problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
105*. Prove that for any odd $n$ the expression $n^{12}-n^{8}-n^{4}+1$ is divisible by 512. | Instruction. Let's factor the given polynomial:
$$
\begin{aligned}
& n^{12}-n^{8}-n^{4}+1=\left(n^{8}-1\right)\left(n^{4}-1\right)= \\
& =\left(n^{4}+1\right)\left(n^{2}+1\right)^{2}(n+1)^{2}(n-1)^{2}
\end{aligned}
$$
For an odd number $n$, $n^{4}+1$ is an even number, i.e., it is divisible by 2, $\left(n^{2}+1\right... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,313 |
106. Prove that if the product of three consecutive integers is increased by the middle one, the resulting sum will be equal to the cube of the middle number.
$2^{*}$ | Let $n, n+1, n+2$ be three consecutive integers. Then, according to the problem:
$$
\begin{gathered}
n(n+1)(n+2)+(n+1)=(n+1)\left(n^{2}+2 n+1\right)= \\
=(n+1)(n+1)^{2}=(n+1)^{3}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,314 |
107*. Prove that
$$
(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(a+c)
$$ | Instruction. The task is to factorize the polynomial. To prove the identity, we will transform the given polynomial:
$$
\begin{gathered}
(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=(a+b)^{3}+3(a+b)^{2} c+ \\
+3(a+b) c^{2}+c^{3}-a^{3}-b^{3}-c^{3}=(a+b)^{3}+3(a+b)^{2} c+ \\
+3(a+b) c^{2}-\left(a^{3}+b^{3}\right)=(a+b)\left(a^{2}+2 a ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,315 |
108*. Prove that $\left(a^{2}+b^{2}\right)(a b+c d)-a b\left(a^{2}+b^{2}-c^{2}-d^{2}\right)=$ $=(a c+b d)(a d+b c)$. | Instruction. The proof reduces to factoring the polynomial. Transform the polynomial standing in the left-hand side of the equality to be proven:
$$
\begin{gathered}
\left(a^{2}+b^{2}\right)(a b+c d)-a b\left(a^{2}+b^{2}-c^{2}-d^{2}\right)= \\
=\left(a^{2}+b^{2}\right)(a b+c d)-a b\left(a^{2}+b^{2}\right)+a b\left(c^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,316 |
109*. Prove the identity:
\[
\begin{gathered}
(a+b+c)^{3}+(b-a-c)^{3}+(c-a-b)^{3}+ \\
+(a-b-c)^{3}=24 a b c .
\end{gathered}
\] | Indication. Transform the left side of the equation:
$$
\begin{aligned}
& A=\left[(a+b+c)^{3}+(b-a-c)^{3}\right]+\left[(c-a-b)^{3}+\right. \\
& \left.\quad+(a-b-c)^{3}\right]=2 b\left(3 a^{2}+b^{2}+3 c^{2}+6 a c\right)- \\
& \quad-2 b\left(3 a^{2}+b^{2}+3 c^{2}-6 a c\right)=2 b \cdot 12 a c=24 a b c
\end{aligned}
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,317 |
112*. Prove that if $P=\frac{a+b+c}{2}$, then
\[
\begin{gathered}
{[(P-a)+(P-b)]^{3}=} \\
=(P-a)^{3}+(P-b)^{3}+3(P-a)(P-b) c
\end{gathered}
\] | Instruction. Apply the formula:
$$
\begin{gathered}
(a+b)^{3}=a^{3}+b^{3}+3(a+b) a b \\
{[(P-a)+(P-b)]^{3}=(P-a)^{3}+(P-b)^{3}+} \\
+3(P-a)(P-b)(P-a+P-b)= \\
=(P-a)^{3}+(P-b)^{3}+3(P-a)(P-b)(2 P-a-b)
\end{gathered}
$$
but
$$
P=\frac{a+b+c}{2}
$$
so,
$$
2 P-a-b=a+b+c-a-b=c
$$
i.e.
$$
[(P-a)+(P-b)]^{3}=(P-a)^{3}+(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,318 |
114*. Prove that if $a+b+c=0$, then
$$
2\left(a^{4}+b^{4}+c^{4}\right)=\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$ | Instruction. Transform the left side of the equality using the condition of the problem:
$$
2\left(a^{4}+b^{4}+c^{4}\right)=2\left[a^{4}+b^{4}+(a+b)^{4}\right]
$$
and factor the polynomial in the brackets. For this, we complete the square:
$$
\begin{gathered}
a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}= \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,319 |
115. Find the condition for the divisibility of $a^{n}-x^{n}$ by $a^{p}-x^{p}$. | Let $a^{p}=y, x^{p}=b$, then
$$
a^{n}=y^{\frac{n}{p}}, \quad x^{n}=b^{\frac{n}{p}}
$$
and the required condition will be: $\frac{n}{p}=k$, i.e., $n$ must be a multiple of $p$. | kp | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,320 |
117. Find the condition for the divisibility of $(x+1)^{n}+(x-1)^{n}$ by $x$.
To find the condition for the divisibility of $(x+1)^{n}+(x-1)^{n}$ by $x$, we need to determine when the expression is divisible by $x$. This means that when $x=0$, the expression should equal zero.
Let's substitute $x=0$ into the expressi... | Instruction. Transform the divisor $x$ into the form:
$$
x=\frac{1}{2}[(x+1)+(x-1)]
$$ | n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,321 |
118. Prove that $(a x+b y)^{3}+(b x+a y)^{3}$ is divisible by $(a+b)(x+y)$. | Instruction. Transform the divisor:
$$
\begin{gathered}
(a+b)(x+y)=a x+b x+a y+b y= \\
=(a x+b y)+(b x+a y)
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,322 |
119. Prove that $\left(x^{2}-x y+y^{2}\right)^{3}+\left(x^{2}+x y+y^{2}\right)^{3}$ is divisible by $2 x^{2}+2 y^{2}$. | Instruction. Transform the divisor:
$$
2 x^{2}+2 y^{2}=\left(x^{2}-x y+y^{2}\right)+\left(x^{2}+x y+y^{2}\right)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,323 |
120. Prove that the polynomial $x^{3}+y^{3}+z^{3}-3 x y z$ is divisible by $x+y+z$. | Instruction. The divisor is represented by $x+(y+z)$. Calculate the remainder by setting $x=-y-z$.
Note: The text has been translated while preserving the original line breaks and format. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,324 |
121*. Find the condition for divisibility
$$
x^{n}+x^{n-1}+\ldots+x+1 \text { by } x^{p}+x^{p-1}+\ldots+x+1 \text {. }
$$
22 | Instruction. The task is to transform the dividend and divisor into a form for which the question of divisibility can be solved using already discussed problems. Multiply the dividend and divisor by $(x-1)$, where $x \neq 1$. We get:
$$
\begin{aligned}
& \left(x^{n}+x^{n-1}+\ldots+x+1\right)(x-1)=x^{n+1}-1 \\
& \left(... | n+1=k(p+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,325 |
122*. Prove that the product
$$
\left(x^{n}-1\right)\left(x^{n-1}-1\right)\left(x^{n-2}-1\right)
$$
is divisible by the product
$$
(x-1)\left(x^{2}-1\right)\left(x^{3}-1\right)
$$ | Indication. The exponents of $x$ are three consecutive integers: $n, n-1, n-2$. Among them, there must be numbers divisible by 2 and 3. Therefore, one of the factors of the dividend must be divisible by $x^{2}-1$, another by $x^{3}-1$, and all factors are divisible by $x-1$.
