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|---|---|---|---|---|---|---|---|---|---|
104. A number written in the decimal system is divisible by 2 or 5 because its last digit is divisible by 2 or 5; for which divisors is this divisibility rule applicable in the duodecimal system? | 104. For 2, 3, and 4. | 2,3,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,940 |
106. Given two segments $a$ and $b$; take $b$ as the unit of length, and let the measure of the length of segment $a$ be the number $m$; what will be the measure of the length of segment $b$ if the unit of length is taken to be segment $a$? | 106. $\frac{1}{m}$. | \frac{1}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,942 |
107. What condition must be satisfied for a given integer power to be represented a) as a power of the same base? b) as a power of a different base? | 107. The exponent should not be a prime number, i.e., it should be factorable into two integer factors, both different from one. For example, $5^{6}=\left(5^{4}\right)^{3}=\left(5^{3}\right)^{2}$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,943 |
108. In what case can the root of a known whole and positive number be represented a) as the root of the root of the same number? b) as the root of another number? | 108. a) If the index of the root is not a prime number; b) always; for example, $\sqrt{3}=\sqrt[6]{27}$. | \sqrt{3}=\sqrt[6]{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,944 |
109. Does multiplication always result in the increase of the multiplicand? | 109. The product is greater or less than the multiplicand depending on the sign of the multiplicand, the sign of the multiplier, and, finally, on whether the absolute value of the multiplier is between 0 and 1, or greater than 1. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,945 |
111. What can be said about the numbers $a, b, c, d$, if they, taken in any order, always form a geometric proportion? | 111. Either all four numbers are equal, or all four are equal in absolute value, with 2 being positive and 2 being negative. Indeed: from the positions $\frac{a}{b}=\frac{c}{d}$ and $\frac{a}{b}=\frac{d}{c}$; multiplying the sides, we get: $\frac{a^{2}}{b^{2}}=1$, from which $\frac{a}{b}= \pm 1$, or $a= \pm b$; similar... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,946 |
112. What operations on integers are unconditionally executable within the integers? | 112. Addition, subtraction, multiplication, raising to a power. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,947 |
114. Which operations on rational numbers are unconditionally performable within the rational numbers? | 114. Addition, subtraction, multiplication, division (excluding division by 0, which remains absurd even in the realm of irrational and complex numbers). | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,949 |
118. Let the order of the sides of the triangle be $a>b>\boldsymbol{c}$; indicate the order of the corresponding heights $\boldsymbol{h}_{a}$, $h_{b}, h_{c}$. | 118. The doubled area of the triangle $a h_{a}=b h_{b}=c h_{c}$, and therefore from the inequality $a>b>c$ it follows that $h_{a}<h_{b}<h_{c}$. | h_{}<h_{b}<h_{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,952 |
119. What is the shortest distance on the surface of the given right parallelepiped (edges: $a>b>c$) between one of the vertices and the opposite vertex (the opposite vertex to $M$ is considered to be the vertex $N$, where the three faces parallel to the faces meeting at vertex $M$ meet)? | 119. Unfold the surface of the parallelepiped onto a plane by cutting along some edges; this can be done in three different ways; by then connecting the new positions of point $\boldsymbol{M}$ with $N$, we obtain three straight segments, the shortest of which will be the required shortest distance; the lengths of these... | \sqrt{^{2}+(b+)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,953 |
123. What number does not change when raised to a power? What number retains its absolute value when raised to any positive integer power? | 123. The only common roots of all equations of the type $x^{n}=x(n=1,2,3 \ldots)$ are 1 and 0. Any complex number of the form $\cos \alpha+i \sin \alpha$ (and therefore especially the numbers: $-1, i$ and $-i$) always retains its absolute value when raised to any power. | \cos\alpha+i\sin\alpha | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,955 |
125. What number is equal to its reciprocal? What number is equal to its reciprocal with the opposite sign? | 125. 1 and $-1$; +i and $-i$. | 1-1;+i-i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,957 |
129. Write two complex numbers, a) the sum of which or b) the product, or c) both the sum and the product are real numbers. | 129. a) $a+b i$ and $c-b i$; b) $a+b i$ and $k a+k b i$; c) $a+b i$ and $a-b i$. Show that only pairs of complex numbers of these types satisfy the given requirements. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,960 |
130. Indicate the necessary and sufficient condition for a complex number $a+b i$ to be 1) equal to its conjugate, 2) equal to the reciprocal of its conjugate. | 130. a) $b=0$, b) $a^{2}+b^{2}=1$ | )b=0,b)^{2}+b^{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,961 |
131. The necessary and sufficient condition for the product of two complex numbers $a+b i$ and $c+d i$ to be a) a real number, b) a purely imaginary number. | 131. a) $a d+b c=0, \quad$ b) $a c-b d=0$. | )+=0,\quadb)-=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,962 |
132. The necessary and sufficient condition for the quotient of two complex numbers to be a) a real number, b) a purely imaginary number. | 132. a) $a d-b c=0, \quad$ b) $a c+b d=0$. | )-=0,\quadb)+=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,963 |
140. How to express with one equality sign that among three numbers $a, b, c$ at least one is equal to zero | 140. $a b c=0$.
