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8.45 For what values of $m$ are the roots of the equation
$$
4 x^{2}-(3 m+1) x-m-2=0
$$
confined to the interval between -1 and 2?
Solve the inequalities (8.46 - 8.66):
$8.46\left|\frac{3 x+1}{x-3}\right|<3$. | 8.45 Let \(x_{1}\) and \(x_{2}\) be the roots of the given equation. According to the condition, \(-10\) and \(f(2)>0\). Thus, we obtain the system of inequalities
Fig. 8.15
\[
\left\{\beg... | -\frac{3}{2}<<\frac{12}{7} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,841 |
$8.52 \frac{4}{\sqrt{2 - x}}-\sqrt{2-x}<2$. | 8.52 Assuming $y=\sqrt{2-x}$, we obtain the inequality
$\frac{4}{y}-y>0$ and, therefore, we can multiply both sides of inequality (1) by $y$. Consequently,
$4-y^{2}>0$.
From this, we find $y<\sqrt{5}-1$. However, $y>0$. Thus, $y>\sqrt{5}-1$ and it remains to solve the inequality $\sqrt{2-x}>\sqrt{5}-1$. Since $\sqrt... | x<2\sqrt{5}-4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,844 |
$8.55 \quad \log _{\frac{1}{3}} \log _{\frac{1}{2}} \frac{x+4}{2 x-3}<0$.
$8.55 \quad \log _{\frac{1}{3}} \log _{\frac{1}{2}} \frac{x+4}{2 x-3}<0$. | 8.55 Replacing 0 with $\log _{\frac{1}{3}} 1$ and considering that the base of the first logarithm satisfies the inequality $0<\log _{\frac{1}{3}} 1<1$ (in this case, the condition $\log _{\frac{1}{2}} \frac{x+4}{2 x-3}>0$ is automatically satisfied). Further, since $1=\log _{\frac{1}{2}} \frac{1}{2}$ and $0<\log _{\fr... | (-\infty,-4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,846 |
$8.56 \frac{1}{\log _{2}(x-1)}<\frac{1}{\log _{2} \sqrt{x+1}}$. | 8.56 Let the inequalities $00
$$
From (1) and (2) it follows that for $11$, i.e., $x>2$. Then both logarithms are positive and we arrive at the inequality $\log _{2} \sqrt{x+1}3$.
Answer: $(1,2) \cup(3, \infty)$. | (1,2)\cup(3,\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,847 |
$8.57 \log _{3} \log _{4} \frac{4 x-1}{x+1}-\log _{\frac{1}{3}} \log _{\frac{1}{4}} \frac{x+1}{4 x-1}<0$. | 8.57 Noticing that
$$
\begin{aligned}
& \log _{\frac{1}{4}} \frac{x+1}{4 x-1}=\log _{4} \frac{4 x-1}{x+1} \\
& \log _{\frac{1}{3}} \log _{4} \frac{4 x-1}{x+1}=-\log _{3} \log _{4} \frac{4 x-1}{x+1}
\end{aligned}
$$
we transform the given inequality:
$$
\begin{aligned}
& \log _{3} \log _{4} \frac{4 x-1}{x+1}+\log _{3... | (\frac{2}{3},\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,848 |
$8.58 \log _{4} x+\log _{2}(\sqrt{x}-1)<\log _{2} \log _{\sqrt{5}} 5$ (find integer values of $x$). | 8.58 Given the domain of the logarithmic function and the square root, we conclude that $x>0$ and $\sqrt{x}-1>0$, hence $x>1$. We will convert all logarithms to base 2: $\log _{2} \sqrt{x}+\log _{2}(\sqrt{x}-1)0$ and solve the inequality $y^{2}-y-21$ and, therefore, $1<\sqrt{x}<2$, i.e., $1<x<4$. The integer values of ... | 2;3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,849 |
$8.60 \log _{\frac{1}{2}} \log _{2} \log _{x-1} 9>0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$8.60 \log _{\frac{1}{2}} \log _{2} \log _{x-1} 9>0$. | 8.60 The inequality makes sense if $\log _{x-1} 9>0$, from which $x-1>1$, i.e., $x>2$. From the given inequality, we have
$$
0<\frac{1}{\log _{x-1} 9}<3
$$
which is equivalent to
$$
\left\{\begin{array}{l}
\log _{x-1} 9>0 \\
\log _{x-1} 9>\frac{1}{3}
\end{array}\right.
$$
Solving these, we get $42$.
Answer: $(4,10)$... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,851 | |
$8.65 \frac{\sqrt{3}}{\cos ^{2} x}<4 \tan x$. | 8.65 Here the condition $\cos x \neq 0$, i.e., $x \neq \frac{\pi}{2}+\pi n$ must be satisfied. Let's perform the transformations:
$\frac{\sqrt{3}}{\cos ^{2} x}-\frac{4 \sin x}{\cos x}0$ for $x \neq \frac{\pi}{2}+\pi n$, it is sufficient to solve the inequality $\sqrt{3}-2 \sin 2 x\frac{\sqrt{3}}{2}$. By setting $z=2 x... | (\frac{\pi}{6}+\pin,\frac{\pi}{3}+\pin),n\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,853 |
$8.67 y=\sqrt[6]{4^{\frac{x+1}{x}}-17 \cdot 2^{\frac{1}{x}}+4}$. | 8.67 The function is defined if
$$
4^{\frac{x+1}{x}}-17 \cdot 2^{\frac{1}{x}}+4 \geq 0, \text { i.e. } 2^{2\left(1+\frac{1}{x}\right)}-17 \cdot 2^{\frac{1}{x}}+4 \geq 0
$$
Let $\quad 2^{\frac{1}{x}}=y>0 ; \quad$ then we get the inequality
$4 y^{2}-17 y+4 \geq 0$, from which $y \leq \frac{1}{4}$ or $y \geq 4$. Therefo... | [-0.5;0)\cup(0;0.5] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,855 |
9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
9.1 $\quad A_{x}^{2} C_{x}^{x-1}=48$. | 9.1 According to formula (9.1), we find
$$
A_{x}^{2}=\frac{x!}{(x-2)!}=\frac{x(x-1)(x-2)!}{(x-2)!}=x(x-1)
$$
Further, using formulas (9.6) and (9.4), we have
$$
C_{x}^{x-1}=C_{x}^{x-(x-1)}=C_{x}^{1}=\frac{x!}{1!(x-1)!}=x
$$
Therefore, the given equation will take the form
$x^{2}(x-1)=48$, or $x^{2}(x-1)=4^{2} \cdo... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,858 |
$9.2 \quad C_{x+1}^{x-2}+2 C_{x-1}^{3}=7(x-1)$. | 9.2 Let's use formula (9.4):
$$
\frac{(x+1)!}{(x-2)!3!}+\frac{2(x-1)!}{(x-4)!3!}=7(x-1)
$$
from which, after simplifying the first fraction by $(x-2)!$ and the second fraction by $(x-4)!$, we get
$$
(x+1) x(x-1)+2(x-1)(x-2)(x-3)=42(x-1)
$$
Further, considering that $x \neq 1$, we arrive at the quadratic equation
$... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,859 |
$9.5 \quad C_{n}^{k} C_{n-k}^{m-k}=C_{m}^{k} C_{n}^{m}$. | 9.5 Instruction. Show that the left and right sides of the identity to be proved can be transformed into the form $\frac{n!}{k!(m-k)!(n-m)!}$
保留源文本的换行和格式，这里直接输出翻译结果，但根据要求，实际上已经进行了翻译处理。 | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 49,862 |
9.6 In the expansion of $(1+x)^{n}$, the fourth term is equal to 0.96. Find the values of $x$ and $n$, if the sum of the binomial coefficients is 1024. | 9.6 Since the sum of the binomial coefficients is equal to $2^{n}$, and $1024=2^{10}$, then $n=10$. Using formula (9.9), we find the fourth term of the expansion
$$
T_{4}=C_{n}^{3} x^{3}=\frac{10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3} x^{3}=120 x^{3}
$$
According to the condition, $120 x^{3}=0.96$, from which $x^{3}=0.0... | 0.2;n=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,863 |
9.7 For what value of $x$ is the fourth term of the expansion $\left(\sqrt{2^{x-1}}+\sqrt[3]{2^{-x}}\right)^{m}$ 20 times greater than $m$, if the binomial coefficient of the fourth term is to the binomial coefficient of the second term as $5: 1 ?$ | 9.7 The binomial coefficients of the fourth and second terms are $C_{m}^{3}$ and $m$, respectively. Therefore, $\frac{m(m-1)(m-2)}{3!m}=5$, or $(m-1)(m-2)=30$, from which $m=7$. Then the fourth term of the expansion has the form $T_{4}=C_{7}^{3} 2^{2(x-1)} \cdot 2^{-x}$, and we arrive at the equation $C_{7}^{3} 2^{x-2}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,864 |
9.8 The commission consists of a chairman, his deputy, and five other people. In how many ways can the commission members distribute among themselves the duties of chairman and deputy? | 9.8 When choosing a chairman and his deputy from the given seven people, it matters not only how these two people are chosen from the seven candidates, but also how their positions are distributed between them (i.e., both the composition and the order of the selected elements matter).