The solutions to problems $123-132$ are bas... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,326 |
123. Prove that any positive integer power of any positive integer, decreased by this number, is an even number. | Instruction. Transform the expression:
$$
a^{n}-a=a\left(a^{n-1}-1\right)=a(a-1)\left(a^{n-2}+a^{n-3}+\ldots+a+1\right)
$$
but $a(a-1)$ is an even number. Therefore, under the conditions of the problem, $a^{n}-a$ is an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,327 |
124. Prove that for any integer $n$ the expression
$$
3^{2 n+2}-2^{n+1}
$$
is divisible by 7. | Instruction. Transform the expression:
$$
\begin{gathered}
3^{2 n+2}-2^{n+1}=9^{n+1}-2^{n+1}=(9-2)\left(9^{n}+9^{n-1} \cdot 2+\ldots+2^{n}\right)= \\
=7\left(9^{n}+9^{n-1} \cdot 2+\ldots+2^{n}\right)
\end{gathered}
$$
it is divisible by 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,328 |
125. Prove that for any integer $n$ the expression
$$
3^{4 n+4}-4^{3 n+3}
$$
is divisible by 17. | Instruction. Transform the expression to the form:
$$
3^{4 n+4}-4^{3 n+3}=81^{n+1}-64^{n+1}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,329 |
126. Prove that for any integer $n$ the expression
$$
3^{2 n+2}+2^{6 n+1}
$$
is divisible by 11. | Instruction. Transform the expression:
$$
\begin{aligned}
& 3^{2 n+2}+2^{6 n+1}=9 \cdot 3^{2 n}+2 \cdot 8^{2 n} ; 3^{2 n}=(11-8)^{2 n}= \\
& =11^{2 n}-2 n \cdot 11^{2 n-1} \cdot 8+\ldots+8^{2 n}
\end{aligned}
$$
from which
$$
\begin{gathered}
9 \cdot 3^{2 n}+2 \cdot 8^{2 n}=9\left(11^{2 n}-2 n \cdot 11^{2 n-1} \cdot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,330 |
127. Prove that for $n$ being a positive integer, the expression
$$
A=3^{2(n+1)} \cdot 5^{2 n}-3^{3 n+2} \cdot 2^{2 n}
$$
is divisible by 117. | Instruction. Transform the expression:
$$
\begin{gathered}
A=3^{2(n+1)} \cdot 5^{2 n}-3^{3 n+2} \cdot 2^{2 n}=3^{2 n+2}\left(5^{2 n}-3^{n} \cdot 2^{2 n}\right)= \\
=9 \cdot 3^{2 n}\left(25^{n}-12^{n}\right)=9 \cdot 13 \cdot 3^{2 n} \cdot B=117 \cdot 3^{2 n} B
\end{gathered}
$$
where
$$
B=25^{n-1}+25^{n-2} \cdot 12+\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,331 |
128. Prove that for a natural number $n$, greater than 1, the expression
$$
7^{2 n}-4^{2 n}-297
$$
is divisible by 264. | Instruction. Transform the expression:
$$
\begin{gathered}
7^{2 n}-4^{2 n}-297=\left(49^{n}-16^{n}\right)-33-264= \\
=33\left(49^{n-1}+49^{n-2} \cdot 16+\ldots+16^{n-1}-1\right)-264
\end{gathered}
$$
HO
$$
49^{n-1}+49^{n-2} \cdot 16+\ldots+16^{n-1}-1
$$
is divisible by 8, since all terms of this expression, except ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,332 |
129. Prove that $5 \cdot 7^{2(n+1)} + 2^{3n}$ for natural $n$ is divisible by 41. | Instruction. $7^{2(n+1)}=49^{n+1}=(41+8)^{n+1} ; 2^{3 n}=8^{n} ;$
$$
5 \cdot 7^{2(n+1)}+2^{3 n}=5 \cdot(41+8)^{n+1}+8^{n}=
$$
$=5 \cdot 41^{n+1}+(n+1) \cdot 5 \cdot 8 \cdot 41^{n}+\ldots+(n+1) \cdot 5 \cdot 41 \cdot 8^{n}+$ $+5 \cdot 8^{n+1}+8^{n}$ is divisible by 41, since
$$
5 \cdot 8^{n+1}+8^{n}=8^{n} \cdot 41
$$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,333 |
130*. Prove that for any whole and positive $n$ the number $11^{n+2}+12^{2 n+1}$ is divisible by 133. | Instruction. The task is to transform the given sum in such a way that it would be possible to determine the divisibility of each term by 133. Transform the dividend:
$$
11^{n+2}+12^{2 n+1}=11^{n} \cdot 11^{2}+12^{2 n} \cdot 12=11^{n} \cdot 121+144^{n} \cdot 12=
$$
(to make the first term divisible by 133, add and su... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,334 |
131*. Prove that the expression $5^{2 n+1}+2^{n+4}+2^{n+1}$ is divisible by 23. | Instruction. Transform the dividend to the form: $5 \cdot 5^{2 n}+18 \cdot 2^{n}$. Add and subtract $5 \cdot 2^{n}$. The term $5\left(5^{2 n}-2^{n}\right)$ should be transformed to the form:
$$
5 \cdot 23\left(25^{n-1}+25^{n-2} \cdot 2+\ldots+2^{n-1}\right)
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,335 |
132*. Prove that $3^{3 n+2}+5 \cdot 2^{3 n+1}$ for $n$ being a positive integer is divisible by 19. | Instruction. Transform the dividend to the form: $9 \cdot 27^{n}+10 \cdot 8^{n}$. Add and subtract $9 \cdot 8^{n}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,336 |
133. Prove that the polynomial
$$
x^{2 n}-n^{2} x^{n+1}+2\left(n^{2}-1\right) x^{n}+1-n^{2} x^{n-1}
$$
is divisible by $(x-1)^{3}$. | Instruction. Transform the polynomial to the form:
$\left(x^{n}-1\right)^{2}-n^{2} x^{n-1}(x-1)^{2}$, and find that it is divisible by $(x-1)^{2}$. Performing the division, we get the quotient:
$$
\begin{gathered}
A=\left(\frac{x^{n}-1}{x-1}\right)^{2}-n^{2} x^{n-1}= \\
=\left(x^{n-1}+x^{n-2}+\ldots+x^{2}+x+1\right)^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,337 |
134. Find such proper fractions, each of which, when its numerator and denominator are decreased by 1, becomes $\frac{1}{2}$. | Let the fraction $\frac{a}{b}$ be a proper fraction. According to the problem, $\frac{a-1}{b-1}=\frac{1}{2}$ or $b=2a-1$. The required fractions will be: $\frac{2}{3}, \frac{3}{5}, \frac{4}{7}, \frac{5}{9} \ldots$ and so on. | \frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9}\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,338 |
136. Prove that adding the same natural number to the numerator and denominator of a fraction less than one increases the fraction; for a fraction greater than one, the fraction decreases. | Let $\frac{a}{b}$ be a proper fraction and $n$ a natural number. We form the fraction $\frac{a+n}{b+n}$ and compare it with the given fraction by finding their difference. Similarly, proceed if $\frac{a}{b}$ is an improper fraction greater than 1. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,339 |
137. Prove that when subtracting the same natural number, which is less than the numerator, from the numerator and denominator of a proper fraction, the resulting fraction is smaller than the given one. | Instruction. $\frac{a}{b}-\frac{a-n}{b-n}=\frac{n(b-a)}{b(b-n)}>0$, since $b>a, b>n$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,340 |
138. Prove that when subtracting the same natural number, which is less than the denominator, from the numerator and the denominator of an improper fraction greater than 1, the fraction increases. | Instruction. $\frac{a-n}{b-n}-\frac{a}{b}=\frac{n(a-b)}{b(b-n)}>0$, since $a>b, b>n$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,341 |
139*. Prove that if the sum of two fractions is equal to one, then the square of the first fraction, added to the second fraction, is equal to the square of the second fraction, added to the first fraction. | Given that $\frac{a}{b}+\frac{c}{d}=1$. Let $\frac{a}{b}>\frac{c}{d}$. Then
$$
\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{a}{b}-\frac{c}{d}\right)=\left(\frac{a}{b}\right)^{2}-\left(\frac{c}{d}\right)^{2}=\frac{a}{b}-\frac{c}{d}
$$
from which
$$
\left(\frac{a}{b}\right)^{2}+\frac{c}{d}=\left(\frac{c}{d}\right)^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,342 |
140*. Show that for no integer $n$ the fractions $\frac{n-6}{15}$ and $\frac{n-5}{24}$ can simultaneously be equal to integers. | Assumption. Suppose that the given fractions can simultaneously be equal to integers, i.e.