The above text has been translated into English, maintaining the original text's line breaks and format. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,968 |
141. How can we express with a single inequality sign that the signs of two real numbers $a$ and $b$ are the same? that they are different? | 141. $ab > 0 ; ab < 0$. | >0;<0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39,969 |
143. Any number that is not a prime number is divisible by a prime number less than the square root of the given number (or equal to the square root). | 143. Indeed, let $N = ab$, where $a < \sqrt{N}$ and $b > \sqrt{N}$. Or $a$ is a prime number, or it is a product of prime numbers, each of which is $< \sqrt{N}$. If $N$ is a perfect square of a prime number, then we have the case mentioned in parentheses. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,971 |
144. What checks need to be performed to ensure that the number 397 is a prime number or not? Are these checks sufficient? | 144. The square root of the number 397 is less than 20, and therefore (see question 62) it is sufficient (and necessary) to check whether it is divisible by the initial numbers less than 20, i.e., by $2,3,5$, $7,11,13,17,19$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,972 |
146. Quickly write a twelve-digit number that, at first glance, could be decided not to be a perfect square, and it also cannot be immediately determined whether it is a perfect cube. | 146. It is sufficient for the last digit of the written number (the preceding digits can be arbitrary) to be one of the following four: $2,3,7,8$, because the square of an integer cannot end in any of these digits. However, at first glance, it cannot be determined whether the given number is a perfect cube, since the l... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,973 |
147. Quickly write a twelve-digit number that is not a perfect cube. | 147. It is sufficient for the last three digits not to be a perfect cube of a single-digit number; it is also sufficient for the number to end in a number of zeros not divisible by 3. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,974 |
149. The last digits of the squares of a series of natural numbers will be $0,1,4,9,6,5,6,9,4,1,0,1,4,9, \ldots$ and so on. Justify the periodicity of this sequence and the symmetry of the period. | 149. 1) The periodicity is explained by the fact that the last digit of the square of an integer is determined only by the last digit of the base, and the last digits of subsequent tens of natural numbers are obviously periodic; 2) the symmetry of the period is explained by the fact that $6=10-4,7=10-3, \quad 8=10-2, \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,975 |
152. The cooperative purchased various goods from the trust after the prices, marked in the old price list, were increased by $10 \%$. However, the cooperative receives a $10 \%$ discount from the trust. For the entire purchase, the cooperative paid $N$ rubles in cash. By what number is $N$ divisible without a remainde... | 152. To 99. | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,977 |
155. The root of an integer power of a given positive integer is not an integer; can it be a fractional number? | 155. No. Suppose that $\sqrt[n]{\boldsymbol{A}}=\frac{p}{q}$, where $A$ is an integer, $\frac{p}{q}$ is an irreducible fraction, $(q>1)$. Then $A=\frac{p^{n}}{q^{n}}$. But if $\frac{p}{q}$ is an irreducible fraction, the same can be said about $\frac{p^{n}}{q^{n}}$, and therefore $\frac{p^{n}}{q^{n}}$ cannot equal the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,978 |
156. Is it possible that when taking the square root of two different integers, the mantissas (the sequence of decimal digits) are the same? | 156. Not at all, except in the case when both numbers are perfect squares (i.e., when the mantissas consist of zeros only). Indeed: if the mantissas of the roots $\sqrt{m}$ and $\sqrt{n}$ are identical, then $\sqrt{m}-\sqrt{n}=a$, where $a$ is an integer; or $\sqrt{m}=$ $=\sqrt{n}+a$, from which, squaring: $m=n+a^{2}+2... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,979 |
157. From two positive integers, square roots are extracted; is it possible that all decimal digits of one complement the corresponding digits of the other to 9? | 157. No. Suppose that all pairs of corresponding digits of the mantissas of the roots $\sqrt{m}$ and $\sqrt{n}$ complement each other to 9, then $\sqrt{m}+\sqrt{n}=a$, where $a$ is an integer (indeed, the mantissas of $\sqrt{\bar{m}}$ and $\sqrt{n}$ contain an infinite number of digits, and therefore if these digits co... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,980 |
158. How is it known that no matter how many decimal places of $\sqrt{2}$ we calculate, we will never discover a period? | 158. If there were a period, then by converting a periodic decimal fraction into a simple one, we would obtain a rational number (fraction), which is impossible (cf. question 155). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,981 |
159. Explain the meaning of the saying: « $\sqrt{2}$ cannot be calculated exactly, it can only be calculated with some approximation». | 159. There is neither an integer nor a fractional number whose square would be equal to the number 2; but one can always find a fractional number (certainly an improper fraction) such that the difference between the square of this number and the number 2 is less than any arbitrarily small number given in advance (for e... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,982 |
160. Write two irrational binomials, a) the sum of which, b) the product of which, c) both the sum and the product are rational numbers. | 160. For example, a) $3+\sqrt{5}$ and $7-\sqrt{5}$; b) $\sqrt{11}+\sqrt{7}$ and $2 \sqrt{11}-2 \sqrt{7} ;$ c) $2+\sqrt{3}$ and $2-\sqrt{3}$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,983 | |
161. Does the liberation of fractions from irrationality in the denominator have any other purpose besides providing an excuse for exercises?
Does the removal of irrationality from the denominator of fractions serve any other purpose besides offering a pretext for exercises? | 161. If it is required to calculate the approximate value of an expression containing irrational numbers (radicals) in the denominator, then division by a multi-digit decimal is quite inconvenient; it is also not easy to determine the degree of approximation (error); for example, instead of
$$
\frac{1}{\sqrt{2}}=\frac... | \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,984 |
162. If the number under the square root contains known factors squared, they can be taken out from under the root sign. Is it ever useful, on the contrary, to introduce numbers standing before it under the root sign? | 162. It happens; note: $a \sqrt{\frac{\bar{b}}{a}}=\sqrt{a^{2} \frac{b}{a}}=\sqrt{a \bar{b}}$; another example: it is required to calculate the expression $25 \sqrt{3}$ with an approximation to 0.01. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,985 |
163. What does "action inverse relative to the given one" mean? (Cf. Question 76.) Show that if the direct action does not have the commutative property (Question 73), then there can be t w o actions inverse relative to the given one. Example.
