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1.60 Represent the number 19 as the difference of cubes of natural numbers. Show that such a representation is unique. | 1.60 Let $19=x^{3}-y^{3}$, where $x>y$. Since $x^{3}-y^{3}=(x-y) \times$ $\times\left(x^{2}+x y+y^{2}\right)$, and the number 19 is prime, it can only be represented as a product in the form $1 \cdot 19$. Therefore, we obtain the system
$\left\{\begin{array}{l}x-y=1, \\ x^{2}+x y+y^{2}=19 .\end{array}\right.$
Substit... | 3^{3}-2^{3} | Number Theory | proof | Yes | Yes | olympiads | false | 49,498 |
$2.1 \quad \frac{2}{3-x}+\frac{1}{2}=\frac{6}{x(3-x)}$. | 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 | 49,499 |
$2.2 \frac{x+2}{x+1}+\frac{x+6}{x+3}+\frac{x+10}{x+5}=6$. | 2.2 Rewriting the equation as
$$
1+\frac{1}{x+1}+1+\frac{3}{x+3}+1+\frac{5}{x+5}=6
$$
we get
$$
\frac{1}{x+1}+\frac{3}{x+3}+\frac{5}{x+5}=3
$$
From here, bringing to a common denominator, we have
$$
\begin{aligned}
& 9 x^{2}+46 x+45=3 x^{3}+27 x^{2}+69 x+45 \\
& 3 x^{3}+18 x^{2}+23 x=0 ; x\left(3 x^{2}+18 x+23\rig... | x_{1}=0;x_{2,3}=-3\\frac{2}{3}\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,500 |
$2.9 \quad \frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$.
Solve the equation:
$2.9 \quad \frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$. | 2.9 Task. Use the substitution $\frac{x^{2}+x-5}{x}=z$.
Solution: $x_{1}=-5, x_{2}=1, x_{3,4}=-1 \pm \sqrt{6}$. | x_{1}=-5,x_{2}=1,x_{3,4}=-1\\sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,503 |
$2.10 \frac{1}{x^{3}+2}-\frac{1}{x^{3}+3}=\frac{1}{12}$. | 2.10 Instruction. Use the substitution $x^{3}+2=z$. Answer: $\quad x_{1}=1, x_{2}=-\sqrt[3]{6}$. | x_{1}=1,x_{2}=-\sqrt[3]{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,504 |
$2.11 \frac{21}{x^{2}-4 x+10}-x^{2}+4 x=6$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.11 \frac{21}{x^{2}-4 x+10}-x^{2}+4 x=6$. | 2.11 Instruction. Use the substitution $x^{2}-4 x+10=z$ Answer: $x_{1}=1, x_{2}=3$.
| x_{1}=1,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,505 |
$2.12 x^{2}+x+x^{-1}+x^{-2}=4$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.12 x^{2}+x+x^{-1}+x^{-2}=4$. | 2.12 Rewriting the equation as
$$
\left(x^{2}+\frac{1}{x^{2}}\right)+\left(x+\frac{1}{x}\right)=4
$$
let $x+\frac{1}{x}=z$. Then $x^{2}+2+\frac{1}{x^{2}}=z^{2}$, or
$$
\begin{aligned}
& z^{2}-2=x^{2}+\frac{1}{x^{2}} . \text { The given equation becomes } \\
& z^{2}+z-6=0, \text { from which } z_{1}=2, z_{2}=-3
\end{... | x_{1}=1,x_{2,3}=\frac{-3\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,506 |
$2.13 \sqrt{x-2}=x-4$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.13 \sqrt{x-2}=x-4$. | 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 | 49,507 |
$2.14 \sqrt{15-x}+\sqrt{3-x}=6$. | 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 | 49,508 |
$2.15 \sqrt{x^{2}+9}-\sqrt{x^{2}-7}=2$. | ### 2.15 We have
$$
\begin{aligned}
& \left(x^{2}+9\right)-\left(x^{2}-7\right)=16, \text { or } \\
& \left(\sqrt{x^{2}+9}-\sqrt{x^{2}-7}\right)\left(\sqrt{x^{2}+9}+\sqrt{x^{2}-7}\right)=16
\end{aligned}
$$
Since $\sqrt{x^{2}+9}-\sqrt{x^{2}-7}=2$, then $\sqrt{x^{2}+9}+\sqrt{x^{2}-7}=8$.
Therefore, we obtain the syst... | x_{1,2}=\4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,509 |
$2.16 \sqrt{x+1}-\sqrt{9-x}=\sqrt{2 x-12}$.
$2.171+\sqrt{1+x \sqrt{x^{2}-24}}=x$. | 2.16 Raising both sides of the equation to the square, we have
$$
(x+1)+(9-x)-2 \sqrt{(x+1)(9-x)}=2 x-12, \text{ or }
$$
$\sqrt{(x+1)(9-x)}=11-x$.
Raising to the square again, after transformations we get $x^{2}-15 x+56=0$, from which $x_{1}=7, x_{2}=8$.
By checking, we are convinced that both roots are valid.
Ans... | x_{1}=7,x_{2}=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,510 |
$2.18 \frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$. | 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 | 49,511 |
$2.19 \frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$. | 2.19 To solve the equation, we will use the following statement: if the equality $\frac{a}{b}=\frac{c}{d}$ is true, then the equality (check it!) $\frac{a+b}{a-b}=\frac{c+d}{c-d}$ is also valid. We have
$$
\begin{aligned}
& \frac{(\sqrt{x}+\sqrt[3]{x})+(\sqrt{x}-\sqrt[3]{x})}{(\sqrt{x}+\sqrt[3]{x})-(\sqrt{x}-\sqrt[3]{... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,512 |
$2.20 \sqrt[7]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | 2.20 Instruction. Use the substitution $\sqrt[7]{\frac{5-x}{x+3}}=z$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,513 |
2.35. $\left\{\begin{array}{l}\sqrt{\frac{y}{x}}-2 \sqrt{\frac{x}{y}}=1 \\ \sqrt{5 x+y}+\sqrt{5 x-y}=4 .\end{array}\right.$
2.36 $\left\{\begin{array}{l}\sqrt[3]{x}+\sqrt[3]{y}=3 \\ \sqrt[3]{x^{2}}-\sqrt[3]{x y}+\sqrt[3]{y^{2}}=3\end{array}\right.$
2.37 $\left\{\begin{array}{l}\sqrt[3]{x} \sqrt{y}+\sqrt[3]{y} \sqrt{x... | 2.35 Assuming $\sqrt{\frac{y}{x}}=z \geqslant 0$, we get $z-\frac{2}{z}=1$, or $z^{2}-z-2=0$, and thus, $z_{1}=2, z_{2}=-1$ (this root is not suitable). Therefore, $\sqrt{\frac{y}{x}}=2$, i.e., $y=4 x$. Then $\sqrt{9 x}+\sqrt{x}=4$, or
$3 \sqrt{x}+\sqrt{x}=4$, from which $x=1$, and $y=4$.
