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$5.55 \operatorname{ctg}^{4} x=\cos ^{2} 2 x-1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.55 \operatorname{ctg}^{4} x=\cos ^{2} 2 x-1$. | 5.55 Since $\operatorname{ctg}^{4} x \geqslant 0, \cos ^{2} 2 x-1<0$, we arrive at the system of equations
$\left\{\begin{array}{l}\operatorname{ctg}^{4} x=0 \\ \cos ^{2} 2 x-1=0 .\end{array}\right.$
From this, we find: 1) $\left.x=\frac{\pi}{2}(2 k+1) ; 2\right) x=\frac{\pi k}{2}$, but in this case, for $k=21$, $\op... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,721 | |
$5.56 \sin 2 z-\sin 6 z+2=0$. | 5.56 Since $|\sin \alpha| \leqslant 1$, the equation $\sin 2 z-\sin 6 z=-2$ is possible only when $\sin 2 z=-1, \sin 6 z=1$. If $\sin 2 z=-1$, then
$2 z=-\frac{\pi}{2}+2 \pi k$, from which $z=-\frac{\pi}{4}+\pi k$. For these values of $z$
the equation $\sin 6 z=1$ is satisfied.
Answer: $\quad z=\frac{\pi}{4}(4 k-1),... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,722 |
$5.57 \sin ^{3} x(1+\operatorname{ctg} x)+\cos ^{3} x(1+\operatorname{tg} x)=2 \sqrt{\sin x \cos x}$. | 5.57 We have $\sin x \cos x>0$, but for $\sin x>0, \cos x>0$. Transform the left side of the equation:
$$
\begin{aligned}
& \sin ^{3} x \frac{\sin x+\cos x}{\sin x}+\cos ^{3} x \frac{\sin x+\cos x}{\sin x}= \\
& =(\sin x+\cos x)\left(\sin ^{2} x+\cos ^{2} x\right)=\sin x+\cos x
\end{aligned}
$$
## Then we get
$\sin ... | \frac{\pi}{4}+2\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,723 |
$5.58 \sin 3 x=a \sin x$. | 5.58 Transform the left side of the equation:
$\sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\cos 2 x \sin x=$
$=2 \sin x \cos ^{2} x+\cos ^{2} x \sin x-\sin ^{3} x=3 \sin x\left(1-\sin ^{2} x\right)-$ $-\sin ^{3} x=3 \sin x-4 \sin ^{3} x$.
It remains to solve the equation $3 \sin x-4 \sin ^{3} x=a \sin x$. We have:
1) $\s... | x_{1}=\pikforanyx_{2}=\\frac{1}{2}\arccos\frac{-1}{2}+\pikfor-1\leqslant\leqslant3\(k\inZ) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,724 |
5.59 Find the angles $\alpha, \beta$ and $\gamma$ of the first quadrant, if it is known that they form an arithmetic progression with a common difference of $\frac{\pi}{12}$, and their tangents form a geometric progression. | 5.59 Let $\alpha=\beta-\frac{\pi}{12}, \gamma=\beta+\frac{\pi}{12}$. Since by condition $\operatorname{tg} \beta=$
$=q \operatorname{tg} \alpha, \operatorname{tg} \gamma=q^{2} \operatorname{tg} \alpha=q \operatorname{tg} \beta$, then
$\frac{\operatorname{tg} \beta}{\operatorname{tg}\left(\beta-\frac{\pi}{12}\right)}=... | \alpha=\frac{\pi}{6},\beta=\frac{\pi}{4},\gamma=\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,725 |
5.60 Solve the system of equations
$$
\left\{\begin{array}{l}
\sin x \sin y=0.75 \\
\operatorname{tg} x \operatorname{tg} y=3
\end{array}\right.
$$ | 5.60 We have
$$
\left\{\begin{array}{l}
\sin x \sin y=0.75 \\
\frac{\sin x}{\cos x} \cdot \frac{\sin y}{\cos y}=3
\end{array}\right.
$$
From the second equation, we get $\frac{0.75}{\cos x \cos y}=3$, hence $\cos x \cos y=0.25$. Then we arrive at the system
$\left\{\begin{array}{l}\sin x \sin y=0.75 \\ \cos x \cos y... | \\frac{\pi}{3}+\pi(k+n),\\frac{\pi}{3}+\pi(n-k),n,k\in\boldsymbol{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,726 |
6.1 In a shooting competition, for each miss in a series of 25 shots, the shooter received penalty points: for the first miss - one penalty point, and for each subsequent miss - 0.5 points more than for the previous one. How many times did the shooter hit the target, receiving 7 penalty points? | 6.1 Let $x$ be the number of misses in a series of 25 shots. Then, using formula (6.3), we get the equation
$$
\frac{2 \cdot 1+0.5(x-1)}{2} x=7, \text{ or } x^{2}+3 x-28=0
$$
from which $x_{1}=4, x_{2}=-7$ (does not fit). Therefore, the shooter hit the target 21 times.