$$
\frac{n-6}{15}=A ; \quad \frac{n-5}{24}=B
$$
where $A$ and $B$ are integers. Then $n-6=15 A$ and $n-5=24 B$, hence $n=15 A+6$ and $n=24 B+5$, i.e., $15 A+6=24 B+5$ or $24 B-15 A=1$, or $3(8 B-5 A)=1$, which cannot be true fo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,343 |
141*. Prove that if a proper fraction is irreducible, then the fraction that complements it to one is also irreducible. | Proof. Let the given fraction be $\frac{a}{b}$. The fraction that complements it to one is $\frac{b-a}{b}$. Suppose this fraction is reducible. Then $b$ and $b-a$ must have a common divisor, i.e., $a$ and $b$ must have a common divisor, which contradicts the condition of the problem. Therefore, the assumption is incorr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,344 |
142*. Prove that the sum of two irreducible fractions can only equal a whole number when the fractions have the same denominator. | Let $\frac{a}{b}$ and $\frac{c}{d}$ be two irreducible fractions and let $\frac{a}{b}+\frac{c}{d}=A$. Then $a d+b c=A b d$, i.e., $b c$ is divisible by $d^{*}$, but $c$ and $d$ are coprime by condition, so $b$ is divisible by $d$. Moreover, $a d$ must be divisible by $b$, since $b c$ and $A b d$ are divisible by $b$, w... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,345 |
143*. Find the condition under which the difference of two irreducible fractions is equal to their product. | Let $\frac{a}{b}$ and $\frac{c}{d}$ be two irreducible fractions, such that $\frac{a}{b}-\frac{c}{d}=\frac{a}{b} \cdot \frac{c}{d}$, from which $a d-b c=a c$, or $a d=c(a+b)$, or $b c=a(d-c)$. Given that these fractions are irreducible, we conclude that $c$ is a divisor of $a$ and $a$ is a divisor of $c$, i.e., $a=c$ a... | \frac{}{b},\frac{}{+b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,346 |
145. Prove the identity:
$$
\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}=1
$$
for $a \neq b, \quad a \neq c, \quad b \neq c$. | Instruction. Reduce the given fractions to a common denominator, add them, and transform the numerator to the form:
$$
(a-b)(a-c)(b-c)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,347 |
146. Prove that the sum of the fractions $\frac{b-c}{1+b c}, \frac{c-a}{1+c a}, \frac{a-b}{1+a b}$ is equal to their product. Specify the domain of permissible values for the variables involved.[^2] | Instruction. Find the sum of the given fractions and transform the numerator of the obtained fraction to the form:
$$
(b-c)(a-b)(c-a)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,348 |
147. Prove that the expression
$$
\left(\frac{a^{3 n}}{a^{n}-1}+\frac{1}{a^{n}+1}\right)-\left(\frac{a^{2 n}}{a^{n}+1}+\frac{1}{a^{n}-1}\right)
$$
is an integer if $a$ is an integer, and $n \neq 1$.
Specify the domain of permissible values for $a$. | Instruction. After performing the actions, transform the numerator of the obtained fraction to the form:
$$
\left(a^{2 n}-1\right)\left(a^{2 n}+2\right)
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,349 |
149. Prove that
$$
b^{2}-x^{2}=\frac{4}{c^{2}}[p(p-a)(p-b)(p-c)]
$$
if
$$
a+b+c=2 p, \quad x=\frac{b^{2}+c^{2}-a^{2}}{2 c} ; c \neq 0
$$ | $$
\begin{gathered}
b^{2}-x^{2}=(b-x)(b+x)=\left(b-\frac{b^{2}+c^{2}-a^{2}}{2 c}\right)\left(b+\frac{b^{2}+c^{2}-a^{2}}{2 c}\right)= \\
=\frac{(a+b-c)(a-b+c)(b+c-a)(b+c+a)}{4 c^{2}}= \\
=\frac{2 p(2 p-2 a)(2 p-2 b)(2 p-2 c)}{4 c^{2}}=\frac{4}{c^{2}}[p(p-a)(p-b)(p-c)]
\end{gathered}
$$
The translation is as follows:
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,351 |
150*. Prove that if $a+b+c=0$, then
$$
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9
$$
where $a \neq 0, \quad b \neq 0, c \neq 0, a \neq b, \quad a \neq c, b \neq c$. | Proof.
