$
\begin{aligned}
& \left(\frac{a+2}{\sqrt{2} i}-\frac{a}{\sqrt{2 a}+2}+\frac{2}{a-\sqrt{2 a}}\right) \cdot \frac{\sqrt{a}-\sqrt{2}}{a+2}= \\
& =\left(\frac{a+2}{\sqrt{2 a}}-\frac{a}{\sqrt{2}(\sqrt{a}+\sqrt{2})}+\frac{2}... | \frac{1}{\sqrt{}+\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,404 |
2.037. $\frac{1-x^{-2}}{x^{1/2}-x^{-1/2}}-\frac{2}{x^{3/2}}+\frac{x^{-2}-x}{x^{1/2}-x^{-1/2}}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq 1 .\end{array}\right.$
$$
\begin{aligned}
& \frac{1-x^{-2}}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}+\frac{x^{-2}-x}{x^{1 / 2}-x^{-1 / 2}}=\frac{1-\frac{1}{x^{2}}}{\sqrt{x}-\frac{1}{\sqrt{x}}}+\frac{\frac{1}{x^{2}}-x}{\sqrt{x}-\frac{1}{\sqrt{x... | -\sqrt{x}(1+\frac{2}{x^{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,405 |
2.038. $\left(\frac{\sqrt{a}}{2}-\frac{1}{2 \sqrt{a}}\right)^{2} \cdot\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)$. | Solution.
Domain of definition: $0<a \neq 1$.
$$
\begin{aligned}
& \left(\frac{\sqrt{a}}{2}-\frac{1}{2 \sqrt{a}}\right)^{2} \cdot\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)=\left(\frac{(\sqrt{a})^{2}-1}{2 \sqrt{a}}\right)^{2} \cdot \frac{(\sqrt{a}-1)^{2}-(\sqrt{a}+1)^{2}}{(\sqrt{a}+1)(\sq... | \frac{1-}{\sqrt{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,406 |
2.039. $\frac{9 b^{4 / 3}-\frac{a^{3 / 2}}{b^{2}}}{\sqrt{a^{3 / 2} b^{-2}+6 a^{3 / 4} b^{-1 / 3}+9 b^{4 / 3}}} \cdot \frac{b^{2}}{a^{3 / 4}-3 b^{5 / 3}} ; \quad b=4$. | Solution.
$$
\begin{aligned}
& \frac{9 b^{4 / 3}-\frac{a^{3 / 2}}{b^{2}}}{\sqrt{a^{3 / 2} b^{-2}+6 a^{3 / 4} b^{-1 / 3}+9 b^{4 / 3}}} \cdot \frac{b^{2}}{a^{3 / 4}-3 b^{5 / 3}}=\frac{\frac{9 b^{4 / 3} \cdot b^{2}-a^{3 / 2}}{b^{2}}}{\sqrt{\frac{a^{3 / 2}}{b^{2}}+\frac{6 a^{3 / 4}}{b^{+1 / 3}}+9 b^{4 / 3}}} \times \\
& \... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,407 |
2.041. $\frac{1}{2(1+\sqrt{a})}+\frac{1}{2(1-\sqrt{a})}-\frac{a^{2}+2}{1-a^{3}}$. | ## Solution.
Domain of definition: $0 \leq a \neq 1$.
$$
\begin{aligned}
& \frac{1}{2(1+\sqrt{a})}+\frac{1}{2(1-\sqrt{a})}-\frac{a^{2}+2}{1-a^{3}}=\frac{1-\sqrt{a}+1+\sqrt{a}}{2(1+\sqrt{a})(1-\sqrt{a})}-\frac{a^{2}+2}{1-a^{3}}=\frac{2}{2(1-a)}- \\
& -\frac{a^{2}+2}{(1-a)\left(1+a+a^{2}\right)}=\frac{1}{1-a}-\frac{a^{... | \frac{-1}{^{2}++1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,408 |
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