Compare the original text's line breaks and format, and output the translat... | 163. Obviously, the equality $\frac{a}{0}=\infty$, which reads: “$a$ divided by zero equals infinity,” cannot be taken literally. It is merely a brief and, for certain considerations, convenient formulation of the following rather complex mathematical fact: if in the fraction $\frac{a}{b}$ the numerator $a (a \neq 0)$ ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,986 |
165. If the dividend and divisor are simultaneously equal to zero, the quotient, according to the above definition of division (see the previous question), is equal to any number $m$, because $m \cdot 0 = 0$. Despite this, it sometimes happens that when the dividend and divisor simultaneously become equal to zero, the ... | 165. Let the algebraic fraction be $\frac{x^{2}-3 x+2}{x^{2}-4 x+3}$. For $x=1$, both the numerator and the denominator become zero, and the expression takes the indeterminate form $\frac{0}{0}$, i.e., it has no definite value. Let's try to circumvent this difficulty in the following way: the numerator and the denomina... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 39,987 |
169. In the ascending chain of concepts from the particular to the more general, insert intermediate links between the concept of "3" and the concept of "number".
Insert intermediate links between the concept of "3" and the concept of "number" in the ascending chain of concepts from the particular to the more general. | 169. 3 - the initial number - a positive integer (natural number) - an integer - a rational number - a real number - a number. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,989 |
170. In what sense are real numbers a special case of complex numbers? | 170. Each real number can be considered as a complex number, the coefficient of the imaginary part of which is equal to zero. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,990 |
172. Based on the rule for converting a repeating decimal into a simple fraction, we have:
$$
0.999 \ldots=\frac{9}{9}=1
$$
on the other hand, every decimal fraction whose integer part is zero is less than one. Explain this apparent contradiction. | 172. Every decimal finite fraction, the integer part of which is equal to zero, is less than 1. But the infinite fraction $0.999 \ldots$ is the limit of the sequence of finite fractions 0.9; 0.99; 0.999 \ldots$ and the limit of a sequence of numbers less than one can be equal to one. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,991 |
173. On the $O X$ axis, two points $\boldsymbol{A}$ and $\boldsymbol{B}$ are given, the abscissas of which are expressed by two consecutive natural numbers $\boldsymbol{a}$ and $a+1$. Between these numbers, we insert 9 arithmetic means and mark the corresponding points; between each pair of adjacent new numbers, we ins... | 173. No; for by inserting intermediate points, we will only reach points corresponding to numbers with a finite number of non-zero decimal places. For example, we will never reach the point $a+\frac{1}{3}$, since $0.333 \ldots$ (certainly, we will not reach points corresponding to irrational numbers either). | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,992 |
174. In arithmetic, irrational numbers can be calculated, in contrast to rational numbers, only with a known approximation; can certain special incommensurable segments [for example, $\sqrt{2}$ cm, $(\sqrt{3}+\sqrt{5})$ cm] be constructed in geometry with the same accuracy as commensurable ones? Examples. | 174. Yes. Construction using the Pythagorean theorem $\left(\sqrt{2}=\sqrt{1^{2}+1^{2}}, \quad \sqrt{3}=\sqrt{2^{2}-1^{2}}, \quad \sqrt{5}=\sqrt{\left.2^{2}+1^{2}\right)}\right.$ or geometric mean $(\sqrt{2}=\sqrt{2 \cdot 1}, \quad \sqrt{3}=\sqrt{2+1.5})$ and others. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,993 |
176. In direct practical measurements, we always obtain rational numbers (either integers, or in the form of common fractions, or in the form of decimal fractions with a greater or lesser number of decimal places). If this is the case, does calculation with irrational numbers have any practical value? | 176. Yes. For without introducing the concept of irrational numbers, it would not have been possible to formulate concisely and clearly many practical rules of physics and applied geometry, for example, the rule for calculating the diagonal of a square when the side is given ($a \sqrt{2}$), or the definition of the per... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,994 |
177. Can the sum, difference, product, and quotient of two irrational numbers be rational numbers? Provide corresponding examples and a general rule. | 177. $$
\left(7+\sqrt{5}\right)-\left(3+\sqrt{5}\right) ; \quad \sqrt{2} \cdot \sqrt{8} ; \quad \frac{5+\sqrt{3}}{2.5+\sqrt{0.75}}
$$ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,995 | |
178. From every integer, it is possible to take roots of all possible integer powers, which will generally be irrational numbers. With these roots, taken in any quantity, algebraic operations can be performed (addition, subtraction, multiplication, division, raising to rational powers). Will we thus exhaust all possibl... | 178. In higher algebra, it is proven that there exist real algebraic numbers that cannot be represented using a finite number of roots of rational numbers (an algebraic number is a number that satisfies an algebraic equation with integer coefficients). In addition to algebraic irrational numbers, there are transcendent... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,996 |
179. What do the statements mean: 1) "e is an irrational number"? 2) " $\pi$ is a transcendental number"? | 179. 2) This means that $\pi$ does not satisfy any algebraic equation with rational coefficients, in other words: it is impossible for the equality
$$
A_{0} \pi^{n}+A_{1} \pi^{n-1}+A_{2} \pi^{n-2}+\ldots+A_{n-1} \pi+A_{n}=0
$$
where $A_{0}, A_{1}, A_{2} \ldots A_{n-1}, A_{n}$ are integers. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,997 |
181. Show that in the decimal logarithm system, the logarithms of positive integers can be either whole numbers or irrational numbers, - never rational fractions. | 181. Let the number $N$ not be an integer power of the base 10. Suppose that $N=10^{\frac{m}{n}}$, where $\frac{m}{n}$ is a rational fraction. Raising to the power $n$, we get $N^{n}=10^{m}$, but $N$ by assumption is not a power of 10, and therefore $N^{n}$ cannot be a power of 10. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,998 |