Answer: $(1 ; 4)$. | (1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,515 |
2.38 Without solving the equation $a x^{2}+b x+c=0$, find $x_{1}^{-2}+x_{2}^{-2}$, where $x_{1}$ and $x_{2}$ are the roots of this equation. | 2.38 After performing the transformations and using Vieta's theorem, we get
$$
\begin{aligned}
& \frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{x_{1}^{2}+x_{2}^{2}}{\left(x_{1} x_{2}\right)^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{\left(x_{1} x_{2}\right)^{2}}= \\
& =\frac{\frac{b^{2}}{a^{2}}-\frac{2 c}{a... | \frac{b^{2}-2}{^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,516 |
2.39 Form a quadratic equation with roots $\frac{1}{x_{1}}$ and $\frac{1}{x_{2}}$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$. | ### 2.39 We have
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}} ; \frac{1}{x_{1}} \cdot \frac{1}{x_{2}}=\frac{1}{x_{1} x_{2}}
$$
By Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a} ; x_{1} x_{2}=\frac{c}{a}$. Therefore, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c} ; \fr... | ^{2}++=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,517 |
2.40 Form a second-degree equation, one of whose roots is equal to the sum, and the other is equal to the product of the roots of the equation $a x^{2}+b x+c=0$. | 2.40 If the roots of the desired equation are denoted by $\bar{x}_{1}$ and $\bar{x}_{2}$, and those of the given equation by $x_{1}$ and $x_{2}$, then from the condition it follows that $\bar{x}_{1}=x_{1}+x_{2} ; \bar{x}_{2}=x_{1} x_{2}$. Using Vieta's theorem, we get $\bar{x}_{1}=-\frac{b}{a} ; \bar{x}_{2}=\frac{c}{a}... | ^{2}x^{2}+(-)x-=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,518 |
2.41 For what integer value of $p$ do the equations $3 x^{2}-4 x+p-2=0$ and $x^{2}-2 p x+5=0$ have a common root? Find this root. | ### 2.41 We have
\[
\left\{\begin{array}{l}3 x^{2}-4 x+p-2=0 \\ x^{2}-2 p x+5=0\end{array}\right.
\]
Substituting the expression \( p=2-3 x^{2}+4 x \) into the second equation, we get
\[
\begin{aligned}
& 6 x^{3}-7 x^{2}-4 x+5=6 x^{3}-6 x^{2}-x^{2}-4 x+5= \\
& =6 x^{2}(x-1)-(x-1)(x+5)=(x-1)\left(6 x^{2}-x-5\right)= ... | 3,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,519 |
2.42 Find all values of $a$ for which the sum of the roots of the equation $x^{2}-2 a(x-1)-1=0$ is equal to the sum of the squares of the roots. | 2.42 We have $x^{2}-2 a x+(2 a-1)=0$ and $x_{1}+x_{2}=x_{1}^{2}+x_{2}^{2}$, i.e., $x_{1}+x_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$. According to Vieta's theorem, $2 a=4 a^{2}-4 a+2$, or $2 a^{2}-3 a+1=0$, from which $a_{1}=1$, $a_{2}=\frac{1}{2}$.
Answer: $a_{1}=1 ; a_{2}=0.5$. | a_{1}=1;a_{2}=0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,520 |
2.43 For what value of $p$ is the ratio of the roots of the equation $x^{2}+p x-16=0$ equal to -4? | 2.43 We have $x^{2}+p x-16=0$ and $\frac{x_{1}}{x_{2}}=-4$. From the system $\left\{\begin{array}{l}x_{1} x_{2}=-16, \\ \frac{x_{1}}{x_{2}}=-4\end{array}\right.$
we find $x_{1}=8, x_{2}=-2$ or $x_{1}=-8, x_{2}=2$. Next, using the equality $x_{1}+x_{2}=-p$, we get $p_{1}=-6, p_{2}=6$.
Answer: for $p= \pm 6$. | \6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,521 |
$2.46 \frac{x-m}{x-1}+\frac{x+m}{x+1}=\frac{x-2 m}{x-2}+\frac{x+2 m}{x+2}-\frac{6(m-1)}{5}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.46 \frac{x-m}{x-1}+\frac{x+m}{x+1}=\frac{x-2 m}{x-2}+\frac{x+2 m}{x+2}-\frac{6(m-1)}{5}... | 2.46 After performing the transformations (do this on your own), we will obtain $-\frac{3}{5}(m-1)\left(x^{4}+4\right)=0$. It is easy to see that when $m=1$, $x$ can be any number except $\pm 1, \pm 2$; when $m \neq 1$, we get $x^{4}+4=0$, or $x^{4}=-4$ (no real roots).
Answer: if $m=1$, then $x$ is any number except ... | if=1,thenxisanyexcept\1,\2;if\neq1,therenoroots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,523 |
$2.49 \frac{x^{2}+1}{x+1}+\frac{x^{2}+2}{x-2}=-2$.
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 | 49,525 |
$2.50 x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.50 x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=0$. | 2.50 It is easy to see that one of the roots of the given equation is $x_{1}=a$. Then
$$
\begin{aligned}
& {\left[x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c\right]:(x-a)=} \\
& =x^{2}-(b+c) x+b c
\end{aligned}
$$
From the equation $x^{2}-(b+c) x+b c=0$, we find that $x_{2}=b$, $x_{3}=c$.
Answer: $x_{1}=a, x_{2}=b, x_... | x_{1}=,x_{2}=b,x_{3}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,526 |
$2.51 \frac{1}{x^{2}}+\frac{1}{(x+2)^{2}}=\frac{10}{9}$. | 2.51 Assuming $x+1=z$, we get
$$
\frac{1}{(z-1)^{2}}+\frac{1}{(z+1)^{2}}=\frac{10}{9}, \text { or } \cdot 5 z^{4}-19 z^{2}-4=0
$$
From here, $z_{1}^{2}=4$, i.e., $z= \pm 2$, and $z^{2}=-\frac{1}{5}$ (this equation has no roots).