This means we are dealing with pe... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,865 |
9.9 How many three-digit numbers can be formed from the digits 1, 2, $3,4,5,6?$ | 9.9 Each three-digit number composed of the specified digits can be considered as an arrangement with repetitions, made up of three digits taken from the given six digits. The desired number of numbers can be found using the formula (9.2):
$\widetilde{A}_{6}^{3}=6^{3}=216$.
Answer: 216 numbers. | 216 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,866 |
9.10 In how many ways can three duty officers be selected from a group of 20 people? | 9.10 The desired number of ways is equal to the number of combinations (without repetition) that can be formed from 20 elements taken three at a time. According to formula (9.4), we get
$C_{20}^{3}=\frac{20!}{17!3!}=\frac{20 \cdot 19 \cdot 18}{6}=1140$.
Answer: 1140 ways. | 1140 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,867 |
9.11 In a confectionery shop, three types of pastries are sold: Napoleons, eclairs, and sponge cakes. In how many ways can you buy 9 pastries? | 9.11 Here, we need to find the number of all possible combinations of 9 elements that can be formed from the given three elements, where these elements can repeat in each combination, and the combinations differ from each other by at least one element. This means we are looking for the number of combinations with repet... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,868 |
9.12 How many different four-digit numbers can be formed from the digits $0,1,2,3$, if each digit appears only once in the representation of the number? | 9.12 The considered four-digit number can be represented as some permutation of the digits $0,1,2, 3$, where the first digit is not zero. Since four digits can form $P_{4}=4$! permutations and of these, $P_{3}=3$! start with zero, the desired number is $4!-3!=3 \cdot 3!=1 \cdot 2 \cdot 3 \cdot 3=18$.
Answer: 18 number... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,869 |
9.13 How many different sound combinations can be taken from ten selected piano keys, if each sound combination can contain from three to ten sounds? | 9.13 For each sound combination, the Slavic keys are pressed simultaneously, so for $k$ sounds we have $C_{10}^{k}$ combinations. Thus, the desired number is $C_{10}^{3}+C_{10}^{4}+\ldots+C_{10}^{10}=C_{10}^{0}+C_{10}^{1}+C_{10}^{2}+\ldots+C_{10}^{10}-$ $-C_{10}^{0}-C_{10}^{1}-C_{10}^{2}=2^{10}-1-10-45=968$.
Answer: 9... | 968 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,870 |
9.14 The lock opens only if a specific three-digit number is entered. An attempt consists of randomly selecting three digits from the five given. The number was guessed only on the last of all attempts. How many attempts preceded the successful one? | 9.14 I method. The first digit could be chosen in five ways, the second also in five ways, i.e., for a two-digit number there were $5^{2}$ options. Thus, the total number of three-digit numbers under these conditions was $5^{3}=125$, and 124 unsuccessful attempts preceded the successful one.
II method. The total numbe... | 124 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,871 |
9.15 In how many ways can two rooks be placed on a chessboard so that one cannot capture the other? (One rook can capture another if it is on the same row or the same column of the chessboard).
## Group 5
Solve the equations (9.16-9.17): | 9.15 The first rook can be placed on any of the 64 squares. In this case, 14 squares are under threat, so for the second rook, there are any of $64-15=49$ squares left. Thus, the total number of options is $64 \cdot 49=3136$.
Answer: 3136 ways. | 3136 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,872 |
$9.17 C_{x}^{x-1}+C_{x}^{x-2}+C_{x}^{x-3}+\ldots+C_{x}^{x-9}+C_{x}^{x-10}=1023$. | 9.17 Obviously, when $x=10$ the left side of the equation is 1 less than the sum of the binomial coefficients in the expansion of the binomial $(a+b)^{10}$. This sum is 1024. Therefore, $x=10$ is a solution to the equation. There are no other solutions, as for $x>10$ the left side of the equation is greater than 1023.
... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,874 |
9.18 Solve the system of equations
$$
\left\{\begin{array}{l}
A_{y}^{x}: P_{x-1}+C_{y}^{y-x}=126 \\
P_{x+1}=720
\end{array}\right.
$$ | 9.18 From the second equation, it follows that $(x+1)!=720$; since $720=6!$, then $x=5$. Considering that $C_{y}^{y-x}=C_{y}^{x}$ (according to formula (9.6)), we rewrite the first equation as: $A_{y}^{5}: P_{4}+C_{y}^{5}=126$. Further, using formula (9.4), we have $A_{y}^{5}: P_{4}=P_{5} C_{y}^{5}: P_{4}=5 C_{y}^{5}$,... | (5;7) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,875 |
9.19 For what values of $x$ and $y$ is the equality possible
$$
C_{y}^{x}: C_{y+2}^{x}: A_{y}^{x}=1: 3: 24 ?
$$ | 9.19 Applying formula (9.4), we get
$$
\frac{y!}{x!(y-x)!}: \frac{(y+2)!}{x!(y-x+2)!}=\frac{1}{3} ; C_{y}^{x}:\left(x!C_{y}^{x}\right)=\frac{1}{24}
$$
From the second equation, we have $x!=24$, i.e., $x=4$ (since $24=1 \cdot 2 \cdot 3 \cdot 4$), and from the first equation, we find
$$
\frac{(y-x+1)(y-x+2)}{(y+1)(y+2... | (4;8) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,876 |
9.20 Find the largest binomial coefficient in the expansion of $\left(n+\frac{1}{n}\right)^{n}$, if the product of the fourth term from the beginning and the fourth term from the end is 14400. | 9.20 The fourth term from the beginning has the form $T_{4}=C_{n}^{3} n^{n-3} \frac{1}{n^{3}}$, and the fourth term from the end has the form $T_{n-2}=C_{n}^{n-3} n^{3} \frac{1}{n^{n-3}}$.
