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1.26 Given the expression
$$
\begin{aligned}
& \frac{x^{3}-a^{-\frac{2}{3}} \cdot b^{-1}\left(a^{2}+b^{2}\right) x+b^{\frac{1}{2}}}{b^{\frac{3}{2}} x^{2}} \\
& \text { Substitute } x=a^{\frac{2}{3} b}-\frac{1}{2} \text { and simplify the result }
\end{aligned}
$$
Eliminate the irrationality in the denominator of the ... | 1.26 After performing the specified substitution, we find
$$
\begin{aligned}
& \frac{\left(a^{\frac{2}{3}} b^{-\frac{1}{2}}\right)^{3}-a^{-\frac{2}{3}+\frac{2}{3}} b^{-1-\frac{1}{2}}\left(a^{2}+b^{2}\right)+b^{\frac{1}{2}}}{b^{\frac{3}{2}}\left(a^{\frac{2}{3}} b^{\left.-\frac{1}{2}\right)^{2}}\right.}= \\
& =\frac{a^{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.53 \frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}-\sqrt{3}=2$. | 1.53 Consider the equality
$$
\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}=2+\sqrt{3}
$$
Obviously, if this equality is true, then the given equality is also true. Let
$$
a=\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}, b=2+\sqrt{3}
$$
It is easy to establish that $a>0$ and ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$1.54 \sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=4$. | ### 1.54 Let
$\sqrt[3]{38+\sqrt{1445}}+\sqrt[3]{38-\sqrt{1445}}=x$.
Cubing both sides of this equation, and using formula (1.11), we get
$38+\sqrt{1445}+38-\sqrt{1445}+3 \sqrt[3]{(38+\sqrt{1445})(38-\sqrt{1445})} x=x^{3}$,
or $x^{3}+3 x-76=0$. By substitution, we verify that $x=4$ is one of the roots of the resulti... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.1 \quad \frac{2}{3-x}+\frac{1}{2}=\frac{6}{x(3-x)}$. | 2.1 Let's move all terms of the equation to the left side and transform the obtained equation to the form
$$
\frac{x^{2}-7 x+12}{2 x(3-x)}=0
$$
From the equation $x^{2}-7 x+12=0$, we find $x_{1}=3, x_{2}=4$. When $x=3$, the denominator becomes zero; therefore, 3 is not a root. Thus, $x=4$.
Answer: $x=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.13 \sqrt{x-2}=x-4$.
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$2.13 \sqrt{x-2}=x-4$. | 2.13 Let's square both sides of the equation:
$$
\begin{aligned}
& (\sqrt{x-2})^{2}=(x-4)^{2} ; x-2=x^{2}-8 x+16 \\
& x^{2}-9 x+18=0 ; x_{1}=3 ; x_{2}=6
\end{aligned}
$$
Let's check the found roots by substituting them into the original equation. If $x=3$, we get $1=-1$ - an incorrect equality; if $x=6$, we get $2=2$... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.14 \sqrt{15-x}+\sqrt{3-x}=6$. | 2.14 We have $\sqrt{15-x}=6-\sqrt{3-x}$. Squaring both sides of the equation, we get $15-x=36-12 \sqrt{3-x}+3-x$; $\sqrt{3-x}=2 ; 3-x=4 ; x=-1$. Checking shows that this value is a root of the equation.
Omвem: $x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.18 \frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$. | 2.18 Reducing the fractions in the left part of the equation, we get
$\sqrt[3]{x^{2}}+1-\sqrt[3]{x}+1=4$, or $\sqrt[3]{x^{2}}-\sqrt[3]{x}-2=0$.
From this, $\sqrt[3]{x}=-1$ or $\sqrt[3]{x}=2$. Since $\sqrt[3]{x}$ cannot be equal to -1 (as this would make the denominators of the original fractions zero), then $x=8$.
O... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.20 \sqrt[7]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | 2.20 Instruction. Use the substitution $\sqrt[7]{\frac{5-x}{x+3}}=z$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$. | 2.49 We have
$$
\left(x^{2}+1\right)(x-2)+\left(x^{2}+2\right)(x+1)=-2(x+1)(x-2)
$$
Performing the transformations, we get $2 x^{3}+x^{2}+x-4=0$. Since the sum of all coefficients is zero, then $x=1$. Let's write the last equation in the form
$2 x^{3}-2 x^{2}+3 x^{2}-3 x+4 x-4=0$,
or $(x-1)\left(2 x^{2}+3 x+4\right... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$.
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$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$. | 2.55 By squaring both sides of the equation, we get
$$
\begin{aligned}
& x+1+4 x+13+2 \sqrt{(x+1)(4 x+13)}=3 x+12 \\
& \sqrt{(x+1)(4 x+13)}=-(x+1)
\end{aligned}
$$
Another squaring would eliminate the irrationality, but there is no need for this transformation here. We notice that the derived equation can have a solu... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$.
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$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$. | ### 2.58 We have
$\sqrt{x}=2-\frac{2 x+1}{x+2}$, or $\sqrt{x}=\frac{3}{x+2}$.
After squaring and performing transformations, the equation will take the form $x^{3}+4 x^{2}+4 x-9=0$. Notice that the sum of the coefficients of the last equation is zero, which means it has a root $x=1$. Therefore,
$$
x^{3}-x^{2}+5 x^{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
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$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$. | 2.59 We have $\sqrt{x+2}=\sqrt[3]{3 x+2}$. After raising the equation to the sixth power, we get $(x+2)^{3}=(3 x+2)^{2}$, or $x^{3}-3 x^{2}+4=0$, or $(x+1)\left(x^{2}-4 x+4\right)=0$.
From this, $x_{1}=-1$ (extraneous root), $x_{2}=2$.