182. What can be said about two numbers, the logarithms of which in the decimal logarithm system have a) the same mantissas, b) the same characteristics? | 182. a) The ratio of both numbers is an integer (positive or negative) power of the number 10 (base of logarithms), b) the ratio of the larger number to the smaller one is less than the number 10. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,999 |
184. Given $\log 2$ and $\log 3$ with five decimal places. For which integers from 1 to 100 can the logarithms be calculated without tables and with how many accurate decimal places? | 184. For all numbers of the form $2^{m} \cdot 3^{n} \cdot 5^{p}$, less than 100 ( $m, n$, $p$ can be zero). The first four decimal places are accurate. | The\first\four\decimal\places\\accurate | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,001 |
187. On which rule of exponent theory is the theorem of the logarithm of a product based? | 187. $a^{m} \cdot a^{n}=a^{m+n}$. | ^{}\cdot^{n}=^{+n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,003 |
189. Can a) a negative number, b) a positive fraction be taken as the base of a logarithmic system? | 189. a) No. b) Yes. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,005 |
191. If two real numbers $a$ and $b$ (greater than 0) are given, then the logarithm of $a$ to the base $b$ equals the reciprocal of the logarithm of $b$ to the base $a$. | 191. By the definition of logarithm:
$$
a=b^{\log _{b} a}
$$
Taking the logarithm of this identity with respect to base $a$, we get:
$$
1=\log _{b} a \cdot \log _{a} b
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,007 |
192. What is the base of the logarithm system in which a known number $a$ is equal to its logarithm? | 192. Let the desired base be $x$; then it should be
$$
\log _{x} a=a \quad \text { or } \quad x^{a}=a,
$$
from which
$$
x=\sqrt[a]{a}
$$ | \sqrt[]{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,008 |
194. Why is the concept of a power with fractional, irrational, and negative exponents necessary for the theory of logarithms? What numbers would have logarithms if we only knew about integer exponents? Which numbers would not have logarithms if we only knew about positive exponents? | 194. If we were only familiar with integer exponents, then only the integer powers of the base would have logarithms. Not introducing negative exponents would deprive logarithms of all real positive numbers less than one. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,009 |
195. What conveniences do calculations using logarithms offer? | 195. The convenience of computations using logarithms is based on the fact that the operations required to be performed on numbers correspond to simpler operations on their logarithms. Indeed, multiplication corresponds to addition, division to subtraction, exponentiation to multiplication, and root extraction to divis... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,010 |
199. The identity $\left(a+\frac{1}{2}\right)^{2}=a(a+1)+\frac{1}{4}$ allows for the quick calculation of the square of a number consisting of an integer part and $\frac{1}{2}$ (for example, $7 \frac{1}{2}, 29 \frac{1}{2}$ ). Formulate the rule derived from this formula in words and apply it to several examples. | 199. Rule: multiply the integer part of the given number by the next natural number and append $\frac{1}{4}$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,012 |
200. Modify the formula provided in the previous question so that it allows for the quick calculation of the square of an integer ending in 5. (Use the fact $\frac{1}{2}=0.5$ ). | 200.
$$
\quad(10 a+5)^{2}=100 a(a+1)+25
$$
Rule: multiply the part of the number representing the tens by the next natural number and append 25. | (10+5)^{2}=100(+1)+25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,013 |
203. Why are the following approximate equalities valid:
$$
\begin{aligned}
& 1.0023 \times 1.0057 = 1.008 \\
& 1.05 \times 9.97 = 10.02 \\
& 1.03 \times 0.99 \times 0.98 = 1 ?
\end{aligned}
$$
Provide the general formulas underlying this approximate calculation and justify them. What is the approximation in each of ... | 203. $(1+\alpha)(1+\beta)=1+(\alpha+\beta)+\alpha \beta$ (if $\alpha$ and $\beta$ are very small, the product $\alpha \beta$ can be neglected),
$$
\begin{aligned}
& (1+\alpha)(1-\beta)=1+(\alpha-\beta)-\alpha \beta \\
& (1+\alpha)(1-\beta)(1-\gamma)=1+\alpha-\beta-\gamma-\alpha \gamma+\beta \gamma+\alpha \beta \gamma
... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,014 |
205. Why is $\sqrt{1.02} \approx 1.01 ; \sqrt{0.98} \approx 0.99$ ? | 205. 206) $\sqrt{1+x}=1+\frac{x}{2}$ (approximately). Indeed: $1+\frac{x}{2}-\sqrt{1+x}=\frac{\left(1+\frac{x}{2}-\sqrt{1+x}\right)\left(1+\frac{x}{2}+\sqrt{1+x}\right)}{1+\frac{x}{2}+\sqrt{1+x}}=$ $=\frac{1+x+\frac{x^{2}}{4}-1-x}{1+\frac{x}{2}+\sqrt{1+x}}=\frac{x^{2}}{4\left(1+\frac{x}{2}+\sqrt{1+x}\right)} ;$ if $x \... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,016 |
207. In the class, there are about 40 students; in the hall, there are approximately one and a half thousand people; the publication costs about a hundred rubles; in a printed sheet, there are approximately 40,000 letters. What does the expression "approximately" mean in each of these cases? Explain the concept of abso... | 207. An estimate of 40 students in a class has value only if the assumed error does not exceed a known number, for example, $5; 5$ in this case is called the absolute error (or uncertainty), $\frac{5}{40}=0.125$ is the relative error. "Approximately 1,500 people in the hall": absolute error, for example, 100, relative ... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 40,017 |
208. How does the accuracy of an approximate number change if it is: a) multiplied by 10 (or 35)? b) divided by 10 (or 35)? | 208. a) The accuracy decreases by one decimal digit (by two decimal digits when multiplying by 35); b) the accuracy increases by one decimal digit (by one decimal digit when dividing by 35). | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,018 |
209. The sum of several approximate numbers should be determined with an accuracy of 0.01; with what accuracy should each addend be taken? | 209. If there are no more than five addends, then with an accuracy of 0.001. If the number of addends is greater, then for each, even incomplete, group of five, the number of accurate decimal places in the addends should be increased by one. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,019 |
210. The absolute error of the sum of several approximate numbers will be the greatest if all the addends are taken with an excess or all with a deficiency. The absolute error of the difference of two approximate numbers will be the least if the minuend and subtrahend are both taken with an excess or both with a defici... | 210.