Therefore, $T_{4} T_{n-2}=\left(C_{n}^{3}\right)^{2}=14400$, from which $C_{n}^{3}=120$. Further, we have
$$
n(n-1)(n-2)=720 ; n(... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,877 |
9.21 The sum of the third from the beginning and the third from the end binomial coefficients in the expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{n}$ is 9900. How many rational terms are contained in this expansion? | 9.21 The coefficients specified in the condition are equal to $C_{n}^{2}$. We have $2 \cdot \frac{n(n-1)}{2}=9900$, or $n(n-1)=100 \cdot 99$,
from which $n=100$. Then $T_{k+1}=C_{100}^{k} 3^{\frac{100-k}{4}} 4^{\frac{k}{3}} ;$ according to the condition, $\frac{k}{3}$ and $\frac{100-k}{4}$ are integers, i.e., $k$ is d... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,878 |
9.22 Thirty people are divided into three groups (I, II, and III) with 10 people in each. How many different group compositions are possible? | 9.22 Group I of 10 people can be formed in $C_{30}^{10}$ ways. From the remaining 20 people, Group II can be formed in $C_{20}^{10}$ ways, and each time the remaining 10 people automatically form Group III. Since each selection of Group I is paired with each selection of Group II, the total number of group compositions... | \frac{30!}{(10!)^{3}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,879 |
9.23 How many four-digit numbers, composed of the digits $0,1,2,3,4,5$, contain the digit 3 (the digits in the numbers do not repeat)? | 9.23 From the six digits $0,1,2,3,4,5$, one can form $A_{6}^{4}$ different four-digit combinations (including those that start with the digit 0). Among these combinations, there are $A_{5}^{3}$ combinations with zero as the first digit, so we get $A_{6}^{4}-A_{5}^{3}$ four-digit numbers composed of the given six digits... | 204 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,880 |
9.24 How many four-digit numbers divisible by 5 can be formed from the digits $0,1,3,5,7$, if each number must not contain identical digits? | 9.24 The desired numbers end with five or zero. If the last digit is 0, then there are $A_{4}^{3}=4 \cdot 3 \cdot 2=24$ such numbers. If the last digit is 5, then the first digit cannot be 0 (the desired numbers are four-digit); in this case, there are $A_{4}^{3}-A_{3}^{2}=18$ such numbers. Therefore, 42 numbers satisf... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,881 |
9.25 On a bookshelf, 30 volumes can fit. In how many ways can they be arranged so that the first and second volumes do not stand next to each other? | 9.25 Removing the first volume, we get 29! permutations of the books. The first volume can be placed next to the second in two ways; therefore, in 2 * 29! cases, the first and second volumes are next to each other. Since there are a total of 30! permutations, in 30! - 2 * 29! of them, the first and second volumes are n... | 28\cdot29! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,882 |
9.26 In a tournament, 16 chess players are participating. Determine the number of different first-round schedules (schedules are considered different if they differ in the participants of at least one game; the color of the pieces and the board number are not taken into account). | 9.26 The participants of the first party can be chosen in $C_{16}^{2}$ ways, and the participants of the second party in $C_{14}^{2}$ ways. Since the order of selecting pairs does not matter, the participants of two parties can be chosen in $\frac{C_{16}^{2} C_{14}^{2}}{2!}$ ways, the participants of three parties in $... | 2027025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,883 |
9.27 In the metro train at the initial stop, 100 passengers entered. How many ways are there to distribute the exit of all these passengers at the next 16 stops of the train? | 9.27 The first passenger can exit at any of the 16 stops, as can the second, i.e., for two passengers there are $16^{2}$ possibilities. Therefore, for 100 passengers, there are $16^{100}$ ways.
Answer: $16^{100}$ ways. | 16^{100} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,884 |
9.28 Eight authors must write a book consisting of 16 chapters. In how many ways can the material be distributed among the authors if two people will write three chapters each, four will write two chapters each, and two will write one chapter each? | 9.28 The desired number is:
$$
\begin{aligned}
& C_{16}^{3} \cdot C_{13}^{3} \cdot C_{10}^{2} \cdot C_{8}^{2} \cdot C_{6}^{2} \cdot C_{4}^{2} \cdot C_{2}^{1}= \\
& =\frac{16!}{3!13!} \cdot \frac{13!}{3!10!} \cdot \frac{10!}{2!8!} \cdot \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot 2=\frac{16!}{2^{6... | \frac{16!}{2^{6}\cdot3^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,885 |
9.29 The letters of the Morse alphabet consist of symbols (dots and dashes). How many letters can be represented if we require that each letter contain no more than five symbols? | 9.29 Letters containing exactly $k$ characters are $2^{k}$; therefore, in total, we can represent
$2^{1}+2^{2}+2^{3}+\ldots+2^{5}=\frac{2\left(2^{5}-1\right)}{2-1}=62$ letters.
Answer: 62 letters. | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,886 |
9.30 A gardener has to plant 10 trees over the course of three days. In how many ways can he distribute the work over the days, if he will plant no less than one tree per day | 9.30 Let's separate the trees designated for planting on each day with boundaries. There are two such boundaries, and there are 9 possible places for them (the number of intervals between ten trees). We have $C_{9}^{2}=36$.
Answer: 36 ways. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 49,887 |
10.1 Given complex numbers $z_{1}=-2+5 i$ and $z_{2}=3-4 i$.
Find: a) $z_{1}+z_{2}$; b) $z_{2}-z_{1}$; c) $z_{1} z_{2}$; d) $\frac{z_{1}}{z_{2}}$. | 10.1 a) Using formula (10.3), we get
$$
z_{1}+z_{2}=(-2+5 i)+(3-4 i)=(-2+3)+(5-4) i=1+i
$$
b) Similarly, according to formula (10.4), we find
$$
z_{2}-z_{1}=(3-4 i)-(-2+5 i)=(3+2)+(-4-5) i=5-9 i
$$
c) Using formula (10.5):
$$
\begin{aligned}
& z_{1} z_{2}=(-2+5 i)(3-4 i)= \\
& =(-2 \cdot 3-5(-4))+((-2)(-4)+5 \cdot... | )1+i;b)5-9i;)14+23i;)-\frac{26}{25}+\frac{7}{25}i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,888 |
10.2 Find the number conjugate to the number $z=\frac{2 \sqrt{3}-i}{\sqrt{3}+i}+3$. | 10.2 Multiplying the numerator and the denominator of the fraction by the expression $\sqrt{3}-i$, the conjugate of the denominator, we get
$$
\begin{aligned}
& z=\frac{(2 \sqrt{3}-i)(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}+3=\frac{6-2 \sqrt{3} i-\sqrt{3} i-1}{4}+3= \\
& =\frac{5-3 \sqrt{3} i}{4}+3=\frac{17}{4}-\frac{3... | 4.25+0.75\sqrt{3}i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,889 |
10.3 Find the sum $A=i+i^{2}+i^{3}+\ldots+i^{15}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 10.3 Let's group the terms as follows:
$$
\begin{aligned}
& A=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+ \\
& +\left(i^{9}+i^{10}+i^{11}+i^{12}\right)+\left(i^{13}+i^{14}+i^{15}\right)
\end{aligned}
$$
from which
$$
A=\left(i+i^{2}+i^{3}+i^{4}\right)\left(1+i^{4}+i^{8}\right)+i^{13}+i^{14... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,890 |
## 10.5 Find:
a) the real part of the complex number $z=\frac{(2-i)^{3}}{3+4 i}$;
b) the imaginary part of the complex number $z=\left(\frac{1+i}{1-i}\right)^{11}$.