Answer: $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.78 For what values of a do the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$ have a common root? | 2.78 Subtracting the second equation from the first, we get
$$
a x - x + 1 - a = 0, \text{ or } (a-1)(x-1) = 0
$$
$$
x^{2} + x + 1 = 0
$$
such an equation has no real roots. If $a \neq 1$, then $x = 1$ and, therefore, $1 + a + 1 = 2$, i.e., $a = -2$.
Answer: when $a = -2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.8 The front wheel of a moving model makes 6 more revolutions than the rear wheel over a distance of 120 m. If the circumference of the front wheel is increased by $\frac{1}{4}$ of its length, and the circumference of the rear wheel is increased by $\frac{1}{5}$ of its length, then over the same distance, the front wh... | 3.8 The circumference of a wheel $C$, the number of revolutions $n$, and the distance $s$ are related by the formula $C n=s$. We will fill in the table of values of these quantities in the order indicated by the numbers (1), (2), .., (12).
| Wheel | Before change | | | After change | | |
| :---: | :---: | :---: | ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.14 A positive integer is thought of. To its representation, the digit 7 is appended on the right, and from the resulting new number, the square of the thought number is subtracted. The remainder is then reduced by $75\%$ of this remainder, and the thought number is subtracted again. In the final result, zero is obtai... | 3.14 Let a number $x$ be thought of. Then, following the text of the condition, we get the numbers
$10 x+7, 10 x+7-x^{2}$ and the remainder $\frac{25}{100}\left(10 x+7-x^{2}\right)$.
Then $\frac{1}{4}\left(10 x+7-x^{2}\right)-x=0$, or $x^{2}-6 x-7=0$. Only the value $x=7$ works.
Answer: 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.18 The distance between points $A$ and $B$ is 270 m. A body moves uniformly from $A$ to $B$; upon reaching $B$, it immediately returns with the same speed. A second body, which leaves $B$ for $A$ 11 s after the first body leaves $A$, moves uniformly but more slowly. On its way from $\boldsymbol{B}$ to $\boldsymbol{A}... | 3.18 A convenient model of the problem is a graph of uniform motion in the coordinate system "path" ($s$ - in meters), "time" ($t$ in seconds). Let $AC$ (Fig. 3.3) be the graph of the motion from $A$ to $B$ of the first body with speed $v_{1}=\operatorname{tg} \alpha$ (time axis $At$); $CD$ be the graph of the motion f... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.22 A pedestrian and a cyclist set off simultaneously towards each other from cities $A$ and $B$, the distance between which is $40 \mathrm{km}$, and meet 2 hours after departure. Then they continue their journey, with the cyclist arriving in $A$ 7 hours and 30 minutes earlier than the pedestrian in $B$. Find the spee... | 3.22 Fill in the table of speed, distance, and time values in the order indicated by the numbers (1), (2), .., (12):
| Tourist | Before Meeting | | | After Meeting | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | speed, km/h | time, h | distance, km | speed, km/h | time, h | distance, km |
| Pe... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.25 Two snow-clearing machines are working on snow removal. The first can clear a street in 1 hour, while the second can do it in $75\%$ of this time. Starting the cleaning simultaneously, both machines worked together for 20 minutes, after which the first machine stopped. How much more time is needed for the second m... | 3.25 Let's accept the entire volume of work as a unit. The productivity of the first machine is 1 (per hour), and the second is $1: \frac{3}{4}=\frac{4}{3}$ (per hour). Working together for $\frac{1}{3}$ hour, they will complete $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{4}{3} = \frac{7}{9}$ of the entire work. The... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.29 The volume of substance $A$ is half the sum of the volumes of substances $B$ and $C$, and the volume of substance $B$ is $\frac{1}{5}$ of the sum of the volumes of substances $A$ and $C$. Find the ratio of the volume of substance $C$ to the sum of the volumes of substances $A$ and $B$. | 3.29 According to the condition, $2 V_{A}=V_{B}+V_{C}$ and $5 V_{B}=V_{A}+V_{C}$. Let $V_{A}=x V_{C}$ and $V_{B}=y V_{C}$. Then we get the system
$\left\{\begin{array}{l}2 x-y=1 \\ -x+5 y=1\end{array}\right.$
from which $x=\frac{2}{3}, y=\frac{1}{3}$. Therefore,
$\frac{V_{C}}{V_{A}+V_{B}}=\frac{1}{x+y}=1$
Answer: 1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.30 It is known that the difference between the variable quantities $z$ and $y$ is proportional to the quantity $x$, and the difference between the quantities $x$ and $z$ is proportional to the quantity $y$. The coefficient of proportionality is the same and is equal to a positive integer $k$. A certain value of the q... | 3.30 I n s t r u c t i o n. Write the condition as a system of three equations. From the first two equations, express $\frac{x}{y}$ and $\frac{y}{z}$ in terms of $k$, and then substitute the obtained expressions for $\frac{x}{y}$ and $\frac{y}{z}$ into the third equation and calculate the desired value of $k$.
A n s w... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.38 It is known that a freely falling body travels 4.9 m in the first second, and in each subsequent second, it travels 9.8 m more than in the previous one. If two bodies start falling from the same height, one 5 s after the other, then after what time will they be 220.5 m apart from each other? | 3.38 Instruction. Use the well-known physics formula $s=4.9 t^{2}$.