Answer: 21 times. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,727 |
6.2 Find the number of terms of an arithmetic progression, the sum of all of which is 112, the product of the second term by the common difference is 30, and the sum of the third and fifth terms is 32. Write the first three terms of this progression. | 6.2 Using the condition and formula (6.1), we have the system
$$
\left\{\begin{array} { l }
{ a _ { 2 } d = 3 0 , } \\
{ a _ { 3 } + a _ { 5 } = 3 2 , }
\end{array} \text { or } \left\{\begin{array}{l}
\left(a_{1}+d\right) d=30 \\
a_{1}+2 d+a_{1}+4 d=32
\end{array}\right.\right.
$$
From here, we find two solutions: ... | 7terms;1,6,11or7,10,13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,728 |
6.3 Find four numbers that form a geometric progression, where the sum of the extreme terms is -49, and the sum of the middle terms is 14. | 6.3 Using the condition and formula (6.6), we obtain the system
$$
\left\{\begin{array} { l }
{ b _ { 1 } + b _ { 4 } = - 49 , } \\
{ b _ { 2 } b _ { 3 } = 14 , }
\end{array} \text { or } \left\{\begin{array}{l}
b_{1}\left(1+q^{3}\right)=-49 \\
b_{1} q(1+q)=14
\end{array}\right.\right.
$$
From this, it follows that
... | 7,-14,28,-56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,729 |
6.4 Find the third term of an infinite geometric progression with a common ratio $|q|<1$, the sum of which is $\frac{8}{5}$, and the second term is $-\frac{1}{2}$. | 6.4 Let's use formulas (6.10) and (6.6). Then we get the system
$\left\{\begin{array}{l}\frac{b_{1}}{1-q}=\frac{8}{5}, \\ b_{1} q=-\frac{1}{2} .\end{array}\right.$
Further, we have $16 q^{2}-16 q-5=0$, from which $q_{1}=-\frac{1}{4}$, $q_{2}=\frac{5}{4}>1$ (does not fit the condition). Therefore, $b_{3}=\left(-\frac{... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,730 |
6.5 The sum of three numbers forming an arithmetic progression is 2, and the sum of the squares of these same numbers is $\frac{14}{9}$. Find these numbers. | 6.5 According to the condition, we write the system
$\left\{\begin{array}{l}a_{1}+a_{2}+a_{3}=2, \\ a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=\frac{14}{9} .\end{array}\right.$
Next, using formula (6.2), we have $\frac{a_{1}+a_{3}}{2} \cdot 3=2$, from which $a_{1}+a_{3}=\frac{4}{3}$ and, therefore, $a_{2}=\frac{2}{3}$. Thus, we a... | \frac{1}{3};\frac{2}{3};1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,731 |
6.6 Calculate
$$
\begin{aligned}
& \left(1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}\right)- \\
& -\left(2^{2}+4^{2}+6^{2}+\ldots+(2 n)^{2}+\ldots+200^{2}\right)
\end{aligned}
$$ | 6.6 Note that $(2 n)^{2}-(2 n-1)^{2}=4 n-1$. The number $n$ can be found from the equation $4 n-1=200^{2}-199^{2}$, or $4 n-1=399$, from which $n=100$. We will show that the sequence $a_{n}=(2 n)^{2}-(2 n-1)^{2}$ is an arithmetic progression. Indeed,
$a_{n}-a_{n-1}=\left((2 n)^{2}-(2 n-1)^{2}\right)-\left((2 n-2)^{2}-... | -20100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,732 |
6.7 The denominator of the geometric progression is $\frac{1}{3}$, the fourth term of this progression is $\frac{1}{54}$, and the sum of all its terms is $\frac{121}{162}$. Find the number of terms in the progression. | 6.7 Since $b_{4}=b_{1} q^{3}$, then $\frac{1}{54}=b_{1} \cdot \frac{1}{27}$, from which $b_{1}=\frac{1}{2}$. Now, using formula (6.7a), we get the equation
$$
\frac{\frac{1}{2}\left(1-\frac{1}{3^{n}}\right)}{\frac{2}{3}}=\frac{121}{162}, \text { or } 1-\frac{1}{3^{n}}=\frac{121 \cdot 4}{162 \cdot 3}, \text { or } 1-\f... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,733 |
6.8 Find the first term and the common ratio of the geometric progression, given that $b_{4}-b_{2}=-\frac{45}{32}$ and $b_{6}-b_{4}=-\frac{45}{512}$. | 6.8 Instruction. Form the system
$$
\left\{\begin{array}{l}
b_{1} q\left(q^{2}-1\right)=-\frac{45}{32} \\
b_{1} q^{3}\left(q^{2}-1\right)=-\frac{45}{512}
\end{array}\right.
$$
from which to find $q$.
Answer: $b_{1}=6, q=\frac{1}{4}$ or $b_{1}=-6, q=-\frac{1}{4}$. | b_{1}=6,q=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,734 |
6.9 An arithmetic progression has the following property: for any $n$, the sum of its first $n$ terms is equal to $5 n^{2}$. Find the common difference of this progression and its first three terms. | 6.9 Using formula (6.3) and the condition of the problem, we get the equality
$$
\frac{2 a_{1}+d(n-1)}{2} n=5 n^{2}, \text{ or } 2 a_{1}-d=(10-d) n
$$
Since in this equality only $n$ can vary, then $d=10$. For $d=10$, we find $a_{1}=5$. Therefore, the first three terms of this progression are: $5 ; 15 ; 25$.