Perform the operations (10.6-10.8): | 10.5 a) We have
$$
\begin{aligned}
& \frac{(2-i)^{3}}{3+4 i}=\frac{(2-i)^{3}(3-4 i)}{(3+4 i)(3-4 i)}=\frac{\left(8-3 \cdot 4 i+3 \cdot 2 i^{2}-i^{3}\right)(3-4 i)}{25}= \\
& =\frac{(2-11 i)(3-4 i)}{25}=\frac{-38-41 i}{25}=-1.52-1.64 i
\end{aligned}
$$
i.e., $\operatorname{Re} z=-1.52$.
b) Since
$\frac{1+i}{1-i}=\fr... | -1.52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,892 |
10.9 Given $x$ and $y$ as real numbers, solve the equation
$$
(-5+2 i) x-(3-4 i) y=2-i
$$ | 10.9 By expanding the brackets in the left part of the equation and grouping the real and imaginary parts, we have $(-5 x-3 y)+(2 x+4 y) i=2-i$.
Now, using the condition of equality of two complex numbers (10.2), we obtain the system
$\left\{\begin{array}{l}5 x+3 y=-2 \\ 2 x+4 y=-1\end{array}\right.$
from which $x=-... | -\frac{5}{14},-\frac{1}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,894 |
10.10 For what real values of $x$ and $y$ are the complex numbers $z_{1}=2 x^{2}-3 i-1+y i$ and $z_{2}=y+x^{2} i-3-2 i$ conjugates of each other? | 10.10 Let's write these numbers as $z_{1}=\left(2 x^{2}-1\right)+(y-3) i$ and $z_{2}=(y-3)+\left(x^{2}-2\right) i$. For these numbers to be conjugates, the following system must hold:
$\left\{\begin{array}{l}2 x^{2}-1=y-3 \\ y-3=2-x^{2}\end{array}\right.$
From this, it follows that $2 x^{2}-1=2-x^{2}$, or $x^{2}=1$, ... | x_{1}=1,y_{1}=4;x_{2}=-1,y_{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,895 |
10.11 For what real values of $x$ and $y$ are the complex numbers $\quad z_{1}=x^{2}(1-3 i)-10 \quad$ and $\quad z_{2}=y(y-i) \quad$ opposites?
Illustrate the sets of points for which the given conditions (10.12-10.16) are satisfied:
$10.12-1 \leq \operatorname{Im} z<2$.
$10.13|z| \leq 2$.
$10.14|z+i|=1.5$. | 10.11 Instruction. Write the numbers $z_{1}$ and $z_{2}$ in algebraic form and use the condition of equality of complex numbers $z_{1}$ and $-z_{2}$.
Answer: $x_{1}=1, y_{1}=-3 ; x_{2}=-1, y_{2}=-3$. | x_{1}=1,y_{1}=-3;x_{2}=-1,y_{2}=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,896 |
$10.17 z=4-4 \sqrt{3} i$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$10.17 z=4-4 \sqrt{3} i$. | 10.17 Using formula (10.8), we find the modulus of the given number:
$$
r=\sqrt{4^{2}+(-4 \sqrt{3})^{2}}=\sqrt{16+48}=8
$$
Next, using formulas (10.9) we have
$\cos \varphi=\frac{4}{8}=\frac{1}{2}, \sin \varphi=\frac{-4 \sqrt{3}}{8}=-\frac{\sqrt{3}}{2}$,
from which it follows that $\arg z=\varphi=-\frac{\pi}{3}$.
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,897 | |
$10.20 z=(\sqrt{3}+1) i$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$10.20 z=(\sqrt{3}+1) i$. | 10.20 Point $z$ belongs to the upper part of the imaginary axis (Fig. 10.9).
We have
$r=\sqrt{3}+1, \varphi=\frac{\pi}{2}$
and, consequently,
$z=\sqrt{3}+1\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$.
$$
But $1+i=\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right)$ and, therefore, the trigonometric form of this number is
$$
z=\sqrt{2} \sin 36^{\circ}\left(\cos 45^{\c... | \sqrt{2}\sin36(\cos45+i\sin45) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,901 |
10.29 Derive the formulas expressing $\cos 5 \alpha$ in terms of $\cos \alpha$ and $\sin 5 \alpha$ in terms of $\sin \alpha$.
Perform the operations (10.30-10.32):
| 10.29 Instruction. Find the degree $(\cos \alpha+i \sin \alpha)^{5}$ in two ways: using de Moivre's formula and using the binomial formula of Newton.