Answer: 7 seconds after the start of the fall of the first body. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.48 A young man was returning home from vacation on a bicycle. At first, after traveling several kilometers, he spent one day more than half the number of days remaining after this until the end of his vacation. Now the young man has two options to travel the remaining distance to arrive home on time: to travel $h$ km... | 3.48 Let the segment $A B$ (Fig. 3.11) represent the entire journey of the young man and the number of days ($x$) it should take him to travel it at a rate of $v$ km per day. According to the problem, the time segment $A C$ is one unit more than half of the time segment $C B$; therefore,
 | 3.59 Impurities constitute $\frac{1}{5}$ of the solution. After the first filtration, $\left(\frac{1}{5}\right)^{2}$ impurities remain, and after the $k$-th filtration, $-\left(\frac{1}{5}\right)^{k+1}$ impurities remain. According to the condition, $\left(\frac{1}{5}\right)^{k+1} \leqslant 10^{-4}$; $-(k+1) \lg 5 \leq... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4.35 \operatorname{tg} 435^{\circ}+\operatorname{tg} 375^{\circ}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$4.35 \tan 435^{\circ}+\tan 375^{\circ}$. | ### 4.35 We have
$$
\begin{aligned}
& \tan 435^{\circ} + \tan 375^{\circ} = \tan 75^{\circ} + \tan 15^{\circ} = \frac{\sin 90^{\circ}}{\cos 75^{\circ} \cos 15^{\circ}} = \\
& = \frac{2}{\cos 90^{\circ} + \cos 60^{\circ}} = 4
\end{aligned}
$$
Here, reduction formulas, as well as (4.23) and (4.26), were used. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4.47 \frac{\sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)\left(\tan^{2} \alpha-1\right) \cot\left(\alpha-\frac{5 \pi}{4}\right)}{\sin ^{2}\left(\frac{5 \pi}{4}+\alpha\right)}=2$. | 4.47 Note that
$$
\begin{aligned}
& \sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)=\cos ^{2} \alpha \\
& \operatorname{tg}^{2} \alpha-1=\frac{\sin ^{2} \alpha-\cos ^{2} \alpha}{\cos ^{2} \alpha}=\frac{(\sin \alpha-\cos \alpha)(\sin \alpha+\cos \alpha)}{\cos ^{2} \alpha} \\
& \sin \alpha-\cos \alpha=\sqrt{2} \sin \left(\... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4.77 \operatorname{tg} 9^{\circ}+\operatorname{tg} 15^{\circ}-\operatorname{tg} 27^{\circ}-\operatorname{ctg} 27^{\circ}+\operatorname{ctg} 9^{\circ}+\operatorname{ctg} 15^{\circ}=8$. | ### 4.77 We have
$$
\begin{aligned}
& \tan 9^{\circ} + \tan 81^{\circ} - (\tan 27^{\circ} + \tan 63^{\circ}) + \tan 15^{\circ} + \tan 75^{\circ} = \\
& = \frac{2}{\cos 90^{\circ} + \cos 72^{\circ}} - \frac{2}{\cos 90^{\circ} + \cos 36^{\circ}} + \frac{2}{\cos 90^{\circ} + \cos 60^{\circ}} = \\
& = \frac{2}{\cos 72^{\c... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.88 Prove that the expression
$$
\frac{1-2 \sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3 \pi}{2}\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}
$$
does not depend on $\alpha$, where $\alpha \neq \frac{\pi n}{2}+\frac{\pi}{12}$. | 4.88 After applying the reduction formulas, we get
\[
\begin{aligned}
& \frac{1-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}=\frac{-2 \cos 2 \alpha+\sqrt{3} \sin 2 \alpha}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}= \\
& =\frac{2\left(\frac{\sqrt{3}}{2} \sin 2 \alpha-\frac{1}{2} ... | -2 | Algebra | proof | Yes | Yes | olympiads | false |
6.7 The denominator of the geometric progression is $\frac{1}{3}$, the fourth term of this progression is $\frac{1}{54}$, and the sum of all its terms is $\frac{121}{162}$. Find the number of terms in the progression. | 6.7 Since $b_{4}=b_{1} q^{3}$, then $\frac{1}{54}=b_{1} \cdot \frac{1}{27}$, from which $b_{1}=\frac{1}{2}$. Now, using formula (6.7a), we get the equation
$$
\frac{\frac{1}{2}\left(1-\frac{1}{3^{n}}\right)}{\frac{2}{3}}=\frac{121}{162}, \text { or } 1-\frac{1}{3^{n}}=\frac{121 \cdot 4}{162 \cdot 3}, \text { or } 1-\f... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.7 $\quad 2^{\log _{3} x^{2}} \cdot 5^{\log _{3} x}=400$. | 7.7 The logarithmic function $y=\log _{3} x$ is defined for $x>0$. Therefore, according to formula (7.6), we have $\log _{3} x^{2}=2 \log _{3} x$. Consequently, $2^{2 \log _{3} x} \cdot 5^{\log _{3} x}=20^{2} ; 4^{\log _{3} x} \cdot 5^{\log _{3} x}=20^{2} ; 20^{\log _{3} x}=20^{2}$. From this, $\log _{3} x=2$, i.e., $x... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$. | 7.8 Since $4^{\sqrt{x}}=2^{2 \sqrt{x}}$ and $2^{\sqrt{x}-1}=2^{\sqrt{x}} \cdot 2^{-1}=\frac{1}{2} \cdot 2^{\sqrt{x}}$, the given equation will take the form
$$
2^{2 \sqrt{x}}-\frac{9}{2} \cdot 2^{\sqrt{x}}+2=0
$$
Let's make the substitution $2^{\sqrt{x}}=y$, where $y>0$ due to the property of the exponential function... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.9 \quad 3 \cdot 5^{2x-1}-2 \cdot 5^{x-1}=0.2$.
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$7.9 \quad 3 \cdot 5^{2x-1}-2 \cdot 5^{x-1}=0.2$. | 7.9 We have $5^{2 x-1}=5^{2 x} \cdot 5^{-1} ; 5^{x-1}=5^{x} \cdot 5^{-1} ; 0.2=5^{-1}$. After dividing all terms of the given equation by $5^{-1}$, it will take the form
$3 \cdot 5^{2 x}-2 \cdot 5^{x}=1$.