Answer: $x_{1}=1, x_{2}=-3$. | x_{1}=1,x_{2}=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,527 |
$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.55 \sqrt{x+1}+\sqrt{4 x+13}=\sqrt{3 x+12}$. | 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 | 49,529 |
$2.57 \sqrt{x^{2}+x+4}+\sqrt{x^{2}+x+1}=\sqrt{2 x^{2}+2 x+9}$. | 2.57 After squaring, we get
$$
2 x^{2}+2 x+5+2 \sqrt{\left(x^{2}+x+4\right)\left(x^{2}+x+1\right)}=2 x^{2}+2 x+9
$$
or
$\sqrt{\left(x^{2}+x\right)^{2}+5\left(x^{2}+x\right)+4}=2$.
From this,
$\left(x^{2}+x\right)^{2}+5\left(x^{2}+x\right)+4=4$, or $\left(x^{2}+x\right)\left(x^{2}+x+5\right)=0$.
Answer: $\quad x_{... | x_{1}=0,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,530 |
$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$2.58 \sqrt{x}+\frac{2 x+1}{x+2}=2$. | ### 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 | 49,531 |
$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$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 | 49,532 |
2.75 Find the coefficients $m$ and $n$ of the quadratic trinomial $x^{2}+m x+n$, if it is known that the remainders when dividing by the binomials $\boldsymbol{x}-\boldsymbol{m}$ and $\boldsymbol{x}-\boldsymbol{n}$ are respectively $\boldsymbol{m}$ and $\boldsymbol{n}$. | 2.75 Performing the division, we get:
According to the condition, $2 m^{2}+n=m, m n+n^{2}+n=n$. If $n=0$, then $2 m^{2}=m$, i.e., $m=0$ or $m=0.5$. If $n \neq 0$, then $n=-m, m(m-1)=0$, i.e... | =0,n=0;=0.5,n=0;=1,n=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,535 |
2.76 Determine for which values of $m$ one of the roots of the equation $z^{3}-\left(m^{2}-m+7\right) z-\left(3 m^{2}-3 m-6\right)=0$ is equal to -1. Find the two remaining roots of the equation for these values of $m$. | 2.76 Assuming $z=-1$, we obtain the equation $m^{2}-m-6=0$ with respect to $m$, the roots of which are $m=3$ and $m=-2$. For each of these values of $m$, the given equation can be written as: $z^{3}-13 z-12=0$. This equation can be represented as $(z+1)(z+3)(z-4)=0$, from which $z_{1}=-1$; $z_{2}=-3 ; z_{3}=4$
Answer:... | z_{2}=-3,z_{3}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,536 |
2.77 Show that if $a$ and $b$ are roots of the equation $x^{2}+p x+1=0$, and $b$ and $c$ are roots of the equation $x^{2}+q x+2=0$, then $(b-a) \times$ $\times(b-c)=p q-6$. | 2.77 Using Vieta's theorem, we have: $a b=1, a+b=-p$ and $b c=2, b+c=-q$. Then $p q=(a+b)(b+c)=b^{2}+$ $+a c+3$, from which $b^{2}+a c=p q-3,(b-a)(b-c)=\left(b^{2}+\right.$ $+a c)-3=p q-6$. Therefore, $(b-a)(b-c)=p q-6$. | (b-)(b-)=pq-6 | Algebra | proof | Yes | Yes | olympiads | false | 49,537 |
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
$$
such an equation has no real roots. If $a \neq 1$, then $x = 1$ and, therefore, $1 + a + 1 = 2$, i.e., $a = -2$.
Answer: when $a = -2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,538 |
2.79 Find the coefficients of the equation $x^{2}+p x+q=0$ given that the difference between the roots of the equation is 5, and the difference between their cubes is 35. | 2.79 We have $x_{1}-x_{2}=5, x_{1}^{3}-x_{2}^{3}=35$, where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$. Moreover,
$$
x_{1}=\frac{-p+\sqrt{p^{2}-4 q}}{2} \text { and } x_{2}=\frac{-p-\sqrt{p^{2}-4 q}}{2}
$$
Thus $x_{1}^{3}-x_{2}^{3}=\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\r... | p=1,q=-6;p=-1,q=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,539 |
2.80 Show that among the roots of the equation $x^{4}+5 x^{3}+15 x-9=0$ there is only one positive and only one negative (it is not necessary to find the roots themselves). | ### 2.80 We have
$$
\begin{aligned}
& x^{4}+5 x^{3}+15 x-9=\left(x^{2}-3\right)\left(x^{2}+3\right)+5 x\left(x^{2}+3\right)= \\
& =\left(x^{2}+3\right)\left(x^{2}+5 x-3\right)=0
\end{aligned}
$$
i.e., either \( x^{2}+3=0 \) (this equation has no roots), or \( x^{2}+5 x-3=0 \). Since the discriminant of the latter qua... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,540 |
3.1 The tractor driver plowed three plots of land. The area of the first is $\frac{2}{5}$ of the total area of all three plots, and the area of the second is to the area of the third as $\frac{3}{2}: \frac{4}{3}$. How many hectares were there in all three plots if the third plot was 16 hectares smaller than the first? | 3.1 The areas of the plots are equal to $x, y, x-16$ (ha). According to the condition, $x=\frac{2}{5}(x+y+x-16)$ and $y:(x-16)=\frac{3}{2}: \frac{4}{3}$, from which $y=\frac{9}{8}(x-16)$. Further, we have $x=\frac{2}{5}\left(2 x-16+\frac{9}{8} x-18\right)$, or $x=\frac{2}{5}\left(\frac{25 x}{8}-34\right)$, from which $... | 136 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,541 |
3.2 Two workers together produced 72 parts per shift. After the first worker increased their labor productivity by $15 \%$, and the second by $25 \%$, together they started producing 86 parts per shift. How many parts does each worker produce per shift after the increase in labor productivity? | 3.2 Initially, workers produced $x$ and $72-x$ parts per shift, and then $1.15 x$ and $1.25(72-x)$ parts. According to the condition, $1.15 x + 90 - 1.25 x = 86 ; 0.1 x = 4 ; x = 40 ; 1.15 \cdot 40 = 46$.
Answer: 46 and 40 parts. | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,542 |
3.3 The areas of three plots of land are in the ratio $2 \frac{3}{4}: 1 \frac{5}{6}: 1 \frac{3}{8}$. It is known that 72 centners of grain more were harvested from the first plot compared to the second. Find the area of all three plots, if the average yield is 18 centners per hectare. | 3.3 Let $x, y, z$ be the areas of the plots. Then $\frac{x}{y}=\frac{6}{4}, \frac{y}{z}=\frac{4}{3}$, from which $x=6 k, y=4 k, z=3 k$. According to the condition, $(6 k-4 k) \cdot 18=72$, from which $k=2$. Therefore, the total area of all plots is $x+y+z=2(6+4+3)=26$ (ha).