Answer: $\cos 5 \alpha=16 \cos ^{5} \alpha-20 \cos ^{3} \alpha+5 \cos \alpha$, $\sin 5 \alpha=16 \sin ^{5} \alpha-20 \sin ^{3} \alpha+5 \sin \alpha$. | \cos5\alpha=16\cos^{5}\alpha-20\cos^{3}\alpha+5\cos\alpha,\sin5\alpha=16\sin^{5}\alpha-20\sin^{3}\alpha+5\sin\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,904 |
$10.32 \frac{16 i\left(\sin \frac{\pi}{6}+i \cos \frac{\pi}{6}\right)^{2}}{(-1+i \sqrt{3})^{4}}$. | 10.32 Find
\[
\begin{aligned}
& \frac{16 i\left(\sin \frac{\pi}{6}+i \cos \frac{\pi}{6}\right)^{2}}{(-1+i \sqrt{3})^{4}}=\frac{16 i\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{2}}{\left(2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)\right)^{4}}= \\
& \frac{16 i\left(\cos \frac{2 \pi}{3}+i \sin \fra... | i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,906 |
10.33 Find $\sqrt[3]{-8 i}$
## Group 5 | 10.33 For the number $z=-8 i$ we have $r=2, \varphi=-\frac{\pi}{2}$. According to formula (10.14), we get
$$
\sqrt[3]{-8 i}=2\left(\cos \frac{-\frac{\pi}{2}+2 \pi k}{3}+i \sin \frac{-\frac{\pi}{2}+2 \pi k}{3}\right), \text { where } k=0,1,2
$$
For $k=0: 2\left(\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}... | \sqrt{3}-i;2i;-\sqrt{3}-i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,907 |
10.34 Find all values of $x$ for which the complex number $z=\left(\log _{0,5} \log _{4} \log _{8}\left(x^{2}+4\right)-1\right) i \quad$ is represented by a point lying in the upper part of the imaginary axis. | 10.34 According to the condition, $\log _{0.5} \log _{4} \log _{8}\left(x^{2}+4\right)-1>0$ for any $x \in \boldsymbol{R}$. Therefore,
$\log _{0.5} \log _{4} \log _{8}\left(x^{2}+4\right)>\log _{0.5} 0.5 \Rightarrow$
$\Rightarrow 0<\log _{4} \log _{8}\left(x^{2}+4\right)<0.5 \Rightarrow$
$\Rightarrow 1<\log _{8}\lef... | -2\sqrt{15}<x<-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,908 |
10.35 For what real values of $x$ and $y$ are the complex numbers $z_{1}=4 i-2 x y-x y i$ and $z_{2}=y^{2} i-x^{2}+3$ equal? | 10.35 For the complex numbers $z_{1}=-2 x y+(4-x y) i$ and $z_{2}=\left(3-x^{2}\right)+y^{2} i$ to be equal, the following system of equations must hold:
$\left\{\begin{array}{l}3-x^{2}=-2 x y, \\ y^{2}=4-x y\end{array} \Rightarrow\left\{\begin{array}{l}x^{2}-2 x y=3, \\ y^{2}+x y=4\end{array} \Rightarrow\right.\right... | x_{1}=3,y_{1}=1;\quadx_{2}=-3,y_{2}=-1;\quadx_{3}=\frac{\sqrt{3}}{3},y_{3}=-\frac{4\sqrt{3}}{3};\quadx_{4}=-\frac{\sqrt{3}}{3},y_{4}=\frac{4\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,909 |
10.36 Find the real number $b$ from the condition that the points representing the complex numbers $3-5 i, 1-i$ and $-2+b i$ lie on the same straight line. | 10.36 The numbers $3-5i, 1-i$ and $-2+bi$ correspond to the points $A(3, -5), B(1, -1)$ and $C(-2, b)$ on the plane (Fig. 10.10). We find the coordinates of the vectors $\overline{A B}$ and $\overline{A C}$; we have $\overline{A B}(-2, 4), \overline{A C}(-5, b+5)$.
+i\sin(-10)) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,911 |
10.38 Form the polynomial of the least degree with real coefficients that has roots $x_{1}=-2$ and $x_{2}=4-3 i$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.38 Since the desired equation must have real coefficients, the number of its complex roots is even. Considering the existence of another complex root $x_{3}=4+3 i$, we conclude that the equation of the lowest degree is a cubic equation:
$$
\begin{aligned}
& (x+2)(x-4+3 i)(x-4-3 i)=0 ;(x+2)\left((x-4)^{2}+9\right)=0... | x^{3}-6x^{2}+9x+50=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,912 |
$10.43 z^{2}-(2+i) z-1+7 i=0$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$10.43 z^{2}-(2+i) z-1+7 i=0$. | 10.43 Using the formula for the roots of a quadratic equation, we get
$$
z=\frac{2+i \pm \sqrt{(2+i)^{2}+4(1-7 i)}}{2}=\frac{2+i \pm \sqrt{7-24 i}}{2}
$$
To find all values of $\sqrt{7-24 i}$, let $\sqrt{7-24 i}=x+y i(x \in \boldsymbol{R}, y \in \boldsymbol{R})$.
Then $7-24 i=x^{2}+2 x y i-y^{2}$, i.e., the numbers ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,914 | |
10.47 Given complex numbers $z_{1}=1+a i$ and $z_{2}=2^{\frac{3}{4}}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)$. Find all real values of $a$ for which $z_{1}^{3}=z_{2}^{2}$. | 10.47 We find
\[
\begin{aligned}
& z_{1}^{3}=(1+a i)^{3}=\left(1-3 a^{2}\right)+\left(3 a-a^{3}\right) i \\
& z_{2}^{2}=\left(2^{\frac{3}{4}}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right)\right)^{2}= \\
& =2^{\frac{3}{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)=2^{\frac{3}{2}}\left(-\frac{\sq... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,916 |
10.48 The sum of two roots of the equation $x^{3}-3 x^{2}+4 x+\lambda=0$ is 2. Find $\lambda$ and solve this equation. | 10.48 Hint. Use the fact that $x_{1}+x_{2}+x_{3}=3$.
Answer: $\lambda=-2, x_{1}=1, x_{2,3}=1 \pm i$. | \lambda=-2,x_{1}=1,x_{2,3}=1\i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,917 |
10.49 Find the modulus of the complex number $z^{3}+z^{5}$, if $z=\cos \alpha+i \sin \alpha, \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$. | 10.49 We have
$$
\begin{aligned}
& \left|z^{3}+z^{5}\right|=|\cos 3 \alpha+i \sin 3 \alpha+\cos 5 \alpha+i \sin 5 \alpha|= \\
& =|\cos 3 \alpha+\cos 5 \alpha+i(\sin 3 \alpha+\sin 5 \alpha)|= \\
& =\sqrt{4 \cos ^{2} 4 \alpha \cos ^{2} \alpha+4 \sin ^{2} 4 \alpha \cos ^{2} \alpha}= \\
& =\sqrt{4 \cos ^{2} \alpha}=2|\cos... | 2|\cos\alpha| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,918 |
10.50 Among the complex numbers satisfying the condition $|z-5 i| \leq 4$, find the one whose argument has the smallest positive value. | 10.50 The inequality $|z-5 i| \leq 4$ is satisfied by points in the plane lying inside and on the boundary of a circle with center $M(0 ; 5)$ and radius 4. From Fig. 10.15, it is clear that the number with the smallest positive argument corresponds to the point $K$, where the line drawn from the origin is tangent to th... | 2.4+1.8i | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,919 |
11.1 Simplify the expression
$$
\frac{\sqrt{a^{2}-2 a b+b^{2}}}{\sqrt[4]{(b-a)^{3}}}
$$ | 11.1 The denominator of the fraction makes sense when $b-a>0$. Taking this into account, let's simplify the numerator:
$\sqrt{a^{2}-2 a b+b^{2}}=\sqrt{(a-b)^{2}}=|a-b|=b-a$.
Finally, we get $\frac{b-a}{\sqrt[4]{(b-a)^{3}}}=\sqrt[4]{b-a}$, where $b>a$. | \sqrt[4]{b-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,920 |
11.2 Solve the equation $|x+1|+|x-1|=2 x^{3}$. | 11.2 We have $|x+1|=0$ when $x=-1$, and $|x-1|=0$ when $x=1$. We will solve the given equation in the intervals $(-\infty,-1)$, $[-1,1)$, $[1, \infty)$.
1) If $x<0$ for any $x$, then $x=0$; this value does not belong to the interval $(-\infty,-1)$.
2) If $-1 \leq x<1$, then $x+1-x+1=2 x^{3} ; 2 x^{3}=2 ; x^{3}=1$; $x=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,921 |
## 11.3 Solve the system of equations
$$
\left\{\begin{array}{l}
6.751 x + 3.249 y = 26.751 \\
3.249 x + 6.751 y = 23.249
\end{array}\right.
$$
Plot the graphs of the functions (11.4 - 11.6): | 11.3 Instruction. Add and subtract the given equations.