Let $5^{x}=y$, where $y>0$. Then we get the equation $3 y^{2}-2 y-1=0$, from which $y_{1}=1, y_{2}=-\frac{1}{3}$ ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$.
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$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$. | 7.25 From the definition of the logarithm, it follows that $\sqrt{x}>0, \sqrt{x} \neq 1$, i.e., $x>0, x \neq 1$. Applying formulas (7.8) and (7.6), we get $\lg (\sqrt{6+x}+6)=\lg x$. According to the hint $4^{0}$, we transition to an equivalent system of equations:
$\left\{\begin{array}{l}x>0, x \neq 1, \\ \sqrt{6+x}+... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.27 \lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$. | 7.27 We have $\lg (3-x)^{3}=\lg \left(27-x^{3}\right)$, which is equivalent to the system
$$
\left\{\begin{array} { l }
{ 3 - x > 0 , } \\
{ ( 3 - x ) ^ { 3 } = 2 7 - x ^ { 3 } , }
\end{array} \text { or } \left\{\begin{array}{l}
x<3, \\
(3-x)^{3}=(3-x)\left(9+3 x+x^{2}\right)
\end{array}\right.\right.
$$
Further, w... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.28 \frac{2-\lg 4+\lg 0.12}{\lg (\sqrt{3 x+1}+4)-\lg 2 x}=1$.
$7.29 0.5\left(\lg \left(x^{2}-55 x+90\right)-\lg (x-36)\right)=\lg \sqrt{2}$. | 7.28 Given the domains of the logarithmic function and the square root, we have the system of inequalities $3 x+1 \geq 0$ (under this condition $\sqrt{3 x+1}+4>0$), $x>0$, the solution of which is $\boldsymbol{x}>0$. This equation is equivalent to the system
$$
\left\{\begin{array}{l}
x>0 \\
\lg (\sqrt{3 x+1}+4)-\lg 2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.
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$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$. | 7.30 For the existence of logarithms, it is necessary that the inequalities $x>0$, $\lg x>0$, $3 \lg x-2>0$ are satisfied simultaneously, i.e., $x>0, x>1, \lg x>\frac{2}{3}$. Hence, $\lg x>\frac{2}{3}$. Now we transition to the equivalent system of equations given by
$$
\left\{\begin{array}{l}
\lg x>\frac{2}{3} \\
\lg... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.36 \quad 3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$.
$7.37 \quad 25^{\log _{2} \sqrt{x+3}-0.5 \log _{2}\left(x^{2}-9\right)}=\sqrt{2(7-x)}$. | 7.36 Here
$3^{x}-5^{2-x}>0$.
Let's use the fact that $2-x=\log _{5} 5^{2-x}$, and write the equation in the form
$\log _{5}\left(2^{3} \cdot 5^{2-x}\right)=\log _{5}\left(3^{x}-5^{2-x}\right)$
Then we have
$2^{3} \cdot 5^{2-x}=3^{x}-5^{2-x} ; 5^{2-x}\left(2^{3}+1\right)=3^{x} ; 5^{2-x}=3^{x-2}$.
Finally, multiply... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.43 \quad 5^{x} \sqrt[x]{8^{x-1}}=500$.
$7.44 \quad 5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$. | 7.43 According to the definition of the root, $x \neq 0$. Let's write the equation as
$$
5^{x} \cdot 2^{\frac{3 x-3}{x}}=5^{3} \cdot 2^{2}
$$
Dividing both sides by $5^{3} \cdot 2^{2} \neq 0$, we get
$$
5^{x-3} \cdot 2^{\frac{3 x-3}{x}-2}=1, \text { or } 5^{x-3} \cdot 2^{\frac{x-3}{x}}=1, \text { or }\left(5 \cdot 2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$.
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9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$. | 9.1 According to formula (9.1), we find
$$
A_{x}^{2}=\frac{x!}{(x-2)!}=\frac{x(x-1)(x-2)!}{(x-2)!}=x(x-1)
$$
Further, using formulas (9.6) and (9.4), we have
$$
C_{x}^{x-1}=C_{x}^{x-(x-1)}=C_{x}^{1}=\frac{x!}{1!(x-1)!}=x
$$
Therefore, the given equation will take the form
$x^{2}(x-1)=48$, or $x^{2}(x-1)=4^{2} \cdo... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$9.2 \quad C_{x+1}^{x-2}+2 C_{x-1}^{3}=7(x-1)$. | 9.2 Let's use formula (9.4):
$$
\frac{(x+1)!}{(x-2)!3!}+\frac{2(x-1)!}{(x-4)!3!}=7(x-1)
$$
from which, after simplifying the first fraction by $(x-2)!$ and the second fraction by $(x-4)!$, we get
$$
(x+1) x(x-1)+2(x-1)(x-2)(x-3)=42(x-1)
$$
Further, considering that $x \neq 1$, we arrive at the quadratic equation
$... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.7 For what value of $x$ is the fourth term of the expansion $\left(\sqrt{2^{x-1}}+\sqrt[3]{2^{-x}}\right)^{m}$ 20 times greater than $m$, if the binomial coefficient of the fourth term is to the binomial coefficient of the second term as $5: 1 ?$ | 9.7 The binomial coefficients of the fourth and second terms are $C_{m}^{3}$ and $m$, respectively. Therefore, $\frac{m(m-1)(m-2)}{3!m}=5$, or $(m-1)(m-2)=30$, from which $m=7$. Then the fourth term of the expansion has the form $T_{4}=C_{7}^{3} 2^{2(x-1)} \cdot 2^{-x}$, and we arrive at the equation $C_{7}^{3} 2^{x-2}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$9.17 C_{x}^{x-1}+C_{x}^{x-2}+C_{x}^{x-3}+\ldots+C_{x}^{x-9}+C_{x}^{x-10}=1023$. | 9.17 Obviously, when $x=10$ the left side of the equation is 1 less than the sum of the binomial coefficients in the expansion of the binomial $(a+b)^{10}$. This sum is 1024. Therefore, $x=10$ is a solution to the equation. There are no other solutions, as for $x>10$ the left side of the equation is greater than 1023.
... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.21 The sum of the third from the beginning and the third from the end binomial coefficients in the expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{n}$ is 9900. How many rational terms are contained in this expansion? | 9.21 The coefficients specified in the condition are equal to $C_{n}^{2}$. We have $2 \cdot \frac{n(n-1)}{2}=9900$, or $n(n-1)=100 \cdot 99$,
from which $n=100$. Then $T_{k+1}=C_{100}^{k} 3^{\frac{100-k}{4}} 4^{\frac{k}{3}} ;$ according to the condition, $\frac{k}{3}$ and $\frac{100-k}{4}$ are integers, i.e., $k$ is d... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3 Find the sum $A=i+i^{2}+i^{3}+\ldots+i^{15}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 10.3 Let's group the terms as follows:
$$
\begin{aligned}
& A=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+ \\
& +\left(i^{9}+i^{10}+i^{11}+i^{12}\right)+\left(i^{13}+i^{14}+i^{15}\right)
\end{aligned}
$$
from which
$$
A=\left(i+i^{2}+i^{3}+i^{4}\right)\left(1+i^{4}+i^{8}\right)+i^{13}+i^{14... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.36 Find the real number $b$ from the condition that the points representing the complex numbers $3-5 i, 1-i$ and $-2+b i$ lie on the same straight line. | 10.36 The numbers $3-5i, 1-i$ and $-2+bi$ correspond to the points $A(3, -5), B(1, -1)$ and $C(-2, b)$ on the plane (Fig. 10.10). We find the coordinates of the vectors $\overline{A B}$ and $\overline{A C}$; we have $\overline{A B}(-2, 4), \overline{A C}(-5, b+5)$.
$. Find all real values of $a$ for which $z_{1}^{3}=z_{2}^{2}$. | 10.47 We find
\[
\begin{aligned}
& z_{1}^{3}=(1+a i)^{3}=\left(1-3 a^{2}\right)+\left(3 a-a^{3}\right) i \\
& z_{2}^{2}=\left(2^{\frac{3}{4}}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)\right)^{2}= \\
& =2^{\frac{3}{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)=2^{\frac{3}{2}}\left(-\frac{\sq... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2 Solve the equation $|x+1|+|x-1|=2 x^{3}$. | 11.2 We have $|x+1|=0$ when $x=-1$, and $|x-1|=0$ when $x=1$. We will solve the given equation in the intervals $(-\infty,-1)$, $[-1,1)$, $[1, \infty)$.
1) If $x<0$ for any $x$, then $x=0$; this value does not belong to the interval $(-\infty,-1)$.
2) If $-1 \leq x<1$, then $x+1-x+1=2 x^{3} ; 2 x^{3}=2 ; x^{3}=1$; $x=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.16 For what value of $m$ does the system of equations
$$
\left\{\begin{array}{l}
2 x+(m-1) y=3 \\
(m+1) x+4 y=-3
\end{array}\right.
$$
have an infinite number of solutions? No solutions? | 11.16 The system has an infinite number of solutions or no solutions if the coefficients of $x$ and $y$ are proportional. Therefore, $\frac{2}{m+1}=\frac{m-1}{4}$ (here $m+1 \neq 0$; it is easy to verify that when $m=-1$, the system has a unique solution). Further, we have $(m+1)(m-1)=8 ; m^{2}-1=8$, from which $m= \pm... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.36 Solve graphically the equation
$$
|x-1|+2x-5=0
$$ | 11.36 Let's write the equation as $|x-1|=5-2 x$. Since $|x-1| \geq 0$ for any $x$,
Fig. 11.11
and $\quad 5-2 x \geq 0$, i.e., $x \leq 2.5$.
Consider the functions $y_{1}=|x-1|, y_{2}=5-2 ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.40 How many real solutions does the system of equations have
$\left\{\begin{array}{l}x^{2}+y=5 \\ x+y^{2}=3 ?\end{array}\right.$ | 11.40 The solutions of the system are the coordinates of the points of intersection of the graphs of the functions given by these equations. Let's write the original system in the form
$\left\{\begin{array}{l}y=5-x^{2}, \\ x=3-y^{2} .\end{array}\right.$
The graph of the first equation is a parabola with the axis of s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.41 For what value of a is the sum of the squares of the roots of the equation $x^{2}+a x+a-2=0$ the smallest? | 11.41 Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then $x_{1} x_{2}=a-2$, $x_{1}+x_{2}=-a$. From this, we can express the sum of the squares of the roots:
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-2 a+4=f(a)
$$
Since $f(a)$ is a quadratic function with a positive leading coeffi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.43 Solve the equation $x+\lg \left(1+4^{x}\right)=\lg 50$. | 11.43 Let's write the equation as $x+\lg \left(1+4^{x}\right)=1+\lg 5$. It is not difficult to establish that the value $x=1$ satisfies the equation. The left side of the equation is $f(x)=x+\lg \left(1+4^{x}\right)$ - an increasing function; therefore, if $x<1$, then $f(x)<1+\lg 5$, and if $x>1$, then $f(x)>1+\lg 5$. ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12.1 $\lim _{x \rightarrow 2} \frac{x^{3}-8}{2 x-4}$. | 12.1 The function $f(x)=\frac{x^{3}-8}{2 x-4}$ is undefined at the point $x=2$. By factoring the numerator using formula (1.14), we can represent this function as
$f(x)=\frac{(x-2)\left(x^{2}+2 x+4\right)}{2(x-2)}$.
In the domain of the function $f(x)$, the expression $x-2 \neq 0$, so the fraction can be simplified b... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
$12.3 \lim _{x \rightarrow-1} \frac{\sqrt{2 x+3}-1}{\sqrt{5+x}-2}$.