Answer:... | 5;15;25;=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,735 |
6.11 Solve the equations:
a) $2 x+1+x^{2}-x^{3}+x^{4}-x^{5}+\ldots=\frac{13}{6}$, where $|x|<1$;
b) $\frac{1}{x}+x+x^{2}+\ldots+x^{n}+\ldots=\frac{7}{2}$, where $|x|<1$. | 6.11 Let's use formula (6.10), valid under the condition $|x|<1$:
a) $2 x+1+\frac{x^{2}}{1+x}=\frac{13}{6} \Rightarrow x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9}$;
b) $\frac{1}{x}+\frac{x}{1-x}=\frac{7}{2} \Rightarrow x_{1}=\frac{1}{3}, x_{2}=\frac{2}{3}$.
Answer: a) $x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9} ;$ b) $x_{1}=\f... | )x_{1}=\frac{1}{2},x_{2}=-\frac{7}{9};b)x_{1}=\frac{1}{3},x_{2}=\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,737 |
6.12 The sum of an infinite geometric progression with a common ratio $|q|<1$ is 16, and the sum of the squares of the terms of this progression is 153.6. Find the fourth term and the common ratio of the progression. | 6.12 Applying formula (6.10) to both progressions, we get
$$
\begin{aligned}
& b_{1}+b_{1} q+b_{1} q^{2}+\ldots=\frac{b_{1}}{1-q} \\
& b_{1}^{2}+\left(b_{1} q\right)^{2}+\left(b_{1} q^{2}\right)^{2}+\ldots=\frac{b_{1}^{2}}{1-q^{2}}
\end{aligned}
$$
According to the condition, we have the system
$\left\{\begin{array}... | \frac{3}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,738 |
6.13 Find the natural numbers forming an arithmetic progression if the products of the first three and the first four of its terms are 6 and 24, respectively. | 6.13 From the condition we have
$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)=6 \\
a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)\left(a_{1}+3 d\right)=24
\end{array} \Rightarrow\right. \\
& \Rightarrow a_{1}+3 d=4 \Rightarrow a_{1}=4-3 d
\end{aligned}
$$
Thus, we arriv... | 1;2;3;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,739 |
6.14 The product of the third and sixth terms of an arithmetic progression is 406. When the ninth term of this progression is divided by its fourth term, the quotient is 2, and the remainder is 6. Find the first term and the common difference of the progression. | 6.14 According to the condition, we have the system
$$
\left\{\begin{array} { l }
{ a _ { 3 } a _ { 6 } = 4 0 6 , } \\
{ a _ { 9 } = 2 a _ { 4 } + 6 , }
\end{array} \text { or } \left\{\begin{array}{l}
\left(a_{1}+2 d\right)\left(a_{1}+5 d\right)=406 \\
a_{1}+8 d=2\left(a_{1}+3 d\right)+6
\end{array}\right.\right.
$$... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,740 |
6.15 The sum of the first three terms of an increasing arithmetic progression is 15. If 1 is subtracted from the first two terms of this progression, and 1 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first ten terms of the arithmetic progression. | 6.15 Let $a_{1}$ be the first term of an arithmetic progression, and $d$ its difference. According to the condition, $\frac{\left(2 a_{1}+2 d\right) 3}{2}=15$, from which $a_{1}+d=5$.
Further, since the numbers $a_{1}-1, a_{1}+d-1, a_{1}+2 d+1$ form a geometric progression, by formula (6.8) we have
$$
\left(a_{1}+d-1... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,741 |
6.16 The sum of three consecutive terms of a geometric progression is 65, and the sum of their logarithms to the base 15 is 3. Find these terms of the progression.
## Group B | 6.16 Due to the condition, we have the system
$$
\left\{\begin{array}{l}
a_{1}+a_{1} q+a_{1} q^{2}=65 \\
\log _{15} a_{1}+\log _{15}\left(a_{1} q\right)+\log _{15}\left(a_{1} q^{2}\right)=3
\end{array}\right.
$$
Writing the second equation as $\log _{15}\left(a_{1} \cdot a_{1} q \cdot a_{1} q^{2}\right)=3$, we get $a... | 5;15;45or45;15;5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,742 |
6.17 The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 189. Find the first term and the common ratio of this progression. | 6.17 According to the condition, we have
$$
\left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}\right)=21 \\
b_{1}^{2}\left(1+q^{2}+q^{4}\right)=189
\end{array}\right.