$$
\begin{aligned}
& (a \pm \alpha)+(b \pm \beta)+(c \pm \gamma)+\ldots= \\
& =(a+b+c) \pm \alpha \pm \beta \pm \gamma \pm \ldots
\end{aligned}
$$
The greatest error is $+(\alpha+\beta+\gamma+\ldots)$ or $-(\alpha+\beta+\gamma+\ldots)$; In the case of elements, math. If there are both + and - signs in the paren... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,020 |
211. To obtain the product of two numbers 4.27 and 11.56 with an accuracy of 1, both factors are taken with one decimal place; what approximate values of the factors provide the smallest error in the product? | 211. Generally, one factor should be taken with a deficiency, the other with an excess; here it is most appropriate to take $4.3 \times 11.5$ (why not the other way around?). | 4.3\times11.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,021 |
212. How can one calculate, without tables, using the binomial theorem, with a certain approximation, the final capital formed from an initial capital $K$, lent at a compound interest rate of $p \%$ annually for $n$ years? | 212. $\boldsymbol{K}_{n}=\boldsymbol{K}\left(1+\frac{p}{100}\right)^{n}$. Applying the binomial formula of Newton, we have
$$
\begin{aligned}
K_{n} & =K\left[1+n \cdot \frac{p}{100}+\frac{n(n-1)}{1 \cdot 2}\left(\frac{p}{100}\right)^{2}+\right. \\
& \left.+\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}\left(\frac{p}{100}\right... | K_{n}=K[1+n\cdot\frac{p}{100}+\frac{n(n-1)}{1\cdot2}(\frac{p}{100})^{2}+\frac{n(n-1)(n-2)}{1\cdot2\cdot3}(\frac{p}{100})^{3}+\cdots] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,022 |
216. Why is it known in advance that the coefficients of the expansion $(a+b)^{n}$ are symmetric, i.e., that the coefficients of terms equally distant from the ends are the same? What is known in advance about the coefficients of the expansion $(a-b)^{n}?$ | 216. Because $(a+b)^{n}=(b+a)^{n}$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,024 |
217. How many terms will there be in the expansion of $(x+a)^{n}$? Can some of these terms disappear? | 217. $n+1$; no. | n+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,025 |
218. How many terms will there be in the expansion of 1) $(a+b)^{n}+(a-b)^{n}$, 2) $(a+b)^{n}-(a-b)^{n}$? Which terms disappear, and which are doubled? | 218. Let $n$ be odd: $n=2 k+1$; in the expanded $(a \pm b)^{n}$ the number of terms will be $2 k+2$, i.e., an even number. In case 1), the terms in which $b$ appears in an odd power are eliminated, the remaining $k+1$ terms are doubled. If $n$ is even, $n=2 k$, then out of the $2 k+1$ terms of the expanded binomial $(a... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,026 |
222. If $\ a^{p} x^{q}$ is one of the terms in the expansion of $(a+x)^{n}$, what condition do $p$ and $q$ satisfy? What is the coefficient $M$? | 222. $p+q=n ; M=C_{p+q}^{p}=C_{p+q}^{q}=\binom{n}{p}=\frac{n!}{p!(n-p)!}=$
$$
=\frac{1 \cdot 2 \cdot 3 \ldots n}{1 \cdot 2 \cdot 3 \ldots p \cdot 1 \cdot 2 \cdot 3 \ldots(n-p)}=\frac{n(n-1)(n-2) \ldots(n-p+1)}{1 \cdot 2 \cdot 3 \ldots p}
$$ | \frac{n(n-1)(n-2)\ldots(n-p+1)}{1\cdot2\cdot3\ldotsp} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,029 |
223. What dependence must exist between the numbers $m, n$ and $k$ so that the expansion of $\left(x^{m}+\frac{1}{x^{n}}\right)^{k}$ contains a term free of $x$? | 223. The number $m k$ must be divisible by $m+n$. For example, $k=15, m=2, n=3$. | kmustbedivisible+n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,030 |
224. How to derive the formulas for $(a-b)^{2},(a-b)^{3}$ from the formulas $(a+b)^{2},(a+b)^{3} ?$ | 224. Use the identity $a-b=a+(-b)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,031 |
225. How to determine the greatest coefficient in the expansion of $(a+b)^{n}$? | 225. If $n$ is an even number, $n=2k$, then the greatest coefficient will be $\underset{2k}{C^{k}}=\underset{n}{C^{\frac{n}{2}}}$. If $n$ is an odd number, $n=2k+1$, then there exist two equal greatest coefficients, namely $C_{2k+1}^{k}=C_{2k+1}^{k+1}$ or $C_{n}^{\frac{n-1}{2}}$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,032 |
226. Calculate the sum of the coefficients 1) of the expansion $(a+b)^{n}$, 2) of the expansion $\left.(a-b)^{n}, 3\right)$ the sum of the coefficients of the terms occupying even positions in the expansion $(a+b)^{n}$. | 226. 1) Let $a=b=1$. The sum of the binomial coefficients will be $(1+1)^{n}=2^{n}$.