$12.3 \lim _{x \rightarrow-1} \frac{\sqrt{2 x+3}-1}{\sqrt{5+x}-2}$.
The above text is already in a mathematical expression format, so the translation is the same as the original text. If you need an explanation or solution for the limit, please let ... | 12.3 Multiplying the numerator and the denominator of the fraction by $(\sqrt{2 x+3}+1)(\sqrt{5+x}+2)$, we get $\lim _{x \rightarrow-1} \frac{(2 x+3-1)(\sqrt{5+x}+2)}{(5+x-4)(\sqrt{2 x+3}+1)}=\lim _{x \rightarrow-1} \frac{2(x+1)(\sqrt{5+x}+2)}{(x+1)(\sqrt{2 x+3}+1)}=$ $\lim _{x \rightarrow-1} \frac{2(\sqrt{5+x}+2)}{\sq... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.8 $f(x)=\sqrt{x^{2}+3}+\frac{2 x}{x+1} ; f^{\prime}(1)=?$ | 12.8 First, we find
$$
f^{\prime}(x)=\frac{x}{\sqrt{x^{2}+3}}+2 \frac{x+1-x}{(x+1)^{2}}=\frac{x}{\sqrt{x^{2}+3}}+\frac{2}{(x+1)^{2}}
$$
Now, setting $x=1$, we get $f^{\prime}(1)=\frac{1}{2}+\frac{2}{4}=1$.
Answer: 1. | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.9 $f(x)=\sin 4 x \cos 4 x ; f^{\prime}\left(\frac{\pi}{3}\right)=$ ? | 12.9 We have
$f^{\prime}(x)=4 \cos 4 x \cdot \cos 4 x+\sin 4 x(-4 \sin 4 x)=$ $=4\left(\cos ^{2} 4 x-\sin ^{2} 4 x\right)=4 \cos 8 x$.
At $x=\frac{\pi}{3}$ we find $f^{\prime}\left(\frac{\pi}{3}\right)=4 \cos \frac{8 \pi}{3}=4 \cos \frac{2 \pi}{3}=-2$.
Answer: -2. | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.11 $f(x)=\frac{(x-2)^{2}}{x^{2}}$. | 12.11 The domain of the function is the entire number line, except for the point $x=0$. Using formula (12.4) and the table of derivatives, we find
$$
f^{\prime}(x)=\frac{2(x-2) x^{2}-(x-2)^{2} 2 x}{x^{4}}=\frac{4(x-2)}{x^{3}}
$$
$f^{\prime}(x)=0$ only at $x=2$. We will construct a table:
| Interval | $(-\infty, 0)$ ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12.18 A pedestrian set out for a walk from point $A$ at a speed of $v$ km/h. After he had walked 6 km from $A$, a cyclist set out after him from $A$ at a speed 9 km/h greater than the pedestrian's speed. When the cyclist caught up with the pedestrian, they turned back and returned to $A$ together at a speed of 4 km/h. ... | 12.18 The time it takes for the cyclist to catch up with the pedestrian is $\frac{6}{9}=\frac{2}{3}$ (hours). Before the meeting, the pedestrian had been walking for $\frac{6}{v}+\frac{2}{3}$ (hours) and covered $v\left(\frac{6}{v}+\frac{2}{3}\right)$ (km). They covered this same distance on the way back at a constant ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$12.28 \int_{8}^{27} \frac{d x}{\sqrt[3]{x^{2}}}$ | 12.28 We have
$$
\int_{8}^{27} \frac{d x}{\sqrt[3]{x^{2}}}=\int_{8}^{27} x^{-2 / 3} d x=\left.\frac{x^{1 / 3}}{1 / 3}\right|_{8} ^{27}=3\left(27^{1 / 3}-8^{1 / 3}\right)=3
$$
Answer: 3. | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.35 $\lim _{x \rightarrow 4} \frac{x+\sqrt{x}-6}{x-5 \sqrt{x}+6}$. | 12.35 The numerator and denominator of the given fraction are quadratic functions of $\sqrt{x}$. We have
$$
\lim _{x \rightarrow 4} \frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\lim _{x \rightarrow 4} \frac{\sqrt{x}+3}{\sqrt{x}-3}=\frac{5}{-1}=-5
$$
Answer: -5. | -5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.59 Find the area of the figure bounded by the lines $y=x^{3}-4 x$ and $y=0$ for $x \geq 0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 12.59 First, let's construct the graph of the function $y=x^{3}-4 x$ for $x \geq 0$. We find the extrema:
$y^{\prime}=3 x^{2}-4$;
$3 x^{2}-4=0 ;$
$x= \pm \frac{2}{\sqrt{3}} \approx \pm 1.1$.
For
^{2}+(c-2)^{2}=c^{2}, \text { or } c^{2}-6 c+5=0
$$
from which $c=5$ (cm) (the second root of the equation does not satisfy the condition).