$$
Squaring both sides of the first equation:
$$
b_{1}^{2}\left(1+q^{2}+q^{4}\right)+2 b_{1}^{2} q\left(1+q+q^{2}\right)=441
$$
Subtracting the second equa... | b_{1}=3,q=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,743 |
6.18 It is known that an arithmetic progression includes terms $a_{2 n}$ and $a_{2 m}$ such that $\frac{a_{2 n}}{a_{2 m}}=-1$. Does this progression have a term equal to zero? If yes, what is the index of this term? | 6.18 We have
$$
\begin{aligned}
& \frac{a_{2 n}}{a_{2 m}}=-1 \Rightarrow a_{2 n}+a_{2 m}=0 \Rightarrow \\
& \Rightarrow\left[a_{1}+(2 n-1) d\right]+\left[a_{1}+(2 m-1) d\right]=0 \Rightarrow \\
& \Rightarrow a_{1}+(n+m-1) d=0
\end{aligned}
$$
This means that the $(n+m)$-th term of the progression is zero. Answer: yes... | yes;the(n+)-term | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,744 |
6.19 Solve the equation
$1+2 x+4 x^{2}+\ldots+(2 x)^{n}+\ldots=3.4-1.2 x$, given that $|x|<0.5$. | 6.19 The left side of the equation is the sum of an infinite geometric progression, where \( b_{1}=1 \) and \( |q|=|2 x|<1 \), since \( |x|<0.5 \). According to formula (6.10), we have
\[
\begin{aligned}
& 1+2 x+4 x^{2}+\ldots=\frac{1}{1-2 x} \\
& \frac{1}{1-2 x}=3.4-1.2 x ;(3.4-1.2 x)(1-2 x)=1
\end{aligned}
\]
\[
2.... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,745 |
6.20 The sum of an infinite geometric progression with a common ratio $|q|<1$ is 4, and the sum of the cubes of its terms is 192. Find the first term and the common ratio of the progression. | 6.20 Since
$$
\begin{aligned}
& b_{1}+b_{1} q+b_{1} q^{2}+\ldots=\frac{b_{1}}{1-q} \\
& b_{1}^{3}+b_{1}^{3} q^{3}+b_{1}^{3} q^{6}+\ldots=\frac{b_{1}^{3}}{1-q^{3}}
\end{aligned}
$$
$$
\left\{\begin{array} { l }
{ \frac { b _ { 1 } } { 1 - q } = 4 , } \\
{ \frac { b _ { 1 } ^ { 3 } } { 1 - q ^ { 3 } } = 192 , }
\end{a... | 6-0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,746 |
6.21 Given two infinite geometric progressions with a common ratio $|q|<1$, differing only in the sign of their common ratios. Their sums are $S_{1}$ and $S_{2}$. Find the sum of the infinite geometric progression formed by the squares of the terms of any of the given progressions. | ### 6.21 We have:
$$
\begin{aligned}
& a_{1}+a_{1} q+a_{1} q^{2}+\ldots=S_{1}=\frac{a_{1}}{1-q} \\
& a_{1}-a_{1} q+a_{1} q^{2}-\ldots=S_{2}=\frac{a_{1}}{1+q} \\
& a_{1}^{2}+a_{1}^{2} q^{2}+a_{1}^{2} q^{4}+\ldots=S=\frac{a_{1}^{2}}{1-q^{2}}
\end{aligned}
$$
From this, it follows that
$$
S=\frac{a_{1}^{2}}{1-q^{2}}=\f... | S_{1}S_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,747 |
6.22 The sum of three numbers is $\frac{11}{18}$, and the sum of their reciprocals, which form an arithmetic progression, is 18. Find these numbers. | 6.22 Let $x, y, z$ be the required numbers. Then, using the conditions, we get the system
$$
\left\{\begin{array}{l}
x+y+z=\frac{11}{18} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=18 \\
\frac{2}{y}=\frac{1}{x}+\frac{1}{z}
\end{array}\right.
$$
Solving it, we find $x=\frac{1}{9}, y=\frac{1}{6}, z=\frac{1}{3}$.
Answer: $\... | \frac{1}{9};\frac{1}{6};\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,748 |
6.23 The sum of the first three terms of an increasing arithmetic progression is 21. If 1 is subtracted from the first two terms of this progression, and 2 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first eight terms of the geometric progression. | 6.23 Let $a_{1}$ be the first term of an arithmetic progression, and $d$ its difference. Then, according to formula (6.2), we have $S_{3}=\frac{2 a_{1}+2 d}{2} \cdot 3=21$, or $a_{1}+d=7$. By the condition, $a_{1}-1$, $a_{1}+d-1, a_{1}+2 d+2$ are three consecutive terms of a geometric progression. Using formula (6.8), ... | 765 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,749 |
6.24 Find four numbers, the first three of which form a geometric progression, and the last three form an arithmetic progression. The sum of the extreme numbers is 21, and the sum of the middle numbers is 18. | 6.24 Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the required numbers. Using the conditions, as well as formulas (6.4) and (6.8), we obtain the system
$$
\left\{\begin{array}{l}
a_{2}^{2}=a_{1} a_{3} \\
2 a_{3}=a_{2}+a_{4} \\
a_{1}+a_{4}=21 \\
a_{2}+a_{3}=18
\end{array}\right.