Omeem: 26 ha. | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,543 |
3.4 The working day has been reduced from 8 to 7 hours. By what percentage does labor productivity need to increase to ensure that, with the same rates, the wage increases by $5 \%$? | 3.4 Let a master produce $a$ parts and earn $b$ rubles for 8 hours of work. Then the rate is $\frac{b}{a}$ rubles per part, and the labor productivity is $\frac{a}{8}$ parts per hour. After increasing productivity by $x \%$, the master started producing $\frac{a}{8}+\frac{x a}{8 \cdot 100}$ parts per hour. Therefore, i... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,544 |
3.5 On the first day of the sports competition, $\frac{1}{6}$ of the boys' team and $\frac{1}{7}$ of the girls' team did not meet the qualifying standards and were eliminated from further competition. Over the remaining period of the competition, an equal number of athletes from both teams were eliminated due to non-co... | 3.5 Let $x$ be the number of boys who met the credit standards and $2x$ be the number of girls. Thus, the initial number of team members consists of $x+48$ boys and $2x+50$ girls. By the end of the first day of the competition, $\frac{1}{6}(x+48)$ boys and $\frac{1}{7}(2x+50)$ girls dropped out. Later, an equal number ... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,545 |
3.6 One farmer obtained an average buckwheat yield of 21 centners per 1 hectare, while another, who had 12 hectares less under buckwheat, achieved an average yield of 25 centners per 1 hectare. As a result, the second farmer collected 300 centners more buckwheat than the first. How many centners of buckwheat were colle... | 3.6 Let's compile the following table:
| Farmer | Area, ha | Yield, units/ha | Mass, centners |
| :---: | :---: | :---: | :---: |
| First | $x$ | 21 | $21 x$ |
| Second | $x-12$ | 25 | $25(x-12)$ |
According to the condition, $25(x-12)-21 x=300$, from which $x=150$. Then $21 x=3150$ (centners), and $25(x-12)=3450$ (c... | 31503450 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,546 |
3.7 Two parks with a total area of 110 hectares are divided into an equal number of plots. The plots of each park are equal in area to each other, but differ from those of the other park. If the first park were divided into plots of the same area as the second, it would have 75 plots, and if the second park were divide... | 3.7 Let $S$ be the area of the park, $n$ be the number of equal-sized plots, and $Q$ be the area of a plot. Then $\frac{S}{n}=Q$. We will fill in the table with the given and required values in the sequence indicated by the numbers (2) and (12).
| Park | Initially | | | With new arrangement | | |
| :---: | :---: |... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,547 |
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 | 49,548 |
3.9 A crystal, during the formation stage, uniformly increases its mass. Observing the formation of two crystals, it was noticed that the first one gained the same mass increment over 3 months as the second one over 7 months. However, after a year, it turned out that the first crystal increased its initial mass by $4 \... | 3.9 Let the annual increase in mass $x$ be $a$; then, according to the condition, the annual increase in mass $y$ is $\frac{3 a}{7}$. We have $a=0.4 x, \frac{3 a}{7}=0.5 y \Rightarrow \frac{4}{5} \cdot \frac{x}{y}=\frac{7}{3} \Rightarrow x: y=35: 12$.
Answer: $35: 12$. | 35:12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,549 |
3.10 Two skins with a total cost of 2250 rubles were sold at an auction with a profit of $40 \%$. What is the cost of each skin if a profit of $25 \%$ was made on the first one, and a loss of $-50 \%$ on the second one? | 3.10 Let $x$ (rubles) be the cost of the first pelt, then $2250-x$ (rubles) is the cost of the second pelt. After selling, the first pelt was sold for $1.25 x$ (rubles), and the second for $1.5(2250-x)$ (rubles). According to the condition, we set up the equation
$1.25 x + 1.5(2250 - x) = 1.4 \cdot 2250$,
from which ... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,550 |
3.11 Apples of the first grade for a total of 228 rubles and apples of the second grade for a total of 180 rubles were delivered to the store. During unloading, the delivered apples were accidentally mixed. It was found that if all the apples are now sold at one price - 90 kopecks lower than the price per kilogram of f... | 3.11 Let's construct the following table:
| Apple Variety | Cost, rub | Quantity, tons | Price, rub/kg |
| :---: | :---: | :---: | :---: |
| First | 228 | $x$ | $\frac{228}{x}$ |
| Second | 180 | $x+5$ | $\frac{180}{x+5}$ |
According to the problem, we have the equation
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,551 |
3.12. The sum of the digits of a two-digit number is 12. If 36 is added to this number, the result is a number written with the same digits but in reverse order. Find the original number. | 3.12 Let the desired number be of the form $10 x+y$. Then, by the condition,
$x+y=12$
and $10 x+y+36=10 y+x$, i.e.,
$x-y+4=0$.
Adding (1) and (2), we get $2 x=8$, hence $x=4$, and $y=8$.
Answer: 48. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,552 |
3.13 A three-digit number ends with the digit 2. If it is moved to the beginning of the number, the resulting number will be 18 more than the original. Find the original number. | 3.13 Let the desired three-digit number be of the form $100x + 10y + 2$; then after moving the digit 2, it will take the form
$200 + 10x + y$.
By the condition,
$200 + 10x + y - (100x + 10y + 2) = 18$
from which $10x + y = 20$. Substituting this expression into (1), we get $200 + 20 = 220$. Thus, the original three... | 202 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,553 |
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 | 49,554 |
3.15 In the direction from $A$ to $B$, the car traveled for some time at a constant speed of $v_{1}=60 \mathrm{km} / \mathrm{h}$. The remaining part of the journey he covered in the same amount of time, but at a speed of $v_{2}=40 \mathrm{km} / \mathrm{h}$. In the opposite direction, the car traveled one half of the di... | 3.15 a) Since the car traveled at each of the specified speeds for equal time intervals, then
$$
v_{\mathrm{cp}}=\frac{v_{1}+v_{2}}{2}=\frac{60+40}{2}=50(\mathrm{KM} / \mathrm{4})
$$
b) The return trip consists of two equal segments of the path (let's assume each is $s$ km), which were covered by the car in different... | )50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,555 |
3.16 A material particle entered the hole of a pipe, and 6.8 minutes later, a second particle entered the same hole. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 m/min, while the second particle covered 3 m in the first minute a... | 3.16 Let $t$ be the time (in minutes) it takes for the second particle to catch up with the first. The distance traveled by the second particle is equal to the sum of $t$ terms of an arithmetic progression, where $a_{1}=3, d=0.5$; therefore,
$s=\frac{2 a_{1}+d(t-1)}{2} t=\frac{6+0.5(t-1)}{2} t$.