Answer: 5 cm. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.6. The legs of a right triangle are 9 and 12 cm. Find the distance between the point of intersection of its angle bisectors and the point of intersection of the medians. | 1.6. In $\triangle A B C$ (Fig. 1.5), we have $A B=\sqrt{9^{2}+12^{2}}=15$ (cm); the median $C D$ is half the hypotenuse, i.e., $\frac{15}{2}$ cm. Let $E$ be the point of intersection of the medians; then $C E=\frac{2}{3} C D=\frac{2}{3} \cdot \frac{15}{2}=5$ (cm) (according to the additional relations, point $1^{\circ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.9. Two circles touch each other externally. A tangent is drawn to the first circle, passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of this circle. How many times greater is the length of the first circle compared to... | 1.9. Let $O_{1}$ and $O_{2}$ be the centers of the circles, and $A$ be the point of tangency (Fig. 1.8). Then $O_{1} A = R_{1}, O_{1} O_{2} = R_{1} + R_{2}, O_{2} A = 3 R_{2}$ (by the condition). We need to find the ratio $2 \pi R_{1} : 2 \pi R_{2} = R_{1} : R_{2}$. In the right triangle $O_{1} A O_{2} \left(\angle A =... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.20. A tangent, parallel to the base, is drawn to the circle inscribed in an isosceles triangle with a base of 12 cm and a height of 8 cm. Find the length of the segment of this tangent, enclosed between the sides of the triangle. | 1.20. Let's find the length of the side $B C$ (Fig. 1.18); $B C=\sqrt{B M^{2}+M C^{2}}=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $\frac{M O}{O B}=\frac{A M}{A B}$, or $\frac{r}{8-r}=\frac{6}{10}$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \s... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.26. The area of an isosceles triangle is equal to $\frac{1}{3}$ of the area of the square constructed on the base of the given triangle. The lengths of the lateral sides of the triangle are shorter than the length of its base by 1 cm. Find the lengths of the sides and the height of the triangle, drawn to the base. | 1.26. By the condition, $B C^{2}=$ $=3 \cdot \frac{1}{2} B C \cdot A H$ (Fig. 1.22), or $A H=$ $=\frac{2}{3} B C$. But $A H^{2}=A B^{2}-\left(\frac{1}{2} B C\right)^{2}$ and, therefore, $\frac{4}{9} B C^{2}=A B^{2}-\frac{1}{4} B C^{2}$, or $A B^{2}=\frac{25}{36} B C^{2}$, i.e., $A B=\frac{5}{6} B C$. Then we get $A B=\... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.28. One of the angles of the parallelogram is $60^{\circ}$, and the shorter diagonal is $2 \sqrt{31}$ cm. The length of the perpendicular dropped from the point of intersection of the diagonals to the longer side is $\frac{\sqrt{75}}{2}$ cm. Find the lengths of the sides and the longer diagonal of the parallelogram. | ### 1.28. Let's conduct
Fig. 1.23 $B N \perp A D$ (Fig. 1.24); since $B N=2 O M$, then $B N=\sqrt{75}$ cm. Considering that in $\triangle A N B \angle A B N=30^{\circ}$, we have $A B=2 A N$... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.30. The lengths of the parallel sides of the trapezoid are 25 and 4 cm, and the lengths of the non-parallel sides are 20 and 13 cm. Find the height of the trapezoid. | 1.30. Given that $B C=$ $=4 \text{ cm}, A D=25 \text{ cm}, A B=20$ cm, $C D=13$ cm (Fig. 1.26). Draw $B E \perp A D$ and $C F \perp A D$. Let $B E=C F=$ $=h, A E=x, F D=y$. Then from $\triangle A B E$ and $\triangle C F D$ we find $h=\sqrt{20^{2}-x^{2}}=\sqrt{13^{2}-y^{2}}$. Considering that $y=25-4-$ $-x=21-x$, we hav... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.44. In a circle with center $O$, a chord $A B$ is drawn, intersecting the diameter at point $M$ and forming an angle of $60^{\circ}$ with the diameter. Find $O M$, if $A M=10$ cm, and $B M=4$ cm. | 1.44. Draw $O P \perp A B$ (Fig. 1.36). Then $A P=B P=7$ cm and, therefore, $M P=3$ cm. Since $\angle P M O=60^{\circ}$, then $\angle M O P=30^{\circ}$ and $O M=2 M P=6 \text{ cm}$.
Answer: 6 cm. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.49. The area of a right triangle is $24 \mathrm{~cm}^{2}$, and the hypotenuse is 10 cm. Find the radius of the inscribed circle. | 1.49. I n d i c a t i o n. Denoting the legs of the triangle by $a$ and $b$, solve the system of equations
$$
\left\{\begin{array}{l}
a b=48 \\
a^{2}+b^{2}=100
\end{array}\right.
$$
and then use the formula $S=p r$.
Answer: 2 cm. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.43. Find the distance between the midpoints of two skew edges of a cube, the total surface area of which is $36 \mathrm{~cm}^{2}$. | 2.43. Since the total surface area of the cube $S_{\text {full }}=36 \mathrm{~cm}^{2}$, the area of one face $S=6 \mathrm{~cm}^{2}$ and the edge of the cube
Fig. 2.37
$A D=a=\sqrt{6}$ cm (... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3.67. In a trapezoid, the smaller base is equal to 2, the adjacent angles are $135^{\circ}$ each. The angle between the diagonals, facing the base, is $150^{\circ}$. Find the area of the trapezoid. | 3.67. By the condition, $A D \| B C, B C=2, \angle A B C=\angle B C D=135^{\circ}$, $\angle A O D=150^{\circ}$ (Fig. 3.69). Since the angles at the base of the trapezoid are equal, the trapezoid is isosceles, i.e., $A B=C D$. We find
. Thus, we obtain the equation $\frac{R^{2}}{2} \sin \frac{360^{\circ}}{n}=\frac{3 R^{2}}{n}$, which holds
$ 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 |
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 |
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 |
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 |
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.
$$
\begin{aligned}
& \frac{\left(13.75+9 \frac{1}{6}\right) \cdot 1.2}{\left(10.3-8 \frac{1}{2}\right) \cdot \frac{5}{9}}+\frac{\left(6.8-3 \frac{3}{5}\right) \cdot 5 \frac{5}{6}}{\left(3 \frac{2}{3}-3 \frac{1}{6}\right) \cdot 56}-27 \frac{1}{6}=\frac{\left(\frac{55}{4}+\frac{55}{6}\right) \cdot \frac{6}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.008. $\left(\frac{3 \frac{1}{3}+2.5}{2.5-1 \frac{1}{3}} \cdot \frac{4.6-2 \frac{1}{3}}{4.6+2 \frac{1}{3}} \cdot 5.2\right):\left(\frac{0.05}{\frac{1}{7}-0.125}+5.7\right)$. | Solution.