$$
By eliminating $a_{1}, a_{2}, a_{4}$ from the... | 3;6;12;18or18.75;11.25;6.75;2.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,750 |
6.25 Three numbers form a geometric progression. If the second number is increased by 2, the progression becomes arithmetic, and if the last number is then increased by 9, the progression becomes geometric again. Find these numbers. | 6.25 According to the condition, the numbers $a, a q, a q^{2}$ form a geometric progression, the numbers $a, a q+2, a q^{2}$ form an arithmetic progression, and the numbers $a, a q+2, a q^{2}+9$ form a geometric progression again. Using formulas (6.4) and (6.8), we obtain the system of equations
$$
\left\{\begin{array... | 4;8;16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,751 |
$7.1 \quad \sqrt{5} \cdot 0.2^{\frac{1}{2 x}}-0.04^{1-x}=0$. | 7.1 Here all powers can be reduced to the same base 5. We have:
$$
\begin{aligned}
& \sqrt{5}=5^{\frac{1}{2}}, 0.2^{\frac{1}{2 x}}=\left(\frac{1}{5}\right)^{\frac{1}{2 x}}=5^{-\frac{1}{2 x}} \\
& 0.04^{1-x}=\left(\frac{1}{25}\right)^{1-x}=5^{-2(1-x)}
\end{aligned}
$$
Then the equation will take the form
$$
5^{\frac{... | x_{1}=1,x_{2}=0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,752 |
7.2
$$
\sqrt{2^{x} \sqrt{4^{x} \cdot 0.125^{\frac{1}{x}}}}=4 \sqrt[3]{2}
$$ | 7.2 Applying the rule of extracting the root from the power and formulas (1.19) and (1.16), we get
$2^{\frac{x}{2}} \cdot 2^{\frac{x}{3}} \cdot 0.5^{\frac{1}{2 x}}=2^{2} \cdot 2^{\frac{1}{3}}$, or $2^{\frac{x}{2}+\frac{x}{3}-\frac{1}{2 x}}=2^{\frac{7}{3}}$.
According to the hint $1^{0}$, we transition to the equivale... | x_{1}=-0.2,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,753 |
$7.4 \quad 2.5^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}} \cdot 0.4^{1-\sqrt{9-x}}=5^{10} \cdot 0.1^{5}$. | 7.4 We have
$\left(\frac{5}{2}\right)^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}} \cdot\left(\frac{2}{5}\right)^{1-\sqrt{9-x}}=5^{10}\left(\frac{1}{10}\right)^{5} ;$
$\left(\frac{5}{2}\right)^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}} \cdot \frac{2}{5}\left(\frac{5}{2}\right)^{\sqrt{9-x}}=\left(\frac{5}{10}\right)^{5} \cdot 5^{5}$.
Mu... | x_{1}=8,x_{2}=-7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,754 |
$7.5 \quad 7^{x}(\sqrt{2})^{2 x^{2}-6}-\left(\frac{7}{4}\right)^{x}=0$. | 7.5 Instruction. Divide both sides of the equation by the positive number $\left(\frac{7}{4}\right)^{x}$.
Omвem: $x_{1}=-3, x_{2}=1$. | x_{1}=-3,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,755 |
7.6 $\quad 2^{x^{2}-1}-3^{x^{2}}=3^{x^{2}-1}-2^{x^{2}+2}$. | 7.6 Let's group the powers with base 2 on the left side and the powers with base 3 on the right side of the equation, and factor each part. We have
$2^{x^{2}-1}\left(1+2^{3}\right)=3^{x^{2}-1}(1+3)$, or $2^{x^{2}-1} \cdot 3^{2}=3^{x^{2}-1} \cdot 2^{2}$.
Dividing both sides of the equation by $3^{x^{2}-1} \cdot 2^{2} ... | x_{1,2}=\\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,756 |
7.7 $\quad 2^{\log _{3} x^{2}} \cdot 5^{\log _{3} x}=400$. | 7.7 The logarithmic function $y=\log _{3} x$ is defined for $x>0$. Therefore, according to formula (7.6), we have $\log _{3} x^{2}=2 \log _{3} x$. Consequently, $2^{2 \log _{3} x} \cdot 5^{\log _{3} x}=20^{2} ; 4^{\log _{3} x} \cdot 5^{\log _{3} x}=20^{2} ; 20^{\log _{3} x}=20^{2}$. From this, $\log _{3} x=2$, i.e., $x... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,757 |
$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$. | 7.8 Since $4^{\sqrt{x}}=2^{2 \sqrt{x}}$ and $2^{\sqrt{x}-1}=2^{\sqrt{x}} \cdot 2^{-1}=\frac{1}{2} \cdot 2^{\sqrt{x}}$, the given equation will take the form
$$
2^{2 \sqrt{x}}-\frac{9}{2} \cdot 2^{\sqrt{x}}+2=0
$$
Let's make the substitution $2^{\sqrt{x}}=y$, where $y>0$ due to the property of the exponential function... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,758 |
$7.9 \quad 3 \cdot 5^{2x-1}-2 \cdot 5^{x-1}=0.2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$7.9 \quad 3 \cdot 5^{2x-1}-2 \cdot 5^{x-1}=0.2$. | 7.9 We have $5^{2 x-1}=5^{2 x} \cdot 5^{-1} ; 5^{x-1}=5^{x} \cdot 5^{-1} ; 0.2=5^{-1}$. After dividing all terms of the given equation by $5^{-1}$, it will take the form
$3 \cdot 5^{2 x}-2 \cdot 5^{x}=1$.