The same distance tra... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,556 |
3.17 A pedestrian walking from home to the railway station, having walked 3 km in the first hour, calculated that he would be 40 minutes late for the train departure if he continued at the same speed. Therefore, he walked the remaining distance at a speed of 4 km/h and arrived at the station 15 minutes before the train... | 3.17 Let's construct the following table:
| The pedestrian would arrive at the station | Distance, km | Speed, km/h | Time, h |
| :--- | :---: | :---: | :---: |
| On time | $x$ | $v$ | $\frac{x}{v}$ |
| Late | $x-3$ | 3 | $\frac{x-3}{3}$ |
| Early | $x-3$ | 4 | $\frac{x-3}{4}$ |
By equating the time intervals recorde... | 14\mathrm{~};3.5\mathrm{~}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,557 |
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 | 49,558 |
3.19 The first tourist, after riding a bicycle for 1.5 hours at a speed of $16 \mathrm{km} / \mathrm{h}$, makes a stop for 1.5 hours, and then continues the journey at the initial speed. After 4 hours from the departure of the first tourist, the second tourist sets off in pursuit on a motorcycle at a speed of 56 km/h. ... | 3.19 An approximate graph of the movement is shown in Fig. 3.4.
Fig. 3.4
Let $t$ be the time (in hours) it takes for the second tourist to catch up with the first. Since
$$
\begin{aligne... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,559 |
3.20 A freight train was delayed on the way for 12 minutes, and then at a distance of 60 km, it made up for the lost time by increasing its speed by 15 km/h. Find the original speed of the train. | 3.20 Let the speed of the train before the delay be $x$ km/h, and after the delay $(x+15)$ km/h. Then (Fig. 3.5)
$A B=\frac{x}{5}, C E=60$,
$C D=60-\frac{x}{5}$,
$B D=\frac{60-\frac{x}{5}}{x}$
.
Fig. 3.6
We have $A C=\frac{120}{x}$, where $x$ is the initial speed;
$C D=1+\frac{1}{6}+\frac{120-x}{x+6}$. Therefore,
$$
\frac{1... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,561 |
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 | 49,562 |
3.23 A team of workers completed a certain task. If the team is reduced by 20 people, then the same task will be completed 5 days later than with the original composition, and if the team is increased by 15 people, then the task will be completed 2 days earlier. How many workers were originally in the team and how many... | 3.23 Let $x$ workers completed the task in $y$ days; then according to the condition $x y=(x-20)(y+5)$ and $x y=(x+15)(y-2)$. We can write both equations as proportions:
$\frac{x-20}{x}=\frac{y}{y+5}$ and $\quad \frac{x+15}{x}=\frac{y}{y-2}$.
Each proportion of the form $\frac{a}{b}=\frac{c}{d}$ can be replaced by an... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,563 |
3.24 One of the two machines processes a batch of parts for 3 days longer than the other. How long would the processing of this batch of parts take on each machine separately, if it is known that when working together, these machines processed a batch of parts three times larger in 20 days? | 3.24 Time ( $t$ ), the amount of work done per unit of time, i.e., productivity ( $W$ ), and the total volume of work ( $V$ ) are related by the equation $V=W t$. Let $V=1$ and fill in the following table:
| Machine | Time, days | Volume of work | Productivity |
| :---: | :---: | :---: | :---: |
| First | $x$ | 1 | $\... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,564 |
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 | 49,565 |
3.26 How many kilograms of water need to be evaporated from 0.5 t of cellulose mass containing $85 \%$ water to obtain a mass with $75 \%$ water content? | 3.26 In the cellulose mass, there is $0.85 \cdot 500 = 425 \mathrm{kg}$ of water. Let $x$ kg of water be evaporated; then we get $425 - x = 0.75(500 - x)$, from which $x = 200(\mathrm{kg})$.
Answer: $200 \mathrm{kg}$. | 200\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,566 |
3.27 An alloy of copper and silver contains 1845 g more silver than copper. If a certain amount of pure silver, equal in mass to $\frac{1}{3}$ of the mass of pure silver originally contained in the alloy, were added to it, a new alloy would be obtained containing $83.5\%$ silver. What is the mass of the alloy and what ... | 3.27 Let's construct the following table:
| Alloy | Mass, g | |
| :--- | :---: | :---: |
| | Silver | Copper |
| Initial | $x$ | $x-1845$ |
| New | $\frac{4 x}{3}$ | $x-1845$ |
Using the condition, we get the equation
$\frac{4 x}{3}=0.835\left(\frac{4 x}{3}+x-1845\right)$, or $4 x=0.835(7 x-5535)$,
from which
$4... | 3165\mathrm{};\approx79 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,567 |
3.28 One of the three barrels is filled with water, while the others are empty. If the second barrel is filled with water from the first barrel, then $\frac{1}{4}$ of the water that was in the first barrel will remain. If the third barrel is then filled from the second, $\frac{2}{9}$ of the water that was in the second... | 3.28 Let $x$ be the capacity of the first barrel; then the capacity of the second is $\frac{3}{4} x$, and the third is $\frac{3}{4} \cdot \frac{7}{9} x=\frac{7}{12} x$. According to the condition, $\frac{7}{12} x + 50 = x$, from which $x = 120$. Thus, the capacities of the barrels are 120, 90, and 70 buckets.
Answer: ... | 120,90,70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,568 |
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 | 49,569 |
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 | 49,570 |
3.31 The marksman at the shooting range was offered the following conditions: each hit on the target is rewarded with five tokens, but for each miss, three tokens are taken away. The marksman was not very accurate. After the last ($n$-th) shot, he had no tokens left. How many shots were in the series and how many were ... | 3.31 Let there be $m$ hits, and thus $n-m$ misses. Then $5 m-3(n-m)=0$, or $8 m=3 n$, from which $m=\frac{3 n}{8}$. Since $m$ and $n$ are natural numbers, $n$ must be a multiple of eight. In the interval $10<n<20$, only the value $n=16$ is suitable. Therefore, 16 shots were fired, of which 6 were successful.