$$
\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 |
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.
$$
\begin{aligned}
& \frac{0.4+8\left(5-0.8 \cdot \frac{5}{8}\right)-5: 2 \frac{1}{2}}{\left(1 \frac{7}{8} \cdot 8-\left(8.9-2.6: \frac{2}{3}\right)\right) \cdot 34 \frac{2}{5}} \cdot 90=\frac{\left(0.4+40-4-5 \cdot \frac{2}{5}\right) \cdot 90}{\left(\frac{15}{8} \cdot 8-\frac{89}{10}+\frac{13}{5} \cdot \fra... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.010. $\frac{\left(5 \frac{4}{45}-4 \frac{1}{6}\right): 5 \frac{8}{15}}{\left(4 \frac{2}{3}+0.75\right) \cdot 3 \frac{9}{13}} \cdot 34 \frac{2}{7}+\frac{0.3: 0.01}{70}+\frac{2}{7}$. | ## Solution.
$$
\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 |
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 |
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 |
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.
$$
\begin{aligned}
& \frac{\left(4.5 \cdot 1 \frac{2}{3}-6.75\right) \cdot \frac{2}{3}}{\left(3 \frac{1}{3} \cdot 0.3+5 \frac{1}{3} \cdot \frac{1}{8}\right): 2 \frac{2}{3}}+\frac{1 \frac{4}{11} \cdot 0.22: 0.3-0.96}{\left(0.2-\frac{3}{40}\right) \cdot 1.6}=\frac{\left(\frac{9}{2} \cdot \frac{5}{3}-\frac{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.016. $\left(16 \frac{1}{2}-13 \frac{7}{9}\right) \cdot \frac{18}{33}+2.2\left(\frac{8}{33}-\frac{1}{11}\right)+\frac{2}{11}$. | ## Solution.
$$
\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 |
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.
$$
\begin{aligned}
& \frac{0.128: 3.2+0.86}{\frac{5}{6} \cdot 1.2+0.8} \cdot \frac{\left(1 \frac{32}{63}-\frac{13}{21}\right) \cdot 3.6}{0.505 \cdot \frac{2}{5}-0.002}=\frac{0.04+0.86}{1+0.8} \cdot \frac{\left(\frac{95}{63}-\frac{39}{63}\right) \cdot \frac{18}{5}}{0.202-0.002}= \\
& =\frac{9}{18} \cdot \f... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.018. $\frac{3 \frac{1}{3}: 10+0.175: 0.35}{1.75-1 \frac{11}{17} \cdot \frac{51}{56}}-\frac{\left(\frac{11}{18}-\frac{1}{15}\right): 1.4}{\left(0.5-\frac{1}{9}\right) \cdot 3}$. | Solution.
$$
\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 |
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.
$$
\begin{aligned}
& \frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3): 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5=\frac{\frac{1}{2}+\frac{5}{8}}{\left(10-\frac{220}{23}\right) \cdot \frac{23}{50}+\frac{8}{5}}+\frac{17}{40}+\frac{19}{20}= \\
& =\frac{\frac{9}{8}}{\frac{1}{5}+\frac{8}{5}}+\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.020. $\left(\left(1 \frac{1}{7}-\frac{23}{49}\right): \frac{22}{147}-\left(0.6: 3 \frac{3}{4}\right) \cdot 2 \frac{1}{2}+3.75: 1 \frac{1}{2}\right): 2.2$. | ## Solution.
$$
\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 |
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 |
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 |
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.
$$
\begin{aligned}
& \left((520 \cdot 0.43): 0.26-217 \cdot 2 \frac{3}{7}\right)-\left(31.5: 12 \frac{3}{5}+114 \cdot 2 \frac{1}{3}+61 \frac{1}{2}\right)= \\
& =\left(223.6: 0.26-217 \cdot \frac{17}{7}\right)-\left(\frac{63}{2} \cdot \frac{5}{63}+114 \cdot \frac{7}{3}+\frac{123}{2}\right)= \\
& =(860-527)-\l... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.026. $\frac{(3.4-1.275) \cdot \frac{16}{17}}{\frac{5}{18} \cdot\left(1 \frac{7}{85}+6 \frac{2}{17}\right)}+0.5\left(2+\frac{12.5}{5.75+\frac{1}{2}}\right)$. | Solution.
$$
\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 |
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 |
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 |
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 |
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.
$$
\begin{aligned}
& \frac{\left(3^{-1}-\sqrt{1 \frac{7}{9}}\right)^{-2}: 0.25}{\frac{37}{300}: 0.0925}+12.5 \cdot 0.64=\frac{\left(\frac{1}{3}-\sqrt{\frac{16}{9}}\right)^{-2} \cdot 4}{\frac{37}{300} \cdot \frac{400}{37}}+8= \\
& =\frac{\left(\frac{1}{3}-\frac{4}{3}\right)^{-2} \cdot 4}{\frac{4}{3}}+8=3(-... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.033. $\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5$. | ## Solution.
$\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5=\frac{\left(\frac{5}{8}+\frac{65}{24}\right) \cdot \frac{2}{5} \cdot \frac{1}{2}}{\left(\frac{13}{10}+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}}=\frac{\frac{10}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.039. $\left(\frac{(3.2-1.7): 0.003}{\left(\frac{29}{35}-\frac{3}{7}\right) \cdot 4: 0.2}-\frac{\left(1 \frac{13}{20}-1.5\right) \cdot 1.5}{\left(2.44+1 \frac{14}{25}\right) \cdot \frac{1}{8}}\right): 62 \frac{1}{20}+1.364: 0.124$. | Solution.
$$
\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 |
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 |
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