$$
\begin{gathered}
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)= \\
=1+\frac{(b-c) c}{a(a-b)}+\frac{(c-a) c}{b(a-b)}+\frac{a(a-b)}{c(b-c)}+ \\
+1+\frac{(c-a) a}{b(b-c)}+\frac{b(a-b)}{c(c-a)}+\frac{(b-c) b}{a(c-a)}+1= \\
=3+\frac{c}{a-b}\left... | 9 | Algebra | proof | Yes | Yes | olympiads | false | 42,352 |
151*. Prove that if
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
$$
To
$$
\begin{gathered}
\frac{1}{a^{2 n+1}}+\frac{1}{b^{2 n+1}}+\frac{1}{c^{2 n+1}}= \\
=\frac{1}{a^{2 n+1}+b^{2 n+1}+c^{2 n+1}}(a \neq 0, b \neq 0, c \neq 0)
\end{gathered}
$$ | Instruction. Based on the condition, prove that
$$
(a+b)(b+c)(a+c)=0
$$
30
i.e., either $a=-b$, or $a=-c$, or $b=-c$. Transform the left side of the equation to be proven using each of the obtained dependencies. Let, for example, $a=-b$. Then
$$
\begin{aligned}
& \frac{1}{a^{2 n+1}}+\frac{1}{b^{2 n+1}}+\frac{1}{c^{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,353 |
152*. Prove that if
then
$$
a b+b c+a c=1
$$
$$
\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}=\frac{4 a b c}{\left(1-a^{2}\right)\left(1-b^{2}\right)\left(1-c^{2}\right)}
$$
where $a \neq \pm 1, b \neq \pm 1, c \neq \pm 1$. | Instruction. Transform the left side to the form:
$$
\frac{a(1-a b-a c)+b(1-b c-a b)+c(1-a c-b c)+a b c(b c+a c+a b)}{\left(1-a^{2}\right)\left(1-b^{2}\right)\left(1-c^{2}\right)}
$$
and note that
$$
\begin{aligned}
& 1-a b-a c=b c \text { by the condition of the problem, } \\
& 1-b c-a b=a c \\
& 1-a c-b c=a b
\end... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,354 |
155*. Prove the identity:
\[
\begin{gathered}
\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-c)(b-a)}+\frac{a-b}{(c-a)(c-b)}= \\
=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a} \\
(a \neq b, a \neq c, b \neq c)
\end{gathered}
\] | Instruction. Transform each term of the left side of the equation by representing each fraction as a sum of two simpler fractions:
$$
\begin{aligned}
& \frac{b-c}{(a-b)(a-c)}=\frac{(a-c)+(b-a)}{(a-b)(a-c)}=\frac{1}{a-b}+\frac{1}{c-a} \\
& \frac{c-a}{(b-c)(b-a)}=\frac{(b-a)+(c-b)}{(b-c)(b-a)}=\frac{1}{b-c}+\frac{1}{a-b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,356 |
156. Prove that $\sqrt{2}$ cannot be a rational number. | Proof. Among the integers, there cannot be a number equal to $\sqrt{2}$, since $1<\sqrt{2}<2$. Suppose that $\sqrt{2}$ is a rational fraction, i.e., $\sqrt{2}=\frac{n}{p}$, and $\frac{n}{p}$ is an irreducible fraction. Then $2=\frac{n^{2}}{p^{2}}$ or $2 p^{2}=n^{2}$, hence $n^{2}$ and $n$ are even numbers, i.e., $n=2 k... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,357 |
157. Prove that if the root of an integer power of a positive integer is not an integer, then it cannot be a fractional number either. | Proof. Suppose that $\sqrt[k]{A}=\frac{n}{p}$, where $A$ is a positive integer, $\frac{n}{p}$ is an irreducible fraction, and $p>1$. Then $A=\frac{n^{k}}{p^{k}}$, but $\frac{n}{p}$ is an irreducible fraction, which means $\frac{n^{k}}{p^{k}}$ is also an irreducible fraction, i.e., the integer $A$ turns out to be equal ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,358 |
158. Prove that the sum, difference, product, and quotient (if the divisor is not zero) of two rational numbers is a rational number.
3 I. V. Baranova and S. E. Lyapin
33 | Consider the sum of two rational numbers: $\frac{a}{b}, \frac{n}{p}$, where $a, b, n, p$ are integers, and $b \neq 0, p \neq 0$. The sum of these numbers
$$
\frac{a}{b}+\frac{n}{p}=\frac{a p+b n}{b p}
$$
is a rational number, since $a p+b n$ is an integer (the sum of products of integers is an integer), and $b p$ is ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,359 |
159. Show that the sum of a rational and an irrational number cannot be expressed as a rational number. | Instruction. The sum of a rational and an irrational number cannot be expressed as a rational number, since if the sum of the numbers is rational and one addend is rational, then the second addend must also be a rational number, as the difference of rational numbers is a rational number (see the previous problem). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,360 |
160. Prove that when taking the square root of two positive rational numbers that are not perfect squares, the resulting decimal sequences cannot be identical. | Proof. Let $p$ and $n$ be two positive rational numbers that are not perfect squares. Suppose that extracting the square root from them results in the same sequence of decimal digits. Then $\sqrt{p}-\sqrt{n}=k ; k$ is an integer, or $\sqrt{p}=\sqrt{n}+k$, from which
$$
p=n+k^{2}+2 \sqrt{n} \cdot k
$$
Since $n$ is not... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 42,361 |
161*. Prove the identity:
$$
\sqrt{A \pm \sqrt{\bar{B}}}=\sqrt{\frac{A+\sqrt{A^{2}-B}}{2}} \pm \sqrt{\frac{A-\sqrt{A^{2}-B}}{2}}
$$
34 | Proof. Let
$$
\sqrt{A+\sqrt{B}}+\sqrt{A-\sqrt{B}}=x
$$
Squaring both sides, we get:
$$
x^{2}=2 A+2 \sqrt{A^{2}-B}
$$
from which
$$
x=\sqrt{2 A+2 \sqrt{A^{2}-B}}
$$
and therefore,
$$
\sqrt{A+\sqrt{B}}+\sqrt{A-\sqrt{B}}=2 \sqrt{\frac{A+\sqrt{A^{2}-B}}{2}}
$$
Similarly:
$$
\sqrt{A+\sqrt{B}}-\sqrt{A-\sqrt{B}}=2 \s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,362 |
163*. Find the condition under which the expression $\sqrt{A \pm \sqrt{\bar{B}}}$ can be represented in the form of a binomial $\sqrt{x} \pm \sqrt{y}$, where $x$ and $y$ are rational numbers. | Solution. Let
$$
\sqrt{A+\sqrt{B}}=\sqrt{x}+\sqrt{y}
$$
then from the previous problem,
$$
\sqrt{A-\sqrt{B}}=\sqrt{x}-\sqrt{y}
$$
Multiplying equations (1) and (2), we get:
$$
\sqrt{A+\sqrt{B}} \cdot \sqrt{A-\sqrt{B}}=(\sqrt{x}+\sqrt{y}) \cdot(\sqrt{x}-\sqrt{y})
$$
or
$$
\sqrt{A^{2}-B}=x-y
$$
Squaring equation ... | A^{2}-B | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,363 |
164. Prove that for $x \geqslant 1$ the expression
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
equals 2 if $x \leqslant 2$; equals $2 \sqrt{x-1}$ if $x>2$. | Instruction. Transform the given expression by applying the formula for the compound radical.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 42,364 | |
165. Prove that
$$
\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}
$$ | Instruction. Check if all expressions under the radicals will be positive. Transform the denominators of the fractions to the form:
$$
\begin{aligned}
& \sqrt{2}+\sqrt{2+\sqrt{3}}=\frac{\sqrt{2}}{2}(3+\sqrt{3}) \\
& \sqrt{2}-\sqrt{2-\sqrt{3}}=\frac{\sqrt{2}}{2}(3-\sqrt{3})
\end{aligned}
$$
Perform the transformations... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,365 |
169*. Prove that
$$
\frac{2 a \sqrt{1+x^{2}}}{x+\sqrt{1+x^{2}}}=a+b
$$
when
$$
x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right) \text { and } \quad a>0, \quad b>0 .
$$ | Instruction. Take into account the expression for $x$, which is to be transformed to the form:
$$
x=\frac{a-b}{2 \sqrt{a b}}
$$ | Algebra | proof | Yes | Yes | olympiads | false | 42,366 | |
173. Prove that with the established definitions of the sum and product of complex numbers, the basic laws of addition and multiplication remain valid. | Let us prove the commutative law of addition for the adopted definition of the sum of two complex numbers:
$$
(a+b i)+(c+d i)=(a+c)+(b+d) i=
$$
(by definition)
$$
=(c+a)+(d+b) i=
$$
(by the commutative law of addition for real numbers)
$$
=(c+d i)+(a+b i)
$$
(by the definition of the sum of complex numbers), i.e.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,368 |
174. Find the necessary and sufficient condition for the sum of two complex numbers $a+bi$ and $c+di$ to be 1) a real number, 2) a purely imaginary number. | 1) The sum of the given complex numbers is expressed as $(a+c)+(b+d) i$. For this complex number to be real, the equality $b+d=0$ or $b=-d$ must hold. If $b=-d$, then the complex number will be real. Therefore, the condition $b=-d$ is necessary and sufficient for the sum of the given complex numbers to be a real number... | - | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,369 |
177. Find the necessary and sufficient condition for the complex number $a+b i$ to be 1) equal to its conjugate, 2) equal to the reciprocal of its conjugate, 3) equal to the opposite of its conjugate number. | Instruction. 2. To
$$
a+b i=\frac{1}{a-b i}
$$
or
$$
(a+b i)(a-b i)=1
$$
the equality $a^{2}+b^{2}=1$ must hold. Conversely, if $a^{2}+b^{2}=1$, then $a+b i=\frac{1}{a-b i}$. Therefore, $a^{2}+b^{2}=1$ is the required condition.