Given: $x=3, y=2$. | 3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,922 |
11.4
a) $x^{2}+5 x+6$; б) $y=x^{2}+5|x|+6$ ；
в) $y=|x^{2}+5 x+6|$; r) $y=|x^{2}+5| x|+6|$. | 11.4 a) We find the roots of the function: $x^{2}+5 x+6=0$ with $x_{1}=-3$ and $x_{2}=-2$; then we find the point of intersection of the parabola with the $O y$ axis: $y=6$ when $x=0$. In addition, we determine the coordinates of the vertex of the parabola: we have $x_{\mathrm{B}}=-\frac{3+2}{2}=-2.5$; $y_{\mathrm{B}}=... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,923 |
11.7 Plot on the same graph the functions $y=\lg x^{2}$ and $y=\lg ^{2} x$. | 11.7 First, let's construct the graph of the function $y=\lg x^{2}=2 \lg |x|$ (Fig. 11.5). Before constructing the graph of the function $y=\lg ^{2} x$, note that it is defined for $x>0$, and its values $y>0$. Next, we find the derivative $y^{\prime}=2 \lg x \cdot \frac{1}{x \ln 10}$ and investigate the function for ex... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,926 |
11.8 Find the values of $x$ for which all values of the function $y=x^{2}+5 x+6$ belong to the interval $[6,12]$. | 11.8 Method I. According to the condition, $6 \leq x^{2}+5 x+6 \leq 12$. Thus, we have the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } + 5 x \geq 0 } \\
{ x ^ { 2 } + 5 x - 6 \leq 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \leq-5 ; x \geq 0 \\
-6 \leq x \leq 1
\end{array}\right.\right.... | -6\leqx\leq-50\leqx\leq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,927 |
11.9 Form a biquadratic equation if the numbers $\sqrt{3}-1$ and $\sqrt{3}+1$ are its roots. | ## 11.9 The general form of a biquadratic equation is: $a x^{4}+b x^{2}+c=0$. According to the condition, it has roots $x_{1}=\sqrt{3}-1$ and $x_{2}=\sqrt{3}+1$. The roots of a biquadratic equation are pairwise opposite; therefore, there are also roots $x_{3}=-(\sqrt{3}-1)$ and $x_{4}=-(\sqrt{3}+1)$. Assuming $a=1$, we... | x^{4}-8x^{2}+4=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,928 |
11.10 Indicate the domain of the function $=\frac{1}{\sqrt{x^{2}-10 x+25}}$. Show that the graph of this function is symmetric with respect to the line $x=5$. | 11.10 We have
$$
y=\frac{1}{\sqrt{x^{2}-10 x+25}}=\frac{1}{\sqrt{(x-5)^{2}}}=\frac{1}{|x-5|}
$$
Let's find the domain of the function: $|x-5| \neq 0$, i.e., $x \neq 5$. The values of the function to the left and right of the point $x=5$ are the same, since $|-a|=|a|$; therefore, the graph of the function is symmetric... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,929 |
11.11 Without transforming the equation $\sqrt{x+1}+\sqrt{3-x}=17$, show that it has no solutions. | 11.11 Let's find the domain of the equation:
$$
\left\{\begin{array} { l }
{ x + 1 \geq 0 , } \\
{ 3 - x \geq 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x \geq-1 \\
x \leq 3
\end{array} \Rightarrow-1 \leq x \leq 3\right.\right.
$$
Therefore, $0 \leq \sqrt{x+1} \leq 2, \quad 0 \leq \sqrt{3-x} \leq 2$, from w... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,930 |
11.12 Find the domain of the function $y=\sqrt{2^{x}-3^{x}}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 11.12 I method. The function is defined under the condition $2^{x}-3^{x} \geq 0$, i.e., $2^{x} \geq 3^{x}$. Dividing both sides of the inequality by $3^{x}>0$, we get $\left(\frac{2}{3}\right)^{x} \geq 1$, from which we obtain $x \leq 0$.
II method. On the same diagram, we plot the graphs of the functions $\quad y=2^{... | (-\infty,0] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,931 |
11.13 Determine for which values of $x$ the equality $\left|x^{2}-8 x+12\right|=x^{2}-8 x+12$ holds.
Solve the equations (11.14-11.15): | 11.13 According to the definition of the modulus, $|a|=a$, if $a \geq 0$. Solving the inequality $x^{2}-8 x+12 \geq 0$, we establish that $x \leq 2$ or $x \geq 6$.
Answer: $(-\infty, 2] \cup[6, \infty)$ | (-\infty,2]\cup[6,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,932 |
$11.14 \sqrt{x^{2}-1}-\frac{6}{\sqrt{x^{2}-1}}=1$. | 11.14 Here $\quad x^{2}-1>0 \Rightarrow|x|>1 \Rightarrow x1$. Let $\sqrt{x^{2}-1}=y>0$. Further, we have $y-\frac{6}{y}=1 ; y^{2}-y-6=0$; $y_{1}=-2<0$ (not suitable), $y_{2}=3$. Thus, we obtain the equation $\sqrt{x^{2}-1}=3$, from which $x^{2}-1=9$, i.e., $x_{1,2}= \pm \sqrt{10}$. Both roots belong to the domain of th... | x_{1,2}=\\sqrt{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,933 |
$11.15 \sqrt{\frac{1+x}{x}}+\frac{1}{x}=5$ | 11.15 Here $\frac{x+1}{x} \geq 0$. Solving this inequality using the interval method, we find $x \leq -1$ or $x > 0$. Let's rewrite the given equation as: $\sqrt{\frac{1}{x}+1}+\frac{1}{x}=5$. By setting $\frac{1}{x}=y$, we get $\sqrt{y+1}+y=5$, or
$\sqrt{y+1}=5-y$.
The domain of the equation (1) is $y \geq -1$: in t... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,934 |
11.16 For what value of $m$ does the system of equations
$$
\left\{\begin{array}{l}
2 x+(m-1) y=3 \\
(m+1) x+4 y=-3
\end{array}\right.