Let $5^{x}=y$, where $y>0$. Then we get the equation $3 y^{2}-2 y-1=0$, from which $y_{1}=1, y_{2}=-\frac{1}{3}$ ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,759 |
$7.10 \quad 10^{\frac{2}{x}}+25^{\frac{1}{x}}=4.25 \cdot 50^{\frac{1}{x}}$. | 7.10 Let's preliminarily perform the transformations:
$$
\begin{aligned}
10^{\frac{2}{x}} & =(2 \cdot 5)^{\frac{2}{x}}=2^{\frac{2}{x}} \cdot 5^{\frac{2}{x}} ; 25^{\frac{1}{x}}=\left(5^{2}\right)^{\frac{1}{x}}=5^{\frac{2}{x}} \\
50^{\frac{1}{x}} & =\left(2 \cdot 5^{2}\right)^{\frac{1}{x}}=2^{\frac{1}{x}} \cdot 5^{\frac... | x_{1,2}=\0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,760 |
$7.13 x^{\frac{\lg x+5}{3}}=10^{\lg x+1}$. | 7.13 Since the logarithmic function is defined for $x>0$, both the left and right sides of the given equation are positive. Taking their logarithms to base 10 and using formulas (7.6) and (7.2), we arrive at the equation
$\frac{\lg x+5}{3} \lg x=\lg x+1$.
Let's make the substitution $y=\lg x$ and solve the equation $... | x_{1}=0.001,x_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,761 |
7.19
$$
\frac{\log _{a} \sqrt{a^{2}-1} \cdot \log _{\frac{1}{a}}^{2} \sqrt{a^{2}-1}}{\log _{a^{2}}\left(a^{2}-1\right) \cdot \log _{\sqrt[3]{a}} \sqrt[6]{a^{2}-1}}
$$ | 7.19 By the definition and property $1^{0}$ of the logarithmic function, we have
$a>0, a \neq 1$
and
$a^{2}-1>0 \Rightarrow|a|>1 \Rightarrow a>1$.
From (1) and (2), it follows that $a>1$. Further, since the denominator of the fraction must be non-zero, then $a^{2}-1 \neq 1$ and, therefore, $\quad a \neq \pm \sqrt{2... | \log_{}\sqrt{^{2}-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,764 |
7.20 Find $\log _{\sqrt{3}} \sqrt[6]{a}$, if it is known that $\log _{a} 27=b$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 7.20 Applying formulas (7.9), (7.8), and again (7.9) sequentially, we get
$\log _{\sqrt{3}} \sqrt[6]{a}=\log _{3} \sqrt[3]{a}=\frac{1}{\log _{\sqrt[3]{a}} 3}=\frac{1}{\log _{a} 27}=\frac{1}{b}$.
Answer: $\frac{1}{b}$. | \frac{1}{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,765 |
7.21 Show that under the condition $x>0$ and $y>0$, from the equality $x^{2}+4 y^{2}=12 x y$ it follows that the equality
$$
\lg (x+2 y)-2 \lg 2=\frac{1}{2}(\lg x+\lg y)
$$
Solve the equations (7.22-7.38): | 7.21 By adding $4xy$ to both sides of the equation $x^{2}+4 y^{2}=12 x y$, we get $(x+2 y)^{2}=16 x y$. Since $x>0, y>0$, then $x+2 y=\sqrt{16 x y}$. Taking the logarithm of the last equation with base 10:
$$
\lg (x+2 y)=\frac{1}{2} \lg (16 x y)
$$
$$
\begin{aligned}
& \lg (x+2 y)=\frac{1}{2} \lg 16+\frac{1}{2}(\lg x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,766 |
$7.22 \quad \lg ^{2}(100 x)+\lg ^{2}(10 x)=14+\lg \frac{1}{x}$. | 7.22 Here $x>0$. After some transformations, we obtain the equivalent equation
$$
(2+\lg x)^{2}+(1+\lg x)^{2}=14-\lg x
$$
Letting $\lg x=y$, we arrive at the quadratic equation $2 y^{2}+7 y-9=0$, which has roots $y_{1}=-9 / 2, y_{2}=1$. Therefore, $x_{1}=10^{-9 / 2}, x_{2}=10$.
Answer: $x_{1}=10^{-9 / 2}, x_{2}=10$. | x_{1}=10^{-9/2},x_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,767 |
$7.23 \quad 7^{\lg x}-5^{\lg x+1}=3 \cdot 5^{\lg x-1}-13 \cdot 7^{\lg x-1}$ | 7.23 Group the powers with base 7 on the left side of the equation, and the powers with base 5 on the right side: $7^{\lg x}+13 \cdot 7^{\lg x-1}=3 \cdot 5^{\lg x-1}+5^{\lg x+1}$.