Answer. 1) $b=0 ; 2) a^{2}+b^{2}=1$; 3) $a=0$. | 1)b=0;2)^{2}+b^{2}=1;3)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,370 |
182. Prove that when two conjugate complex numbers are squared and cubed, the results are again conjugate complex numbers. | Let $a+b i$ and $a-b i$ be conjugate complex numbers. Then
$$
\begin{aligned}
& (a+b i)^{2}=\left(a^{2}-b^{2}\right)+2 a b i \\
& (a-b i)^{2}=\left(a^{2}-b^{2}\right)-2 a b i
\end{aligned}
$$
The complex numbers (1) and (2) are conjugates. Similarly, the second statement can be proven. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,371 |
183. Prove that any positive integer power of a complex number is also a complex number. | Instruction.
$$
\begin{gathered}
(a+b i)^{n}=a^{n}+n a^{n-1} b i-\frac{n(n-1)}{1 \cdot 2} a^{n-2} b^{2}- \\
-\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3} b^{3} i+\ldots+b^{n} i^{n}
\end{gathered}
$$
All terms containing $b$ with an even exponent are real numbers, while those containing $b$ with an odd exponent are i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,372 |
186. Prove that
$$
\frac{\left(1-x^{2}+x i \sqrt{3}\right) \cdot\left(1-x^{2}-x i \sqrt{3}\right)}{1-x^{6}}=\frac{1}{1-x^{2}},|x| \neq 1
$$ | Instruction. Transform the numerator of the fraction on the left side of the equation to the form: $x^{4}+x^{2}+1$, and the denominator to the form:
$$
\left(1-x^{2}\right)\left(x^{4}+x^{2}+1\right), \quad x^{4}+x^{2}+1 \neq 0, \quad|x| \neq 1
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,374 |
189*. Prove that the roots of the binomial equation $x^{n}-1=0$ are determined by the formula:
$$
x_{k}=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}
$$
where $k=0,1,2 \ldots(n-1)$. | Proof. If $x^{n}-1=0$, then $x^{n}=1, x=\sqrt[n]{1}$. But the unit can be represented as:
$$
1=\cos 2 k \pi+i \sin 2 k \pi
$$
from which
$$
\sqrt[n]{1}=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}, \text { where } k=0,1,2 \ldots(n-1)
$$
Therefore, the roots of the binomial equation $x^{n}-1=0$ are determined by ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,376 |
190*. Prove that the sum of the $p$-th powers of the roots of the equation $x^{n}-1=0$, where $p$ is a positive integer, is equal to $n$ if $p$ is divisible by $n$ and is equal to 0 if $p$ is not divisible by $n$. | Proof. Transforming the formula for the roots of a binomial equation obtained in the previous problem, we will have:
$$
x_{k}=\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)^{k}
$$
where
$$
k=0,1,2 \ldots(n-1)
$$
Then
$$
x_{1}=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}
$$
and
$$
x_{k}=\left(x_{1}\right... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,377 |
192. Prove the identities
1) $\cos 2 x=\cos ^{2} x-\sin ^{2} x$, $\sin 2 x=2 \sin x \cdot \cos x$.
2) $\cos 3 x=\cos ^{3} x-3 \cos x \cdot \sin ^{2} x$ $\sin 3 x=3 \cos ^{2} x \cdot \sin x-\sin ^{3} x$. | Proof. It is known that
$$
(\cos x+i \sin x)^{n}=\cos n x+i \sin n x
$$
For $n=2$, we have:
$$
(\cos x+i \sin x)^{2}=\cos 2 x+i \sin 2 x
$$
On the other hand:
$$
(\cos x+i \sin x)^{2}=\cos ^{2} x-\sin ^{2} x+2 i \sin x \cos x
$$
By comparing equations (1) and (2), we get:
$$
\cos 2 x=\cos ^{2} x-\sin ^{2} x ; \q... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,378 |
193*. Prove the identities:
1) $\cos n x=\cos ^{n} x-C_{n}^{2} \cos ^{n-2} x \sin ^{2} x+C_{n}^{4} \cos ^{n-4} \sin ^{4} x-\ldots+$
$$
+\left\{\begin{array}{l}
(-1)^{\frac{n}{2}} \sin ^{n} x \quad(n-\text { even }) \\
(-1)^{\frac{n-1}{2}} n \cos x \sin ^{n-1} x(n-\text { odd })
\end{array}\right.
$$
2) $\sin n x=n \... | Instruction. Raise the complex number $\cos x+i \sin x$ to the $n$-th power, using de Moivre's formula and the binomial theorem; then compare the real and imaginary parts in the obtained expressions.
Example.
$$
\begin{aligned}
\cos 4 x & =\cos ^{4} x-6 \cos ^{2} x \sin ^{2} x+\sin ^{4} x \\
\sin 4 x & =4 \cos ^{3} x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,379 |
194**. Prove that the following sequence of numbers:
$$
\begin{aligned}
& a_{1}=\cos x+i \sin x \\
& a_{2}=\cos 2 x+i \sin 2 x \\
& a_{3}=\cos 3 x+i \sin 3 x \\
& \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
& a_{n}=\cos n x+i \sin n x
\end{aligned}
$$
is a geometric progression and find its sum. | Proof. 1) Transform the form of the numbers in this sequence using de Moivre's formula:
$$
\begin{aligned}
& a_{1}=\cos x+i \sin x \\
& a_{2}=\cos 2 x+i \sin 2 x=(\cos x+i \sin x)^{2}, \\
& a_{3}=\cos 3 x+i \sin 3 x=(\cos x+i \sin x)^{3} \\
& a_{n}=\cos n x+i \sin n x=(\cos x+i \sin x)^{n}
\end{aligned}
$$
Thus, this... | S_{n}=\frac{\sin\frac{nx}{2}\cdot\cos\frac{n+1}{2}x}{\sin\frac{x}{2}}+i\frac{\sin\frac{nx}{2}\cdot\sin\frac{n+1}{2}x}{\sin\frac{x}{2}} | Algebra | proof | Yes | Yes | olympiads | false | 42,380 |
195*. Prove that if
$$
x+\frac{1}{x}=2 \cos \alpha
$$
then
$$
x^{n}+\frac{1}{x^{n}}=2 \cos n \alpha
$$ | Proof. If \( x + \frac{1}{x} = 2 \cos \alpha \), then \( x^2 - 2x \cos \alpha + 1 = 0 \), i.e.,
\[
x = \cos \alpha \pm \sqrt{\cos^2 \alpha - 1} ; \quad x = \cos \alpha \pm i \sin \alpha
\]
From this,
\[
x^n = \cos n \alpha \pm i \sin n \alpha ; \quad \frac{1}{x^n} = x^{-n} ; \quad \cos n \alpha \mp \sin n \alpha = x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,381 |
197**. Prove that if $\left(\frac{x+i}{x-i}\right)^{n}=1$, then $x=\operatorname{ctg} \frac{k \pi}{n}, \quad$ where $k=1,2, \ldots(n-1)$. | Proof. If $\left(\frac{x+i}{x-i}\right)^{n}=1$, then $\frac{x+i}{x-i}=\sqrt[n]{1}$.