$$
have an infinite number of solutions? No solutions? | 11.16 The system has an infinite number of solutions or no solutions if the coefficients of $x$ and $y$ are proportional. Therefore, $\frac{2}{m+1}=\frac{m-1}{4}$ (here $m+1 \neq 0$; it is easy to verify that when $m=-1$, the system has a unique solution). Further, we have $(m+1)(m-1)=8 ; m^{2}-1=8$, from which $m= \pm... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,935 |
11.17 For what values of $p$ does the equation
$x^{2}-\left(2^{p}-1\right) x-3\left(4^{p-1}-2^{p-2}\right)=0$
have equal roots? | 11.17 A quadratic equation has equal roots if its discriminant $D=b^{2}-4 a c$ is equal to zero. We find
$$
\begin{aligned}
& D=\left(2^{p}-1\right)^{2}+12\left(4^{p-1}-2^{p-2}\right)= \\
& =2^{2 p}-2 \cdot 2^{p}+1+\frac{12 \cdot 2^{2 p}}{4}-\frac{12 \cdot 2^{p}}{4}=4 \cdot 2^{2 p}-5 \cdot 2^{p}+1
\end{aligned}
$$
Us... | p=-2p=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,936 |
11.18 Graph the set of values of variables $x$ and $y$ for which the inequalities $3 x-4 y+12>0$ and $x+y-2<0$ are simultaneously satisfied. | 11.18 Let's construct the lines $3x - 4y + 12 = 0$ and $x + y - 2 = 0$ on the plane (Fig. 11.8). Each of them divides the plane into two regions; the points of one of these regions satisfy the corresponding inequality, while the points of the other do not. To determine which of these two regions satisfies the given ine... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,937 |
11.19 Indicate which of the following functions are even, odd, and which are neither even nor odd:
a) $y=\sin ^{3} x+\operatorname{ctg}^{5} x$; b) $y=\sin 2 x+\cos 3 x$; c) $y=\frac{1-\sin x}{1+\sin x}$;
d) $y=\sin ^{4} x+x^{2}+1$; e) $y=\arcsin \frac{x}{2}$;
f) $y=x+\sqrt{x}$; g) $y=x|x|$; h) $y=\arccos 3 x$;
i) $... | 11.19 a) The domain of the function $(x \neq \pi n)$ is symmetric with respect to the origin. Let $m$ belong to the domain of the function. Then
$$
f(m)=\sin ^{3} m+\operatorname{ctg}^{5} m, f(-m)=\sin ^{3}(-m)+\operatorname{ctg}^{5}(-m)
$$
Due to the oddness of the functions $\sin x$ and $\operatorname{ctg} x$, we h... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,938 |
11.20 Show that the parabola $y=x^{2}-x+5.35$ does not intersect the graph of the function $y=2 \sin x+3$. | 11.20 The graph of the function $y=x^{2}-x+5.35$ is a parabola with branches directed upwards. This function takes its minimum value at the vertex of the parabola, i.e., at the point $x_{\mathrm{B}}=0.5, y_{\mathrm{B}}=y_{\text {min }}=0.25-0.5+5.35=5.1$. Now let's find the range of the function $y=2 \sin x+3$. Since $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,939 |
11.21 Find the range of the function $y=(\sin x+\cos x)^{2}$.
untranslated text:
11.21 Найти область значений функции $y=(\sin x+\cos x)^{2}$. | 11.21 Given that $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right), \quad$ we can write the expression for $y$ as $y=(\sin x+\cos x)^{2}=2 \sin ^{2}\left(x+\frac{\pi}{4}\right)$. But $0 \leq \sin ^{2}\left(x+\frac{\pi}{4}\right) \leq 1$, from which $0 \leq 2 \sin \left(x+\frac{\pi}{4}\right) \leq 2$, i.e., the ... | 0\leqy\leq2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,940 |
11.22 Eliminate $t$ from the equations $x=10^{\cos t}, y=10^{\sin t}$. | 11.22 By logarithmizing the equalities with base 10, we get $\lg x=\cos t, \lg y=\sin t$. Then
$$
\cos ^{2} t+\sin ^{2} t=\lg ^{2} x+\lg ^{2} y, \text { or } \lg ^{2} x+\lg ^{2} y=1
$$ | \lg^{2}x+\lg^{2}1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,941 |
11.23 Compute $\sin \left(\arcsin \frac{3}{5}+\arccos \frac{1}{3}\right)$. | 11.23 According to formula (4.7), we find
$$
\begin{aligned}
& \sin \left(\arcsin \frac{3}{5}\right) \cos \left(\arccos \frac{1}{3}\right)+\cos \left(\arcsin \frac{3}{5}\right) \sin \left(\arccos \frac{1}{3}\right)= \\
& =\frac{3}{5} \cdot \frac{1}{3}+\sqrt{1-\frac{9}{25}} \sqrt{1-\frac{1}{9}}=\frac{1}{5}+\frac{4}{5} ... | \frac{3+8\sqrt{2}}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,942 |
11.24 What is greater: $\operatorname{tg} 1$ or $\operatorname{arctg} 1 ?$ | 11.24 Since $\operatorname{arctg} 1=\frac{\pi}{4}\operatorname{tg} \frac{\pi}{4}=1$, then $\operatorname{tg} 1>\operatorname{arctg} 1$. | \operatorname{tg}1>\operatorname{arctg}1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,943 |
11.27 The price of a certain item was reduced twice - first by $15 \%$, and then by another $20 \%$. What is the overall percentage reduction in price?
## Group B | 11.27 Let's take the original price as $100 \%$. Then after the first reduction, the new price will be $100 \%$ $-15 \%=85 \%$ of the original. The second price reduction of $20 \%$ relative to the first will be $85 \% \cdot 0.2=17 \%$ of the original price. Thus, after the second reduction, the new price will be $85 \... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,945 |
11.28 Simplify the expression
$$
\frac{m}{m^{2}+1} \sqrt{1+\left(\frac{m^{2}-1}{2 m}\right)^{2}}
$$ | 11.28 For simplification, we will use formulas (1.9), (1.10):
$$
\begin{aligned}
& A=\frac{m}{m^{2}+1} \sqrt{1+\frac{m^{4}-2 m^{2}+1}{4 m^{2}}}=\frac{m}{m^{2}+1} \sqrt{\frac{m^{4}+2 m^{2}+1}{4 m^{2}}}= \\
& =\frac{m}{m^{2}+1} \sqrt{\frac{\left(m^{2}+1\right)^{2}}{4 m^{2}}}=\frac{m\left|m^{2}+1\right|}{\left(m^{2}+1\ri... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,946 |
11.29 Find the rational roots of the equation
$$
\frac{\sqrt{x+2}}{|x|}+\frac{|x|}{\sqrt{x+2}}=\frac{4}{3} \sqrt{3}
$$
Solve the inequalities (11.30-11.31): | 11.29 We find the domain of the equation: $x+2>0$, hence $x>-2$, and $x \neq 0$, i.e., $(-2,0) \cup(0, \infty)$. Setting $\frac{\sqrt{x+2}}{|x|}=y$, we get the equation
$$
y+\frac{1}{y}=\frac{4 \sqrt{3}}{3}, \text { or } 3 y^{2}-4 \sqrt{3} y+3=0
$$
From this, $\quad y_{1,2}=\frac{2 \sqrt{3} \pm \sqrt{12-9}}{3}=\frac{... | x_{1}=-\frac{2}{3},x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,947 |
$11.30 x^{2}-4|x|+3>0$ | 11.30 Since $x^{2}=|x|^{2}$, we solve the inequality $|x|^{2}-4|x|+3>0$. The left side of this inequality is a quadratic trinomial in terms of $|x|$, which equals zero when $|x|=1$ and $|x|=3$. Using the interval method, we find $|x|<1$ or $|x|>3$. From this, it follows that $-1<x<1$ or $x<-3$ or $x>3$.
Answer: $(-\in... | (-\infty,-3)\cup(-1,1)\cup(3,\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,948 |
11.31 $|x+1|>2|x+2|$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
11.31 $|x+1|>2|x+2|$. | 11.31 We have $|x+1|=0$ when $x=-1$, and $|x+2|=0$ when $x=-2$. We will solve the given inequality in the intervals $(-\infty,-2)$, $[-2,-1)$, and $[-1, \infty)$.
1) If $x-2 x-4$, then $x>-3$. Therefore, $-32 x+4$, or $3 x2 x+4$, from which $x<-3$. In this case, there are no solutions.