By factoring out the power with the smaller exponent and performing further transformations, we get
$$
\begin{aligned}
& 7^{\lg x-1}(7+13)... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,768 |
$7.24 \log _{3}\left(3^{x^{2}-13 x+28}+\frac{2}{9}\right)=\log _{5} 0.2$. | 7.24 Note that the expression under the logarithm sign in the left part is positive for all $x$. Let's simplify the right part:
$$
\log _{5} 0.2=\log _{5} \frac{1}{5}=\log _{5} 1-\log _{5} 5=-1=\log _{3} 3^{-1}
$$
Thus,
$$
\log _{3}\left(3^{x^{2}-13 x+28}+\frac{2}{9}\right)=\log _{3} 3^{-1}
$$
Next we have
$$
\beg... | x_{1}=3,x_{2}=10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,769 |
$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$. | 7.25 From the definition of the logarithm, it follows that $\sqrt{x}>0, \sqrt{x} \neq 1$, i.e., $x>0, x \neq 1$. Applying formulas (7.8) and (7.6), we get $\lg (\sqrt{6+x}+6)=\lg x$. According to the hint $4^{0}$, we transition to an equivalent system of equations:
$\left\{\begin{array}{l}x>0, x \neq 1, \\ \sqrt{6+x}+... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,770 |
$7.26 \frac{\log _{5}(\sqrt{2 x-7}+1)}{\log _{5}(\sqrt{2 x-7}+7)}=0.5$. | 7.26 Given the domain of the square root, we conclude that $2 x-7 \geq 0$. For these values of $x$, we have $\sqrt{2 x-7} \geq 0 ; \quad \sqrt{2 x-7}+1>0, \quad \sqrt{2 x-7}+7>0 \quad$ and $\sqrt{2 x-7}+7 \geq 7$, i.e., $\log _{5}(\sqrt{2 x-7}+7) \neq 0$. Multiplying both sides of the equation by $\log _{5}(\sqrt{2 x-7... | 5.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,771 |
$7.27 \lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$. | 7.27 We have $\lg (3-x)^{3}=\lg \left(27-x^{3}\right)$, which is equivalent to the system
$$
\left\{\begin{array} { l }
{ 3 - x > 0 , } \\
{ ( 3 - x ) ^ { 3 } = 2 7 - x ^ { 3 } , }
\end{array} \text { or } \left\{\begin{array}{l}
x<3, \\
(3-x)^{3}=(3-x)\left(9+3 x+x^{2}\right)
\end{array}\right.\right.
$$
Further, w... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,772 |
$7.28 \frac{2-\lg 4+\lg 0.12}{\lg (\sqrt{3 x+1}+4)-\lg 2 x}=1$.
$7.29 0.5\left(\lg \left(x^{2}-55 x+90\right)-\lg (x-36)\right)=\lg \sqrt{2}$. | 7.28 Given the domains of the logarithmic function and the square root, we have the system of inequalities $3 x+1 \geq 0$ (under this condition $\sqrt{3 x+1}+4>0$), $x>0$, the solution of which is $\boldsymbol{x}>0$. This equation is equivalent to the system
$$
\left\{\begin{array}{l}
x>0 \\
\lg (\sqrt{3 x+1}+4)-\lg 2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,773 |
$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$. | 7.30 For the existence of logarithms, it is necessary that the inequalities $x>0$, $\lg x>0$, $3 \lg x-2>0$ are satisfied simultaneously, i.e., $x>0, x>1, \lg x>\frac{2}{3}$. Hence, $\lg x>\frac{2}{3}$. Now we transition to the equivalent system of equations given by
$$
\left\{\begin{array}{l}
\lg x>\frac{2}{3} \\
\lg... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,774 |
7.31 $\log _{3}(x-3)^{2}+\log _{3}|x-3|=3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.31 $\log _{3}(x-3)^{2}+\log _{3}|x-3|=3$. | 7.31 Here $x \neq 3$. Writing the right-hand side of the equation as $3=3 \log _{3} 3=\log _{3} 3^{3}$, we transition to the equivalent equation $(x-3)^{2}|x-3|=3^{3}$ under the condition $x \neq 3$. Since $a^{2}=|a|^{2}$, we get $|x-3|^{3}=3^{3}$, or $|x-3|=3$, from which $x_{1}=0, x_{2}=6$.
Answer: $x_{1}=0, x_{2}=6... | x_{1}=0,x_{2}=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,775 |
$7.32 \log _{x}\left(9 x^{2}\right) \cdot \log _{3}^{2} x=4$. | 7.32 Here $x>0, x \neq 1$. After transformations, we bring the equation to the form
$\left(2 \log _{x} 3+2\right) \log _{3}^{2} x=4$,
from which, considering that $\log _{x} 3 \cdot \log _{3} x=1$, we get
$2 \log _{3} x+2 \log _{3}^{2} x=4$.
Solving this equation using the substitution $\log _{3} x=y$, we find $x_{... | x_{1}=\frac{1}{9},x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,776 |
$7.33 \log _{2} x+\log _{4} x+\log _{8} x=11$. | 7.33 Using formula (7.7), we get the equation
$$
\log _{2} x+\frac{\log _{2} x}{\log _{2} 4}+\frac{\log _{2} x}{\log _{2} 8}=11, \text { or } \log _{2} x\left(1+\frac{1}{2}+\frac{1}{3}\right)=11
$$
from which $\log _{2} x=6$, i.e., $x=64$.
Answer: $x=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,777 |
$7.36 \quad 3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$.
$7.37 \quad 25^{\log _{2} \sqrt{x+3}-0.5 \log _{2}\left(x^{2}-9\right)}=\sqrt{2(7-x)}$. | 7.36 Here
$3^{x}-5^{2-x}>0$.