2) Let $a=b=1$. The sum of the coefficients $(1-1)^{n}=0$.
3) The sum of the binomial coefficients in even positions, denote it as $S_{p}$, the sum of the coefficients in odd positions as $S_{i}$; from 1) and 2) it follows that $S_{p}+... | 2^{n-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,033 |
227. Prove that any odd power of the number 7, increased by 1, is divisible by $8\left(7^{3}=343 ; 343+1\right.$ is divisible by 8$)$. What can be said about any even power of 7? | 227. When expanding the binomial $(8-1)^{n}$, all terms up to and including the second to last are divisible by 8, the last term equals -1 (when $n$ is odd). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,034 |
228. If a number is not divisible by 7, then its cube, increased or decreased by 1, is divisible by 7. | 228. A number not divisible by 7 has one of the forms $7 n \pm 1$, $7 n \pm 2, 7 n \pm 3$; its cube, calculated by the formula $(a \pm b)^{3}$, consists of terms divisible by 7, and the last term of one of the forms $\pm 1$, or $\pm 8$, i.e., $\pm 7 \pm 1$, or $\pm 27$, i.e., $7 \cdot 4 \pm 1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 40,035 |
229. Generalize the identities:
\[
\begin{gathered}
(a-b)(a+b)=a^{2}-b^{2} \\
\left(a^{3}-a^{2} b+a b^{2}-b^{3}\right)(a+b)=a^{4}-b^{4}
\end{gathered}
\] | 229. If $n$ is an even number: $n=2 k$, then the following general formulas are valid:
$$
\begin{array}{r}
a^{2 k}-b^{2 k}=(a-b)\left(a^{2 k-1}+a^{2 k-2} b+a^{2 k-3} b^{2}+\cdots\right. \\
\left.\ldots+a^{2} b^{2 k-3}+a b^{2 k-2}+b^{2 k-1}\right) \cdot \cdots \\
a^{2 k}-b^{2 k}=(a+b)\left(a^{2 k-1}-a^{2 k-2} b+a^{2 k-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,036 |
231. Generalize the identity:
$$
\left(a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}\right)(a+b)=a^{5}+b^{5}
$$ | 231. General formula for $n$ odd:
$$
\begin{gathered}
a^{2 k+1}+b^{2 k+1}=(a+b)\left(a^{2 k}-a^{2 k-1} b+a^{2 k-2} b^{2} \ldots\right. \\
\left.\ldots+a^{2} b^{2 k-2}-a b^{2 k-1}+b^{2 k}\right) . . . . . .(1)
\end{gathered}
$$
The proof is similar to the answer to the previous question.
Substituting $-b$ for $b$ in ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,037 |
232. If the equality $a x+b=c x+d$ is to be an identity, it is necessary (and sufficient) that $a=c$ and $b=d$. Why | 232. In the identity $a x+b=c x+d$, valid for any value of $x$, let us first set $x=0$; we get
$$
b=d . . . . . . . . . . . . . .(1)
$$
then let $x=1$, which gives
$$
a+b=c+d
$$
subtracting (1) from (2), we get $a=c$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,038 |
234. Determine the necessary and sufficient condition for the value of the fraction $\frac{a x^{2}+b x+c}{a_{1} x^{2}+b_{1} x+c_{1}}$ to remain unchanged for arbitrary values of the letter $x$. Generalize. | 234. $\frac{a}{a_{1}}=\frac{b}{b_{1}}=\frac{c}{c_{1}}$ (cf. prob. 232-233). | \frac{}{a_{1}}=\frac{b}{b_{1}}=\frac{}{c_{1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,040 |
236. Can the quadratic trinomial $a x^{2}+b x+c$ be equal to the binomial $\boldsymbol{A} \boldsymbol{x}+\boldsymbol{B}$ for any value of $x$? | 236. No. In $a x^{2}+b x+c=\boldsymbol{A} x+\boldsymbol{B}$, let $x=0$, then
$$
c=\boldsymbol{B}
$$
Therefore, $a x^{2}+b x=A x$; let $x=1$, we get
$$
a+b=A
$$
let $x=2$, which gives $4 \boldsymbol{a}+2 b=2 \boldsymbol{A}$
or
$$
2 a+b=A
$$
But equations (2) and (3) are incompatible, since $a \neq 0$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,042 |
237. I am not sure if I remember the divisibility theorem for $a^{n} \pm b^{n}$ by $a \pm b$ depending on whether $n$ is even or odd, and whether the dividend or divisor is a sum or a difference. What simple checks can I use to quickly dispel my doubts? | 237. If the exponents $n$ are even, we perform a check for $n=2: a^{2}+b^{2}$ does not divide by $a+b$, nor by $a-b$; generalizing: $a^{2 k}+b^{2 k}$, i.e., the sum of identical even powers of two numbers does not divide by either the sum or the difference of the first powers of these numbers. However, $a^{2}-b^{2}=(a-... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 40,043 |
238. Let $P_{n}(x)$ be a rational function of degree $n$ with respect to $x$:
$$
P_{n}(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}\left(a_{0} \neq 0\right)
$$
When dividing $\boldsymbol{P}_{n}(x)$ by the binomial $x-\alpha$, the quotient is $Q_{n-1}(x)$, which is an entire function of degree $n-1$ (why?) and,... | 238. Substitute into equation (1) the value $\alpha$ for $x$; we get:
$$
P_{n}(\alpha)=(\alpha-\alpha) Q_{n-1} \alpha+R
$$
but by assumption $P_{n} \alpha=0 ; \alpha-\alpha$ is also zero, so
$$
0=0 \cdot Q_{n-1}(\alpha)+R
$$
or
$$
\boldsymbol{R}=0
$$
Conversely, if $\boldsymbol{R}=0$, then equation (1) becomes
$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,044 |
239. What property do the coefficients of an equation of degree $n$ have if it has a root of 1? The same question for an equation, one of whose roots is -1? | 239. In the first case, the sum of all coefficients is zero; in the second, the sum of the coefficients of the even powers of the unknown is equal to the sum of the coefficients of the odd powers. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,045 |