Answer: 1... | 166 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,571 |
3.32 For the transportation of cargo from one place to another, a certain number of trucks of the same capacity were required. Due to road damage, each truck had to be loaded with 0.5 tons less than planned, which is why 4 additional trucks of the same capacity were required. The mass of the transported cargo was no le... | 3.32 Let the mass of the transported cargo be $x$ tons, and the number of trucks be $n$. According to the problem,
$$
x=\left(\frac{x}{n}-\frac{1}{2}\right)(n+4), \text { or } n^{2}+4 n-8 x=0
$$
from which $n=2(-1+\sqrt{1+2 x})$ ( $n$ - natural). In the interval $55<x<64$, only $x=60$ fits. Then $n=2(-1+11)=20$. Ther... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,572 |
3.33 From milk with a fat content of $5 \%$, cottage cheese with a fat content of $15.5 \%$ is produced, leaving behind whey with a fat content of $0.5 \%$. How much cottage cheese is obtained from 1 ton of milk?
## Geometric and Physical Problems | 3.33 Let $x$ tons of cottage cheese with a fat content of $15.5\%$ be obtained; then $1-x$ tons of whey with a fat content of $0.5\%$ will remain. Therefore, in 1 ton of milk, there is
$\frac{15.5 x}{100}+\frac{0.5(1-x)}{100}=\frac{15.5 x+0.5}{100}$ tons of fat.
According to the condition, $\frac{15.5 x+0.5}{100}=\fr... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,573 |
3.35 A material point moves straight-line within the interior of a $60^{\circ}$ angle. After leaving the vertex of this angle, it found itself at a distance $a$ from one side of the angle and at a distance $b$ from the other side after some time. Then it changed its direction of motion and fell onto the side to which i... | 3.35 The length of the path traveled by the point is equal to the sum of the lengths of segments \(O A\) and \(A B = a\) (Fig. 3.7). We have
$$
O A=\frac{a}{\sin \alpha}=\frac{b}{\sin \left(60^{\circ}-\alpha\right)}
$$
, (2), we find $x=30, y=20$, $z=10$.
Answer: 30, 20, and 10 teeth. | 30,20,10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,576 |
3.37 The magnitudes of two forces acting on a material point at a right angle, and the magnitude of their resultant, form an arithmetic progression. Determine the ratio of the magnitudes of the forces. | 3.37 According to the condition,
$a^{2}+(a+d)^{2}=(a+2 d)^{2}$, i.e., $3 d^{2}+2 a d-a^{2}=0$.
Let $d=a k(k>0)$; then
$3 a^{2} k^{2}+2 a^{2} k-a^{2}=0$, from which $3 k^{2}+2 k-1=0$.
Only the root $k=\frac{1}{3}$ is suitable. We find the required ratios: $\frac{a+d}{a}=1+\frac{d}{a}=\frac{4}{3} ; \quad \frac{a+2 d}... | 3:4:5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,577 |
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 | 49,578 |
3.39 The ball is rolling perpendicular to the sideline of a football field. Suppose, moving uniformly decelerated, the ball rolled 4 m in the first second, and 0.75 m less in the next second. A footballer, initially 10 m away from the ball, started running in the direction of the ball's movement to catch it. Moving uni... | 3.39 Instruction. Use the formula $s=v t+a t^{2}$, where $v$ and $a$ are constants. Their values can be calculated from the following conditions: for the ball $s(1)=4, s(2)=7.25$; for the footballer $s(1)=3.5, s(2)=7.5$. Instead of the given formula, the formula for the sum of $n$ terms of an arithmetic progression can... | 5\mathrm{~}; | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,579 |
3.40 A red pencil costs 27 kopecks, a blue one - 23 kopecks. No more than 9 rubles 40 kopecks can be spent on purchasing pencils. It is necessary to purchase the maximum possible total number of red and blue pencils. At the same time, the number of red pencils should be as few as possible, but the number of blue pencil... | 3.40 Let $x$ red and $y$ blue pencils be bought. According to the conditions, $27 x+23 y \leqslant 940$ and $y-x \leqslant 10$. Let's construct the lines
\[
\begin{aligned}
& 27 x+23 y=940 ... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,580 |
3.41 Several students decided to buy an imported tape recorder priced between 170 and 195 dollars. However, at the last moment, two of them refused to participate in the purchase, so each of the remaining had to contribute 1 dollar more. What was the price of the tape recorder?
## Ratios between natural numbers | 3.41 Let $x$ be the original number of students, and $p$ be the cost of the tape recorder. Then $\frac{p}{x}$ is the amount each student was originally supposed to contribute, and $\frac{p}{x-2}$ is the amount each of the tape recorder buyers contributed. Using one of the conditions of the problem, we get the equation
... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,581 |
3.42 The digits of a certain three-digit number form a geometric progression. If in this number the digits of the hundreds and units are swapped, the new three-digit number will be 594 less than the desired one. If, however, in the desired number the digit of the hundreds is erased and the digits of the resulting two-d... | 3.42 The desired number has the form $100 x+10 y+z$, where $x, y, z$ form a geometric progression, i.e.
$x z=y^{2}$.
According to the condition, we have:
$100 z+10 y+x=100 x+10 y+z-594$, from which
$x-z=6$
$10 y+z=10 z+y+18$, from which
$y-z=2$.
The system of equations (1), (2), (3) is satisfied by $x=8$, $\bold... | 842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,582 |
3.43 When multiplying two numbers, one of which is 10 more than the other, a student made a mistake by reducing the tens digit of the product by 4. When dividing the resulting product by the smaller multiplier to check the answer, he got a quotient of 39 and a remainder of 22. Find the multipliers. | 3.43 According to the condition, $x(x+10)-40-22=39 x$, where $x-$ is the sought smaller factor. From this, we find $x=31$ and $x+10=41$.
Answer: 31 and 41. | 3141 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,583 |
3.44 A positive integer is thought of. To its digital representation, a digit was appended on the right. From the resulting new number, the square of the thought number was subtracted. The difference turned out to be 8 times the thought number. What number was thought of and which digit was appended? | 3.44 Let the number $x$ be thought of, then by appending the digit $y$ to the right of it, we get the number $10x + y$. According to the condition, $10x + y - x^2 = 8x$, from which we get $x^2 - 2x - y = 0$, i.e., $x = 1 \pm \sqrt{1 + y}$. Possible values of $y$ are: $0; 3; 8$. Therefore, the thought-of number is 2, or... | 2,3,4;0,3,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,584 |
3.45 The recording of a six-digit number starts with the digit 2. If this digit is moved from the first position to the last, keeping the order of the other five digits, the newly obtained number will be three times the original number. Find the original number.
## Distance: path, speed, time | 3.45 The original six-digit number has the form $2 \cdot 10^{5}+x$. After moving the digit 2 to the last place, we get the number $10 x+2$. According to the condition, $10 x+2=3\left(2 \cdot 10^{5}+x\right)$, from which $x=85714$. Therefore, the original number is 285714.