Combining the solutions (1) and... | (-3,-\frac{5}{3}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,949 |
### 11.32 Solve the system of equations
$$
\left\{\begin{array}{l}
2 x-3|y|=1 \\
|x|+2 y=4
\end{array}\right.
$$
Graph the functions (11.33-11.35): | 11.32 Depending on various combinations of the signs of \(x\) and \(y\), the given system splits into four systems:
\[
-7 y = -7 ; y = 1 > 0, x = 4 - 2 y = 2 > 0
\]
solution of the system ... | (2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,950 |
$11.33 y=\sqrt{(x+2)^{2}}+\sqrt{(x-2)^{2}}$.
The above text is translated into English as follows, preserving the original text's line breaks and format:
$11.33 y=\sqrt{(x+2)^{2}}+\sqrt{(x-2)^{2}}$. | 11.33 Method I. We have $y=|x+2|+|x-2|$. Using the definition of the modulus, we will construct the graph on different segments of the number line: if $x<-2$, then $y=-x-2-x+2=-2 x ; \quad$ if $-2 \leq x<2$, then $y=x+2-x+2=4$; if $x \geq 2$, then $y=x+2+x-2=2 x$ (Fig. $11.9, a$ ).
Method II. We will construct on the ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,951 |
11.36 Solve graphically the equation
$$
|x-1|+2x-5=0
$$ | 11.36 Let's write the equation as $|x-1|=5-2 x$. Since $|x-1| \geq 0$ for any $x$,
Fig. 11.11
and $\quad 5-2 x \geq 0$, i.e., $x \leq 2.5$.
Consider the functions $y_{1}=|x-1|, y_{2}=5-2 ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,953 |
11.37 Depict in the coordinate plane the set of values of the variables $x$ and $y$, satisfying the equation $|x|+|y|=1$ | 11.37 Let's rewrite the given relation as $|y|=1-|x|$. Since $|y| \geq 0$, we have $1-|x| \geq 0$, which implies $|x| \leq 1$, i.e., $-1 \leq x \leq 1$. The desired graph
is symmetric with r... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,954 |
11.39 Find the coefficients of the quadratic function $y=a x^{2}+b x+c$, knowing that at $x=-0.75$ it takes the maximum value of 3.25, and at $x=0$ it takes the value 1.
Fig. 11.1 | 11.39 According to the condition, when $x=0$, the value of $y=1$, from which $c=1$. Further, since the abscissa of the vertex of the parabola $x_{\mathrm{s}}=-\frac{b}{2 a}$ and at $x=-0.75$ we have $y_{\text {max }}=3.25$, then $-\frac{b}{2 a}=-\frac{3}{4}$, from which $b=\frac{3}{2} a$. Then the desired function will... | -4x^{2}-6x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,956 |
11.40 How many real solutions does the system of equations have
$\left\{\begin{array}{l}x^{2}+y=5 \\ x+y^{2}=3 ?\end{array}\right.$ | 11.40 The solutions of the system are the coordinates of the points of intersection of the graphs of the functions given by these equations. Let's write the original system in the form
$\left\{\begin{array}{l}y=5-x^{2}, \\ x=3-y^{2} .\end{array}\right.$
The graph of the first equation is a parabola with the axis of s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,957 |
11.41 For what value of a is the sum of the squares of the roots of the equation $x^{2}+a x+a-2=0$ the smallest? | 11.41 Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then $x_{1} x_{2}=a-2$, $x_{1}+x_{2}=-a$. From this, we can express the sum of the squares of the roots:
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-2 a+4=f(a)
$$
Since $f(a)$ is a quadratic function with a positive leading coeffi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,958 |
11.42 Does the equation $x^{3}+2 x-3=0$ have negative roots? | 11.42 Method I. Factoring the left side of the equation, we get $(x-1)\left(x^{2}+x+3\right)=0$. Since $x^{2}+x+3>0$ for any $x$, then $x-1=0$, from which $x=1$. Therefore, the equation has no negative roots.
Method II. We will solve the given equation graphically. Writing it in the form $x^{3}=-2 x+3$, we will plot t... | no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,959 |
11.43 Solve the equation $x+\lg \left(1+4^{x}\right)=\lg 50$. | 11.43 Let's write the equation as $x+\lg \left(1+4^{x}\right)=1+\lg 5$. It is not difficult to establish that the value $x=1$ satisfies the equation. The left side of the equation is $f(x)=x+\lg \left(1+4^{x}\right)$ - an increasing function; therefore, if $x<1$, then $f(x)<1+\lg 5$, and if $x>1$, then $f(x)>1+\lg 5$. ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,960 |
11.44 Solve the equation $\sqrt{3 a-2 x}+x=a$ and investigate for which values of the parameter $a$ it has roots (and how many) and for which values it does not have roots. | 11.44 Let $\sqrt{3 a-2 x}=t \geq 0$. Then $3 a-2 x=t^{2}$, from which $x=\frac{3 a}{2}-\frac{t^{2}}{2}$; the given equation will take the form $t^{2}-2 t=a$. Let's construct the parabola $y=t^{2}-2 t$ and the line $y=a$ (in Fig. 11.16, the part of the parabola corresponding to $t \geq 0$ is highlighted with a bold line... | x_{1,2}=-1\\sqrt{+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,961 |
11.45 For what values of $m$ does the equation $2 \sqrt{1-m(x+2)}=x+4$ have one root? | 11.45 If \( m=0 \), then the equation will take the form \( 2=x+4 \) and, therefore, has a single root \( x=-2 \). Let \( m \neq 0 \). We set
\[
\sqrt{1-m(x+2)}=y \geq 0, \quad \text { from which after }
\]
transformations we find \( x=\frac{1-y^{2}}{m}-2 \). Then the given equation can be written as:
\[
2 y=\frac{1... | -1-\frac{1}{2}<<\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,962 |
11.48 Which is greater: $3^{400}$ or $4^{300} ?$ | 11.48 Let's bring the given numbers to the same exponent, equal to 100. We have
$$
3^{400}=3^{4 \cdot 100}=\left(3^{4}\right)^{100} ; 4^{300}=4^{3 \cdot 100}=\left(4^{3}\right)^{100}
$$
Since of the two powers with the same positive exponent 100, the one with the larger base is greater, then $3^{400}>4^{300}$. | 3^{400}>4^{300} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 49,964 |
11.49 For what values of $\alpha$ and $\beta$ is the equality $\sin \alpha + \sin \beta = \sin (\alpha + \beta)$ possible? | 11.49 We have
\[
\begin{aligned}
& 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}-2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}=0 \\
& \sin \frac{\alpha+\beta}{2}\left(\cos \frac{\alpha-\beta}{2}-\cos \frac{\alpha+\beta}{2}\right)=0 \\
& -2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha}{2} \sin ... | \alpha+\beta=2\pin;\alpha=2\pin,\beta\in\mathbb{R};\beta=2\pin,\alpha\in\mathbb{R}(n\in\mathbb{Z}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,965 |
11.50 What is the value of the fraction $\frac{1-\sin \alpha}{\cos \alpha}$, if $\operatorname{ctg} \frac{\alpha}{2}=m$? | 11.50 Denoting the given expression by $A$, we will transition to the argument $\frac{\alpha}{2}$ and perform the following transformations:
$$
\begin{aligned}
& A=\frac{\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}-2 \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alph... | \frac{-1}{+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,966 |
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