241. Show that if an equation of degree $n$ with integer coefficients has integer roots, then they are divisors of the constant term of the equation. | 241. Let $\alpha$ be a root of the equation
$$
a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}=0
$$
then
$$
a_{0} \alpha^{n}+a_{1} \alpha^{n-1}+\ldots+a_{n-1} \alpha=-a_{n}
$$
or
from which
$$
\begin{gathered}
a\left(a_{0} \alpha^{n-1}+a_{1} \alpha^{n-2}+\cdots+a_{n-1}\right)=-a_{n} \\
a_{0} \alpha^{n-1}+a_{1} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,046 |
242. When solving a certain equation with one unknown, I obtained two roots. Substituting these roots into the given equation showed that one of the roots satisfies the equation, while the other does not. And yet, I did not make any mistakes in my calculations. What is the matter? | 242. When solving equations, it is often necessary to multiply both sides of the equation by expressions containing the unknown.
Such actions can introduce new ("alien") roots into the equations that do not satisfy the original equation. By substituting the obtained roots, one can determine which of them are indeed ro... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,047 |
247. Show that the equation $\sqrt{x+1}-\sqrt{x-1}=0$ cannot have a root. Show the same for the equation $\sqrt{x}-\sqrt{x-\sqrt{1-x}}=1$ | 247. It is easy to see that the first equation cannot have roots. The second equation can be freed from the square root signs by several squarings. But the solutions 0 and $\frac{16}{25}$ obtained subsequently do not satisfy the original equation, since, as usual, we will take the square root of a positive number with ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,052 |
248. By multiplying the equation by $x$, we introduce an extraneous root $x=0$. Conversely, by dividing the equation by $x$, we lose the root $x=0$. Nevertheless, we calmly divide the reciprocal equation
$x^{4}+x^{3}+x^{2}+x+1=0$ by $x^{2}$ and do not lose any root. Explain this apparent contradiction. | 248. By dividing the equation by $x^{2}$, we could lose the root $x=0$, if the given equation allowed it, but the equation $x^{4}+x^{3}+x^{2}+x+1=0$ does not have such a root. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,053 |
251. Are the equations equivalent:
$$
\begin{array}{r}
\frac{-x-2}{x-3}=\frac{x+1}{x-3} \cdots \cdots \\
-x-2=x+1 . \cdots \\
(-x-2)(x-3)=(x+1)(x-3)
\end{array}
$$ | 251. Equations (1) and (2) are equivalent, whereas equations (2) and (3) [and therefore (1) and (3)] are not equivalent, since equation (3) has the root $\boldsymbol{x}=3$, which does not satisfy equation (2). | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,055 |
253. Given 2 equivalent systems of 2 linear equations with two unknowns $[(1)(2)]$ and $[(3)(4)]$. What can be said about the systems of equations $[(1)(3)],[(1)(4)]$, $[(2)(3)],[(2)(4)]$? | 253. The equivalence of systems $[(1)(2)]$ and $[(3)(4)]$ proves that the points of intersection of two pairs of lines, representing these equations geometrically, coincide ${ }^{1}$ ), i.e., that all four lines represented by equations (1), (2), (3), (4) pass through one point. From this it directly follows that the p... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,056 |
255. Prove that if a number $\alpha$ is a root of a biquadratic equation, then the equation also has the root - $\alpha$. Generalize. | 255. A biquadratic equation includes only terms with even powers of the unknown, and therefore, if the number $\alpha$ satisfies the equation, then the number -$\alpha$ also satisfies it. This is true for equations of any degree that contain only even powers of the unknown.[^0] | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,057 |
256. How can a cubic equation be quickly solved if one of its roots is known? Provide numerical examples. | 256. If given a cubic equation $x^{3}+a x^{2}+b x+c=0$, and if $\alpha$ is a root of this equation, then, as is known, the left side of the equation is divisible by $x-\alpha$. If one of the roots $\alpha$ is known, then by dividing the left side of the equation by $x-\alpha$, we obtain a quadratic equation, from which... | x_{2}=2,x_{3}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,058 |
257. If it is known that the equation
$$
12 x^{5}-8 x^{4}-45 x^{3}+45 x^{2}+8 x-12=0
$$
admits the roots $+1 ; 1.5 ; -2$, write, without any calculations, the two missing roots. | 257. It is easy to notice that if $\alpha$ is a root of the given equation, then $\frac{1}{\alpha}$ will also be a root, and therefore the required roots will be $\frac{1}{1.5}=\frac{10}{15}=\frac{2}{3}$ and $\frac{1}{-2}=-\frac{1}{2}$. | \frac{2}{3},-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,059 |
258. Why does every equation of odd degree with real coefficients have an odd number of real roots, at least one? | 258. The left side of the equation is a polynomial of degree $n$ in $x$ with real coefficients. If the equation has roots $\alpha_{1}, \alpha_{2}, \alpha_{3}, \ldots \alpha_{n}$, then it can be written in the form $\left(x-\alpha_{1}\right)\left(x-\alpha_{2}\right)\left(x-\alpha_{3}\right) \ldots\left(x-\alpha_{n}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,060 |