Answer: 285714. | 285714 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,585 |
3.46 The path from $A$ to $B$ is traveled by a passenger train 3 hours and 12 minutes faster than by a freight train. In the time it takes the freight train to travel from $A$ to $B$, the passenger train travels 288 km more. If the speed of each train is increased by $10 \mathrm{km} / \mathrm{h}$, the passenger train w... | 3.46 In Fig. 3.9, the graphs of train movements before the speed changes are shown.
Fig. 3.9
We find $\operatorname{tg} \angle M C N=v^{(0)}{ }_{\text {пас }}=288: 3.2=90$ (km/h). Let $v^... | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,586 |
3.47 Two athletes start running simultaneously - the first from $A$ to $B$, the second from $B$ to $A$. They run at different but constant speeds and meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the len... | 3.47 In Fig. 3.10, the graphs of the runs of two athletes are shown.
Fig. 3.10
Let $A B=x$ m, $C$ and $D$ be the points of the first and second meetings. If $v_{1}$ and $v_{2}$ are the spe... | 500\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,587 |
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,
. At the moment when the first car arrived in ... | 3.49 Consider a coordinate system "path" ($s$ - in kilometers), "time" ($t$ - in hours). Let $AC$ (Fig. 3.12) be the graph of the motion of the first car with the desired speed $v = \tan \alpha$; $DE$ and $EF$ be the graph of the "there and back" motion of the second car with speed $\tan \beta = 45$; $AD = \frac{1}{3}$... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,589 |
3.50 Two cyclists started simultaneously from the same place in the same direction. A third cyclist started from the same place 10 minutes after them. First, he overtook the first cyclist, and then continued his journey for another 20 minutes before catching up with the second cyclist. From the start to the end of the ... | 3.50 The graphs of the cyclists' movements are shown in Fig. 3.13.
Fig. 3.13
Let's express the distances they have traveled: for the first $A B=\left(\frac{1}{6}+x\right) a$; for the thir... | \frac{1}{4}(+3b+\sqrt{^{2}-10+b^{2}})\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,590 |
3.51 On a river with a current speed of $5 \mathrm{km} / \mathrm{h}$, there are piers $A, B$, and $C$ in the direction of the current, with $B$ located midway between $A$ and $C$. From pier $B$, a raft and a boat depart simultaneously, with the raft moving downstream to pier $C$ and the boat heading to pier $A$. The sp... | 3.51 Instruction. Take $A B=B C$ as a unit and solve the inequality $\frac{1}{v-5}+\frac{2}{v+5}>\frac{1}{5}$, where $v>5$.
Answer: $\quad 5<\nu<15$. | 5<\nu<15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,591 |
3.52 A passenger on a train knows that the speed of this train on this section of the track is 40 km/h. As soon as a passing train started to go by the window, the passenger started a stopwatch and noticed that the passing train went by the window for 3 seconds. Determine the speed of the passing train, given that its ... | 3.52 Instruction. The condition of the problem is equivalent to the oncoming train standing still, while the passenger rushes past it at a speed equal to $(v+40)$ km/h, where $v$ is the sought speed.
Answer: $50 \mathrm{km} / \mathrm{h}$. | 50\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,592 |
3.53 Usually, two mechanisms are involved in performing a certain task simultaneously. The productivity of these mechanisms is not the same, and when working together, they complete the task in 30 hours. Once, the joint operation of the two mechanisms lasted only 6 hours, after which the first mechanism was stopped, an... | 3.53 Let's accept the entire volume of work as a unit. Let the first mechanism complete the task in $x$ hours, and the second in $y$ hours. Since when working together, the capacities (productivities) add up, we have the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{1}{30} \\
1-6\left(\f... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,593 |
3.54 Three pumps, pumping water for irrigation, started working simultaneously. The first and third pumps finished working simultaneously, while the second one finished 2 hours after the start. As a result, the first pump pumped out $9 \mathrm{~m}^{3}$ of water, and the second and third together pumped out $28 \mathrm{... | 3.54 Let the first and second pumps extract $x$ and $y$ m$^{3}$ per hour, respectively, then the third pump extracts $(x+3)$ m$^{3}$ per hour. The second and third pumps extracted $2 y$ and $(28-2 y)$ m$^{3}$ of water, respectively. The first pump worked for $\frac{9}{x}$ hours, and the third pump for $\frac{28-2 y}{x+... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,594 |
3.55 During the unloading of a barge, four lifting cranes of the same power initially worked for 2 hours. Then, two additional cranes of lesser but equal power were put into operation. After this, it took another 3 hours to complete the unloading. If all these cranes had started working simultaneously, the unloading wo... | 3.55 Let's accept the entire volume of work as a unit. Let the more powerful crane need $x$ hours, the less powerful crane need $y$ hours, and the time for one more powerful and one less powerful crane working together be $-z$ hours. Then the corresponding productivities are $\frac{1}{x}, \frac{1}{y}$, and $\frac{1}{z}... | 14.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,595 |
3.56 Two pipes of different cross-sections are connected to a swimming pool, one uniformly supplying water, the other uniformly draining it, with the first pipe filling the pool 2 hours longer than the second pipe takes to empty it. When the pool was filled to $\frac{1}{3}$ of its capacity, both pipes were opened, and ... | 3.56 Let's write down the values of the required and given quantities in the form of a table:
| Pipe | Time, h | Capacity | Productivity |
| :---: | :---: | :---: | :---: |
| Supply | $x$ | 1 | $\frac{1}{x}$ |
| Discharge | $x+2$ | 1 | $\frac{1}{x+2}$ |
| Both together | 8 | 1 | $\frac{1}{24}$ |
According to the cond... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,596 |
3.59 Impurities constitute $20 \%$ of the total volume of the solution. What is the smallest number of filters through which the solution must be passed so that the final impurity content does not exceed $0.01 \%$, if each filter absorbs $80 \%$ of the impurities? (It is known that $\lg 2 \approx 0.30$.) | 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 | 49,597 |
4.1 $\left(1+\frac{1}{\cos 2 \alpha}+\operatorname{tg} 2 \alpha\right)\left(1-\frac{1}{\cos 2 \alpha}+\operatorname{tg} 2 \alpha\right)=2 \operatorname{tg} 2 \alpha$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
4.1 $\left(1+\fr... | 4.1 By regrouping the terms and using formulas (1.8) and (1.9) first, and then (4.5), we get
$\left(1+\frac{1}{\cos 2 \alpha}+\operatorname{tg} 2 \alpha\right)\left(1-\frac{1}{\cos 2 \alpha}+\operatorname{tg} 2 \alpha\right)=$
$=\left((1+\operatorname{tg} 2 \alpha)+\frac{1}{\cos 2 \alpha}\right)\left((1+\operatorname... | 2\tan2\alpha | Algebra | proof | Yes | Yes | olympiads | false | 49,599 |
$4.2 \frac{\tan 2 \alpha+\cot 3 \beta}{\cot 2 \alpha+\tan 3 \beta}=\frac{\tan 2 \alpha}{\tan 3 \beta}$. | 4.2 Let's factor out $\operatorname{tg} 2 \alpha$ in the numerator and $\operatorname{tg} 3 \beta$ in the denominator, and use formula (4.4):
$\frac{\operatorname{tg} 2 \alpha+\operatorname{ctg} 3 \beta}{\operatorname{ctg} 2 \alpha+\operatorname{tg} 3 \beta}=\frac{\operatorname{tg} 2 \alpha(1+\operatorname{ctg} 2 \alp... | \frac{\tan2\alpha}{\tan3\beta} | Algebra | proof | Yes | Yes | olympiads | false | 49,600 |
4.3 $\sin 4 \alpha-\sin 5 \alpha-\sin 6 \alpha+\sin 7 \alpha=-4 \sin \frac{\alpha}{2} \sin \alpha \sin \frac{11 \alpha}{2}$. | 4.3 Denoting the left side of the identity by $A$, we group the terms as follows:
$A=(\sin 4 \alpha-\sin 6 \alpha)+(\sin 7 \alpha-\sin 5 \alpha)$.