$$
\frac{x+i}{x-i}=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}
$$
where $k=1,2, \ldots(n-1)$ or
$$
x+i=(x-i)\left(\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right)
$$
from which
$$
\begin{gathered}
x=\frac{\left(1+\cos \fr... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,382 |
198. Find the conditions that the parameters of the equation $a x+b=0$ must satisfy for the equation to have: 1) a positive root, 2) a negative root, 3) a root equal to zero. | Instruction. 1) The signs of the parameters are different, 2) the signs of the parameters are the same, 3) $b=0 ; a \neq 0$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,383 | |
199. Find the conditions that the parameters $a, b, c$ of the equation $a x=b-c$ must satisfy for the equation to have: 1) a positive root, 2) a negative root, 3) a root equal to zero. | Instruction. 1) when $a>0, b>c$ or when $a>0, c<b$;
2) when $a<0, b>c$ or when $a<0, c>b$;
3) when $a \neq 0, b=c$. | >0,b>or>0,<b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,384 |
201. Prove that from the equality $a x+b y=0$ and $c x+d y=0$ when $a d-c b \neq 0$ it follows that $x=y=0$. | Instruction. Multiply the first of the equalities by $d$, the second by $b$, we get:
$$
\begin{array}{r}
a d x+b d y=0 \\
c b x+b d y=0
\end{array}
$$
Subtracting one equality from the other term by term, we will have:
$$
a d x-c b x=0 \text { or }(a d-c b) x=0 \text {, }
$$
but $a d-c b \neq 0$ by condition, hence... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,386 |
202. Prove that the condition for the compatibility of the equations $a x+b=0$ and $c x+d=0$ is the equality: $a d-b c=0$. | Given. From the condition of the problem, we have: $x=-\frac{b}{a}$ when $a \neq 0$, $x=-\frac{d}{c}$ when $c \neq 0$, from which $\frac{b}{a}=\frac{d}{c}$ or $a d-b c=0$. | -=0 | Algebra | proof | Yes | Yes | olympiads | false | 42,387 |
203. Find the condition for the compatibility of the equations:
$$
a_{1} x+b_{1} y=c_{1}, \quad a_{2} x+b_{2} y=c_{2}, \quad a_{3} x+b_{3} y=c_{3}
$$ | Instruction. Assuming that the equations are consistent, transform the first two into the form:
$$
\left(a_{1} b_{2}-a_{2} b_{1}\right) x=c_{1} b_{2}-c_{2} b_{1}, \quad\left(a_{1} b_{2}-a_{2} b_{1}\right) y=a_{1} c_{2}-a_{2} c_{1}
$$
Determine the form of the values for $x$ and $y$ under the assumption that $a_{1} b_... | a_{1}(b_{2}c_{3}-b_{3}c_{2})+a_{2}(b_{3}c_{1}-c_{3}b_{1})+a_{3}(b_{1}c_{2}-b_{2}c_{1})=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,388 |
204. Prove that the condition for the compatibility of the equations:
$$
\begin{gathered}
y+z=a, \quad x+y=b, \quad x+z=c, \quad x+y+z=d \\
\text { is } a+b+c=2 d .
\end{gathered}
$$ | Instruction. By adding the first three equations term by term, we will have:
$$
2 x+2 y+2 z=a+b+c \quad \text { or } \quad 2(x+y+z)=a+b+c
$$
The fourth equation, according to the problem, is $x+y+z=d$. Therefore, for the consistency of these equations, it must be:
$$
a+b+c=2 d
$$
Note. Problems 205 and 206 deal wit... | +b+=2d | Algebra | proof | Yes | Yes | olympiads | false | 42,389 |
205*. Find the general solution formulas for the equation $a x + b y = c$ under the condition that $a$ and $b$ are coprime numbers. | Instruction. Suppose the equation $a x+b y=c$ has a solution $x_{1}, y_{1}$, then $a x_{1}+b y_{1}=c$. Subtracting the obtained equality from the given equation, we will have:
$$
\left(x-x_{1}\right) a+\left(y-y_{1}\right) b=0 \quad \text { or } \quad a\left(x-x_{1}\right)=-b\left(y-y_{1}\right) .
$$
Since $b$ does n... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,390 |
207. Derive the formula for solving a quadratic equation, $a x^{2}+b x+c=0$, by completing the square. | Instruction. Multiplying the given equation by $4a$ and adding $b^{2}$ to both sides, transform it into the form:
$$
4 a^{2} x^{2}+4 a b x+b^{2}+4 a c=b^{2}
$$
54
from which
$$
(2 a x+b)^{2}=b^{2}-4 a c
$$
or
$$
2 a x+b= \pm \sqrt{b^{2}-4 a c}
$$
or
$$
x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}
$$ | \frac{-\sqrt{b^{2}-4}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,391 |
208. Prove that the quadratic equation $a x^{2}+b x+c=0$ cannot have more than two distinct roots. | Proof. Suppose that the given equation has three distinct roots: $x_{1}, x_{2}, x_{3}$. Then we will have three identities:
$$
\begin{aligned}
& a x_{1}^{2}+b x_{1}+c=0 \\
& a x_{2}^{2}+b x_{2}+c=0 \\
& a x_{3}^{2}+b x_{3}+c=0
\end{aligned}
$$
Subtracting the second identity from the first, and then the third, we get... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,392 |
209. In the equation $x^{2}+p x+q=0$ with rational coefficients, the roots $x_{1}$ and $x_{2}$ are not rational and $x_{1}=x_{2}^{3}$.