260. One of the roots of the biquadratic equation is $2+\sqrt{3}$. Write this equation. | 260.
$$
\begin{gathered}
x_{1}=2+\sqrt{3}, x_{2}=2-v^{3} \\
(x_{3}=-2-\sqrt{3}, x_{4}=-2+\sqrt{3}) \\
x_{1}^{2}=5+4 \sqrt{3}, x_{2}^{2}=5-4 \sqrt{3} \\
p=-10, q=25-48=-23
\end{gathered}
$$
The equation will be $x^{4}-10 x^{2}-23=0$. | x^{4}-10x^{2}-23=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,062 |
262. The roots of the equation $x^{2}+p x+q=0$ with integer coefficients are either integers, or irrational, or complex - never real (rational) fractions. Generalize. | 262. Equation of the $n$-th degree
$$
x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}=0
$$
where $a_{1}, a_{2}, \ldots a_{n}$ are integers, does not have rational roots that are not integers. Proof: suppose that the irreducible fraction $\frac{h}{k}(k>1)$ is a root of the given equation. Then
$$
\left(\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 40,064 |
266. What can be said in advance about the roots of the equation
$$
a x^{2}+b x+a=0 ?
$$ | 266. One root is equal to the reciprocal of the other. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,066 | |
267. What relationship exists between the roots of the equations:
$$
\begin{aligned}
& a x^{2}+b x+c=0 \\
& c x^{2}+b x+a=0
\end{aligned}
$$ | 267. Let $x_{1}, x_{2}$ be the roots of the first equation, and $x_{3}, x_{4}$ the roots of the second. Clearly, $\frac{1}{x_{1}}, \frac{1}{x_{2}}$ satisfy the second equation and that it has no other roots. Thus, $x_{3},{ }_{4}=\frac{1}{x_{1}, 2}$. | x_{3},x_{4}=\frac{1}{x_{1}},\frac{1}{x_{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,067 |
268. If $x_{1}, x_{2}$ denote the roots of the equation $x^{2}+p x+q=0$, then does the equation $x^{2}-p x+q=0$ have such roots? $x^{2}+p x-q=0$? | 268. The equation $x^{2}-p x+q=0$ has roots $-x_{1},-x_{2}$; the roots of the equation $x^{2}+p x-q=0$ cannot be simply expressed in terms of $x_{1}$ and $x_{2}$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,068 |
269. Quickly write two quadratic equations with one unknown of the form $x^{2}+p x+q=0$, having one and only one common root. | 269. Think of a certain number as the common root of these equations, for example, 2, and write the equations, using arbitrary (different) coefficients of the linear terms and selecting the terms free from $\boldsymbol{x}$ accordingly so that the trinomials become zero. That the second roots are not identical follows a... | (x-2)(x+3)=0(x-2)(x-5)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,069 |
270. The equation $x^{2}-4 x+4=0$ is satisfied by the value $x=2$, just like the equation $-4 x+8=0$. Why, then, in the first case, do we say that the equation has 2 equal roots or a double root, while in the second, that it has only one root? | 270. In the first equation, the left side, i.e., the quadratic trinomial, can be factored into linear factors; we get
$$
(x-2)(x-2)=0
$$
from which
$$
x_{1}=2, x_{2}=2
$$
Note that if we slightly change one of the coefficients in the first equation, we would get two different roots; if, however, we do this in the s... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,070 |
271. Explain the precise meaning of the following statement encountered in the study of quadratic equations: "If in the equation $a x^{2}+b x+c=0$ the coefficient $a=0$, then one of the roots is $\infty$, and the other is $=-\frac{c}{b}$." | 271. Of course, this expression cannot be shouted verbatim: for, if in the equation $a x^{2}+b x+c=0$ the coefficient $a=0$, then we simply get a linear equation $b x+c=0$, which has only one root $=-\frac{c}{b}$. Our expression is merely a brief conditional formulation of the following statement: if in the equation $a... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,071 |
272. What are the similarities and differences between all parabolas representing the function $y=x^{2}+p x+q$ for different values of the parameters $p$ and $q$? | 272. All these parabolas are identical and differ only in their position relative to the coordinate axes; any two of these parabolas can be made to coincide by a single parallel translation. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,072 |
273. In what sense can it be said that the parabola $y=a x^{2}(a>0)$ is a flattened or elongated parabola $y=x^{2}$? What does the parabola turn into when $a$ decreases without bound? | 273. The parabola $y=a x^{2}$ is obtained from the parabola $y=x^{2}$ by multiplying all ordinates $y$ by the same number $a$; if $a<1$, the parabola is compressed, and if $a>1$, it is elongated. When $a$ approaches zero, the parabola approaches the x-axis indefinitely. | When\\approaches\zero,\the\parabola\approaches\the\x-axis\indefinitely | Algebra | math-word-problem | Yes | Yes | olympiads | false | 40,073 |
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