Now, we will use formula (4.20) first, and then (4.22):
$A=-2 \sin \alpha \cos 5 \alpha+2 \sin \alpha \cos 6 \alpha=2 \sin \alpha(\cos 6 \alpha-\cos 5 \alpha)=$
$=-4 \si... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,601 |
$4.4 \frac{1}{\sin \alpha}+\frac{1}{\tan \alpha}=\cot \frac{\alpha}{2}$. | 4.4 Method. Using formulas (4.2), (4.18), and (4.4):
$$
\frac{1}{\sin \alpha}+\frac{\cos \alpha}{\sin \alpha}=\frac{1+\cos \alpha}{\sin \alpha}=\frac{1}{\operatorname{tg} \frac{\alpha}{2}}=\operatorname{ctg} \frac{\alpha}{2}
$$
II method. Using formulas (4.28), (4.15), and (4.4), we find
$$
\frac{1}{\sin \alpha}+\fr... | \operatorname{ctg}\frac{\alpha}{2} | Algebra | proof | Yes | Yes | olympiads | false | 49,602 |
4.5 $2 \sin ^{2}(3 \pi-2 \alpha) \cos ^{2}(5 \pi+2 \alpha)=\frac{1}{4}-\frac{1}{4} \sin \left(\frac{5 \pi}{2}-8 \alpha\right)$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
4.5 $2 \sin ^{2}(3 \pi-2 \alpha) \cos ^{2}(5 \pi+2 \alp... | ## 4.5 We have
$$
\begin{aligned}
& 2 \sin ^{2}(3 \pi-2 \alpha) \cos ^{2}(5 \pi+2 \alpha)=2 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha=\frac{1}{2} \sin ^{2} 4 \alpha= \\
& =\frac{1}{4}(1-\cos 8 \alpha)=\frac{1}{4}\left(1-\sin \left(\frac{5 \pi}{2}-8 \alpha\right)\right)=\frac{1}{4}-\frac{1}{4} \sin \left(\frac{5 \pi}{2}-8 ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,603 |
$4.6 \frac{\cos \left(\frac{5 \pi}{2}-6 \alpha\right)+\sin (\pi+4 \alpha)+\sin (3 \pi-\alpha)}{\sin \left(\frac{5 \pi}{2}+6 \alpha\right)+\cos (4 \alpha-2 \pi)+\cos (\alpha+2 \pi)}=\tan \alpha$. | 4.6 First, we will use the reduction formulas and write the left side of the identity as
$\frac{\sin 6 \alpha-\sin 4 \alpha+\sin \alpha}{\cos 6 \alpha+\cos 4 \alpha+\cos \alpha}$.
Now we will use formulas (4.20) and (4.21):
$\frac{2 \sin \alpha \cos 5 \alpha+\sin \alpha}{2 \cos 5 \alpha \cos \alpha+\cos \alpha}=\fra... | \tan\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,604 |
4.7 $(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}=4 \sin ^{2} \frac{\alpha-\beta}{2}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
4.7 $(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}=4 \sin ^{2} \frac{\a... | 4.7 I с по о о о. Applying formulas (4.22), (4.20), and (4.1), we find
$$
\begin{aligned}
& (\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}= \\
& =4 \sin ^{2} \frac{\alpha+\beta}{2} \sin ^{2} \frac{\alpha-\beta}{2}+4 \sin ^{2} \frac{\alpha-\beta}{2} \cos ^{2} \frac{\alpha+\beta}{2}= \\
& =4 \sin ^{2} \frac{\... | 4\sin^{2}\frac{\alpha-\beta}{2} | Algebra | proof | Yes | Yes | olympiads | false | 49,605 |
$4.8 \cos ^{2}\left(\alpha-90^{\circ}\right)+\operatorname{ctg}^{2}\left(\alpha-270^{\circ}\right)=$
$=\frac{1}{\sin ^{2}\left(\alpha+90^{\circ}\right)}-\cos ^{2}\left(\alpha+180^{\circ}\right)$. | 4.8 The identity to be proven (after applying reduction formulas) is equivalent to the following:
$\sin ^{2} \alpha+\operatorname{tg}^{2} \alpha=\frac{1}{\cos ^{2} \alpha}-\cos ^{2} \alpha$.
By transforming the left side of the last equation, we get
$$
\begin{aligned}
& \sin ^{2} \alpha+\frac{\sin ^{2} \alpha}{\cos ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,606 |
$4.9 \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)=\frac{\sin 4 \alpha}{\sqrt{2}}$. | ## 4.9 We have
$$
\begin{aligned}
& \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)= \\
& =\frac{1-\cos \left(\frac{7 \pi}{4}-4 \alpha\right)-1+\cos \left(\frac{9 \pi}{4}-4 \alpha\right)}{2}= \\
& =-\sin (2 \pi-4 \alpha) \sin \frac{\pi}{4}=\frac{\sin 4 \alpha}{\sqrt{2}}
\e... | \frac{\sin4\alpha}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,607 |
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