Prove that the equation has the form: $x^{2}+1=0$. | Proof. If $x_{1}, x_{2}$ are the roots of the equation
$$
x^{2}+p x+q=0
$$
then let
$$
x_{1}=\frac{-p+\sqrt{p^{2}-4 q}}{2}, x_{2}=\frac{-p-\sqrt{p^{2}-4 q}}{2}
$$
By the condition of the problem, $x_{1}=x_{2}^{3}$, i.e.,
$$
\frac{-p+\sqrt{p^{2}-4 q}}{2}=\left(\frac{-p-\sqrt{p^{2}-4 q}}{2}\right)^{3}
$$
or, after ... | x^{2}+1=0 | Algebra | proof | Yes | Yes | olympiads | false | 42,393 |
211. Prove that the sum of the reciprocals of the roots of the equation $x^{2}+p x+q=0$ is equal to $-\frac{p}{q}$. | Instruction. If $x_{1}, x_{2}$ are the roots of the given equation, then
$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=q
$$
Let's find
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{p}{q}
$$ | -\frac{p}{q} | Algebra | proof | Yes | Yes | olympiads | false | 42,395 |
212. Prove that if $x_{1}$ and $x_{2}$ are the roots of the equation
$$
x^{2} + p x + q = 0
$$
then
$$
x_{1}^{2} + x_{2}^{2} = p^{2} - 2 q, \quad x_{1}^{2} - x_{2}^{2} = \pm p \sqrt{p^{2} - 4 q}
$$ | Instruction. Transform the sum of the squares of the roots to the form:
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
From the same equation,
$$
\left(x_{1}-x_{2}\right)^{2}=p^{2}-2 q-2 x_{1} x_{2}=p^{2}-4 q
$$
or
$$
x_{1}-x_{2}= \pm \sqrt{p^{2}-4 q}
$$
Therefore,
$$
x_{1}^{2}-x_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,396 |
215. Given the quadratic equation $x^{2}+p x+q=0$. Find the equation whose roots: 1) differ from the roots of the given equation only in sign, 2) are the reciprocals of the roots of the given equation. | Solution. Let the roots of the given equation be $x_{1}, x_{2}$, then, as is known,
$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=q
$$
If the roots of the desired equation are ( $-x_{1}$ ) and ( $-x_{2}$ ), then
$$
\begin{gathered}
\left(-x_{1}\right)+\left(-x_{2}\right)=-\left(x_{1}+x_{2}\right)=-(-p)=p \\
\left(-x_{1}\righ... | x^{2}-px | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,399 |
217. Prove that the quadratic equation, the roots of which are $n$ times the roots of the equation $a x^{2}+b x+c=0$, has the form:
$$
a x^{2}+n b x+n^{2} c=0
$$ | If $x_{1}$ and $x_{2}$ are the roots of the given equation, then, according to the problem, the roots of the second equation will be: $n x_{1}$ and $n x_{2}$. Therefore,
$$
n x_{1}+n x_{2}=n\left(x_{1}+x_{2}\right)=n\left(-\frac{b}{a}\right)=-\frac{n b}{a} ; n x_{1} \cdot n x_{2}=\frac{n^{2} c}{a},
$$
i.e., the equat... | ^{2}+n+n^{2}=0 | Algebra | proof | Yes | Yes | olympiads | false | 42,401 |
218. Without solving the quadratic equation $x^{2}+p x+q=0$, form a new equation whose roots are the squares of the roots of the given equation. | Solution. Let the roots of the given quadratic equation be: $x_{1}$ and $x_{2}$; the roots of the desired equation: $x_{1}^{2}$ and $x_{2}^{2}$. Then
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q ; \quad x_{1}^{2} \cdot x_{2}^{2}=q^{2}
$$
Thus, the desired equation will have the form:
$... | x^{2}-(p^{2}-2q)x+q^{2}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,402 |
219. Without solving the quadratic equation $x^{2} + p x + q = 0$, form a new equation, one of whose roots is equal to the sum of the cubes of the roots of the given equation, and the other is the cube of their sum. | Instruction. The solution is similar to the solution of the previous problem.
Answer. $x^{2}+p\left(2 p^{2}-3 q\right) x+\left(p^{2}-3 q\right) p^{4}=0$. | x^{2}+p(2p^{2}-3q)x+(p^{2}-3q)p^{4}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,403 |
220. Given the quadratic equation $a x^{2}+b x+c=0$, with roots $x_{1}$ and $x_{2}$. Prove that the equation whose roots are $x_{1}^{2}+x_{2}^{2}$ and $2 x_{1} x_{2}$, will have the form:
$$
a^{3} x^{2}-a b^{2} x+2 c\left(b^{2}-2 a c\right)=0
$$ | Instruction. The solution boils down to formulating expressions for the sum and product of the roots of the new equation, using the expressions for the sum and product of the roots of the given equation through its coefficients. | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,404 |
222. Find the relationship between the coefficients of the equation $a x^{2}+b x+c=0$, if the sum of its roots is twice their difference. | Let $x_{1}, x_{2}$ be the roots of the given equation. Then $x_{1}+x_{2}=-\frac{b}{a} ; x_{1} \cdot x_{2}=\frac{c}{a}$. According to the problem,
$$
x_{1}+x_{2}=2\left(x_{1}-x_{2}\right)
$$
from which $x_{1}=3 x_{2}$. If $x_{1}=3 x_{2}$, then
$$
x_{1}+x_{2}=4 x_{2}=-\frac{b}{a} ; \quad x_{1} x_{2}=3 x_{2}^{2}=\frac{... | 3b^{2}=16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,406 |
223. Find the dependence between the coefficients of the equation $a x^{2}+b x+c=0$, if the ratio of the roots is 2. | Solution. Let one root of the equation $a x^{2}+b x+c=0$ be $x_{1}$, then the other should be $2 x_{1}$. Therefore,
$$
x_{1}+2 x_{1}=-\frac{b}{a}
$$
or
$$
\begin{aligned}
& 3 x_{1}=-\frac{b}{a} \\
& 2 x_{1} \cdot x_{1}=\frac{c}{a}
\end{aligned}
$$
or
$$
2 x_{1}^{2}=\frac{c}{a}
$$
From equations (1) and (2), we ob... | 2b^{2}=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,407 |
224. In the equation $\left(k^{2}-5 k+3\right) x^{2}+(3 k-1) x+2=0$, find the value of $k$ for which the ratio of the roots of the equation would be 2. | Solution. In the equation $a x^{2}+b x+c=0$, the ratio of the roots is 2, which means there is a relationship between the coefficients of the equation: $2 b^{2}=9 a c$ (see problem 223). For the given equation, this relationship will take the form:
$$
2(3 k-1)^{2}=9\left(k^{2}-5 k+3\right) \cdot 2
$$
or
$$
(3 k-1)^{... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,408 |
225. Find the condition under which the difference of the roots in the equation $x^{2}+p x+$ $+q=0$ would be equal to $a$. | Solution. Let the roots of the given equation be denoted by $x_{1}$ and $x_{2}$, then
$$
x_{1}+x_{2}=-p, \quad x_{1} x_{2}=q, \quad x_{1}-x_{2}=a
$$
From the first and third equations, we find $x_{1}$ and $x_{2}$, and substituting the found values for $x_{1}$ and $x_{2}$ into the second equation, we get:
$$
x_{1}=\f... | ^{2}-p^{2}=-4q | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,409 |
226. Find the condition under which the difference of the squares of the roots of the equation $a x^{2}+b x+c=0$ is equal to $\frac{c^{2}}{a^{2}}$. | Instruction. The solution is similar to the solution of the previous problem.
Answer. $b^{4}-c^{4}=4 a b^{2} c$. | b^{4}-^{4}=4^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,410 |
228*. Prove that if the equations
$$
a x^{2}+b x+c=0
$$
and
$$
A x^{2}+B x+C=0
$$
have a common root, and the other roots are different, then
$$
(A c-C a)^{2}=(A b-B a)(B c-C b)
$$
60 | Instruction. If $x_{1}$ is a common root of the given equations, then we will have two identities:
$$
a x_{1}^{2}+b x_{1}+c=0, A x_{1}^{2}+B x_{1}+C=0
$$
Eliminating $x_{1}^{2}$ from these identities, we get:
$$
x_{1}=-\frac{A c-C a}{A b-B a}
$$
Eliminating the free term from these same identities, we will have:
$... | (A-C)^{2}=(Ab-B)(B-Cb) | Algebra | proof | Yes | Yes | olympiads | false | 42,411 |
229. If the value of one of the roots of the equation
$$
x^{2}+p x+q=0
$$
is the reciprocal of the value of one of the roots of the equation
$$
x^{2}+m x+n=0
$$
then there exists the dependence
$$
(p n-m)(q m-p)=(q n-1)^{2}
$$
Prove this. | Let $x_{1}$ be one of the roots of the equation
$$
x^{2}+p x+q=0
$$
then $\frac{1}{x_{1}}$ should be a root of the equation
$$
x^{2}+m x+n=0
$$
We have the identities:
$$
\begin{gathered}
x_{1}^{2}+p x_{1}+q=0 \\
n x_{1}^{2}+m x_{1}+1=0
\end{gathered}
$$
According to problem 228, we will have:
$$
(p n-m)(q m-p)=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,412 |
230. Given two equations
$$
x^{2}+a x+1=0 \text { and } x^{2}+x+a=0
$$
Find all values of $a$ for which these equations have at least one common root. | Instruction. From the second equation, we find $a$ and substitute it into the first:
$$
a=-(x^{2}+x), \quad x^{2}-(x^{2}+x) x+1=0, \quad x^{3}-1=0
$$
or
$$
(x-1)\left(x^{2}+x+1\right)=0
$$
Therefore,
$$
x_{1}=1 \text { and } x_{2,3}=\frac{-1 \pm i \sqrt{3}}{2}
$$
But
$$
a=-(x^{2}+x)
$$
thus,
$$
a_{1}=1, \quad ... | a_1=1,\quada_2=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,413 |
232. If $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2} + b x + c = 0$, then, without solving the equation, prove that
$$
\sqrt[4]{x_{1}^{4} + 4 x_{1}^{3} x_{2} + 6 x_{1}^{2} x_{2}^{2} + 4 x_{1} x_{2}^{3} + x_{2}^{4}} = -\frac{b}{a}
$$ | $$
\begin{gathered}
\sqrt[4]{x_{1}^{4}+4 x_{1}^{3} x_{2}+6 x_{1}^{2} x_{2}^{2}+4 x_{1} x_{2}^{3}+x_{2}^{4}}=\sqrt[4]{\left(x_{1}+x_{2}\right)^{4}}= \\
=x_{1}+x_{2}=-\frac{b}{a}
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,414 |
233. If $x_{1}$ and $x_{2}$ are the roots of the equation
$$
\frac{3 a-b}{c} x^{2}+\frac{c(3 a+b)}{3 a-b}=0
$$
then, without solving it, find $x_{1}^{117}+x_{2}^{117}$. | Solution.
$x_{1}^{117}+x_{2}^{117}=\left(x_{1}+x_{2}\right)\left(x_{1}^{116}-x_{1}^{115} x_{2}+\ldots+x_{2}^{116}\right), \quad$ but $\quad x_{1}+x_{2}=0$, so, $x_{1}^{117}+x_{2}^{117}=0$.
Note. It is useful to solve problem 233 when reviewing, after discussing the equality:
$$
a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,415 |
234. Prove that if $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+b x+b^{2}+a=0$, then the following equality holds:
$$
x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}+a=0
$$ | Instruction. Transform the left side of the given equality to the form:
$$
\begin{gathered}
x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}+a=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}+a= \\
=(-b)^{2}-\left(b^{2}+a\right)+a=0
\end{gathered}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,416 |
235. Let $x_{1}$ and $x_{2}$ be the roots of the quadratic equation
$$
x^{2} + p x + q = 0
$$
with rational coefficients. Prove that the expressions:
$$
x_{1}^{4} + x_{1}^{3} x_{2} + x_{1}^{2} x_{2}^{2} + x_{1} x_{2}^{3} + x_{2}^{4} \text { and } x_{1}^{4} x_{2} + x_{1}^{3} x_{2}^{2} + x_{1}^{2} x_{2}^{3} + x_{1} x_... | Instruction. Transform the given expressions so that they contain the sum and product of the roots of the given equation:
$$
\begin{gathered}
x_{1}^{4}+x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}+x_{1} x_{2}^{3}+x_{2}^{4}=\left(x_{1}+x_{2}\right)^{2}\left[\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right]+ \\
+x_{1}^{2} x_{2}^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,417 |
236. If the roots of the equation $p x^{2}+n x+n=0$ are in the ratio $a: b$, then prove that
$$
\begin{aligned}
& \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{n}{p}}=0 \\
& a \neq 0, b \neq 0, n \neq 0, p \neq 0
\end{aligned}
$$ | Let $x_{1}$ and $x_{2}$ be the roots of the equation $p x^{2} + n x + n = 0$, then
$$
x_{1} + x_{2} = -\frac{n}{p}, \quad x_{1} x_{2} = \frac{n}{p}, \quad x_{1} : x_{2} = a : b
$$
from which
$$
x_{1} = \sqrt{\frac{n a}{p b}}, \quad x_{2} = \sqrt{\frac{n b}{p a}}
$$
Substituting the obtained values for $x_{1}$ and $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,418 |
237*. Prove that if $s_{1}, s_{2}, \ldots, s_{n}$ denote respectively the sums of the first, second, and so on up to the $n$-th powers of the roots of the quadratic equation
$$
a x^{2}+b x+c=0
$$
then
$$
\begin{gathered}
a s_{2}+b s_{1}+2 c=0, a s_{3}+b s_{2}+c s_{1}=0 \\
\ldots a s_{n}+b s_{n-1}+c s_{n-2}=0
\end{ga... | Proof. Let the roots of the given equation be $x_{1}$ and $x_{2}$. For $s_{1}$ we have $s_{1}=x_{1}+x_{2}=-\frac{b}{a}$. To find $s_{2}$, we take the identities:
$$
\begin{aligned}
& a x_{1}^{2}+b x_{1}+c=0 \\
& a x_{2}^{2}+b x_{2}+c=0
\end{aligned}
$$
Adding them, we get:
$$
a\left(x_{1}^{2}+x_{2}^{2}\right)+b\left... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,419 |
239. Show that if the discriminant of the quadratic equation $a x^{2}+b x+c=0$ is zero, then the left-hand side of this equation is a perfect square. | Given. $D=b^{2}-4 a c=0$ - by the condition of the problem, hence $4 a c=b^{2} \quad$ and
$$
c=\frac{b^{2}}{4 a}
$$
5th ed. Baranova and S. E. Lyapin
Substituting into the given quadratic equation the expression obtained for $c$ (1), we will have;
$$
a x^{2}+b x+\frac{b^{2}}{4 a}=0
$$
or
$$
4 a^{2} x^{2}+4 a b x+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,420 |
240. Prove that if a quadratic equation with rational coefficients has an irrational root $a+\sqrt{b}$, then the second root will also be irrational, conjugate to the first, i.e., $a-\sqrt{b}$. | Proof. If the coefficients of a quadratic equation are rational, then the sum and product of the roots, expressed in terms of these coefficients, must also be rational. The sum and product of two expressions, one of which is irrational, can only be rational if the second expression is also irrational and conjugate to t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,421 |
241. Prove that if a quadratic equation with real coefficients has an imaginary root $a+bi$, then it also has the conjugate root, i.e., $a-bi$. | If the coefficients of a quadratic equation are real numbers, then the sum and product of the roots of this equation must also be real numbers. The sum and product of imaginary numbers can only be real numbers if they are conjugates (see problem 176). Indeed,
$$
(a+b i)+(a-b i)=2 a ; \quad(a+b i)(a-b i)=a^{2}+b^{2}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 42,422 |
245. Prove that the roots of the equation $x^{2}-(q+n) x+$ $+(q n-p^{2})=0$ are real and generally different numbers. Find the condition under which the roots of this equation will be equal. | Indication. $D=\left(\frac{q+n}{2}\right)^{2}-\left(q n-p^{2}\right)=\frac{(q-n)^{2}+4 p^{2}}{4} \geqslant 0$, hence, the roots of the equation are real. $D=0$ when $q=n, p=0-$ conditions for the equality of the roots of the equation. | q=n,p=0 | Algebra | proof | Yes | Yes | olympiads | false | 42,426 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.