Let's use the fact that $2-x=\log _{5} 5^{2-x}$, and write the equation in the form
$\log _{5}\left(2^{3} \cdot 5^{2-x}\right)=\log _{5}\left(3^{x}-5^{2-x}\right)$
Then we have
$2^{3} \cdot 5^{2-x}=3^{x}-5^{2-x} ; 5^{2-x}\left(2^{3}+1\right)=3^{x} ; 5^{2-x}=3^{x-2}$.
Finally, multiply... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,779 |
$7.38 x^{\log _{x^{2}}\left(x^{2}-1\right)}=5$.
$7.39\left\{\begin{array}{l}3^{2 x}-2^{y}=725 \\ 3^{x}-2^{\frac{y}{2}}=25\end{array}\right.$
$7.40\left\{\begin{array}{l}(x+y) 2^{y-2 x}=6.25, \\ (x+y)^{\frac{1}{2 x-y}}=5 .\end{array}\right.$
$7.41\left\{\begin{array}{l}2^{\frac{x-y}{2}}+2^{\frac{y-x}{2}}=2.5 \\ \lg (2... | 7.38 According to the indication $2^{0}$,
$x>0$.
Further, taking into account the domain of the logarithmic function, we get $x^{2}-1>0$, or $|x|>1$, from which
$$
x>1
$$
From (1) and (2) it follows that $x>1$ (at the same time, the restriction $x \neq 1$ imposed on the base of the logarithm is automatically satisf... | \sqrt{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,780 |
$7.43 \quad 5^{x} \sqrt[x]{8^{x-1}}=500$.
$7.44 \quad 5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$. | 7.43 According to the definition of the root, $x \neq 0$. Let's write the equation as
$$
5^{x} \cdot 2^{\frac{3 x-3}{x}}=5^{3} \cdot 2^{2}
$$
Dividing both sides by $5^{3} \cdot 2^{2} \neq 0$, we get
$$
5^{x-3} \cdot 2^{\frac{3 x-3}{x}-2}=1, \text { or } 5^{x-3} \cdot 2^{\frac{x-3}{x}}=1, \text { or }\left(5 \cdot 2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,781 |
$7.45 \sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$7.45 \sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$. | 7.45 We have
$|x-3|^{\frac{x+1}{4}}=|x-3|^{\frac{x-2}{3}}$
According to the hint $2^{0}$, the roots of this equation are the solutions of the system
$$
\left\{\begin{array} { l }
{ | x - 3 | > 0 , } \\
{ | x - 3 | \neq 1 , } \\
{ \frac { x + 1 } { 4 } = \frac { x - 2 } { 3 } }
\end{array} \text { , i.e. } \left\{\b... | x_{1}=2,x_{2}=4,x_{3}=11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,782 |
7.54 It is known that $\beta=10^{\frac{1}{1-\lg \alpha}}$ and $\gamma=10^{\frac{1}{1-\operatorname{lg} \beta}}$. Find the dependence of $\alpha$ on $\gamma$. | 7.54 By logarithmizing the equalities with the base, we get $\lg \beta=\frac{1}{1-\lg \alpha}, \lg \gamma=\frac{1}{1-\lg \beta}$. Therefore, $\lg \gamma=\frac{1}{1-\frac{1}{1-\lg \alpha}}=\frac{\lg \alpha-1}{\lg \alpha}$,
from which $\lg \alpha=\frac{1}{1-\lg \gamma}$. According to the definition of the logarithm, we ... | \alpha=10^{\frac{1}{1-\lg\gamma}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,786 |
7.55 Find $\log _{30} 8$, given that $\lg 5=a$ and $\lg 3=b$.
Solve the equations (7.56-7.74): | 7.55 Using formulas (7.6), (7.7), (7.4), $(7.5)$ sequentially, we get
$$
\log _{30} 8=3 \log _{30} 2=\frac{3 \lg 2}{\lg 30}=\frac{3 \lg \frac{10}{5}}{\lg (3 \cdot 10)}=\frac{3(1-\lg 5)}{\lg 3+1}=\frac{3(1-a)}{b+1}
$$
Answer: $\frac{3(1-a)}{b+1}$. | \frac{3(1-)}{b+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,787 |
$7.57 \quad 4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$. | 7.57 Given that $\lg x^{2}=2 \lg x$ (since $x>0$), we will transform the given equation as follows:
$$
\begin{aligned}
& 2^{2 \lg x+2}=2^{\lg x} \cdot 3^{\lg x}-2 \cdot 3^{2 \lg x+2}=0, \text { or } \\
& 4 \cdot 2^{2 \lg x}-2^{\lg x} \cdot 3^{\lg x}-18 \cdot 3^{2 \lg x}=0 .
\end{aligned}
$$
Divide both sides of the l... | 0.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,788 |
7.58 $2 \lg x^{2}-\lg ^{2}(-x)=4$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.58 $2 \log x^{2}-\log ^{2}(-x)=4$. | 7.58 Here it should be $-x>0$, hence $x<0$. Then $\lg x^{2}=2 \lg (-x)$ and the given equation will take the form $4 \lg (-x)-\lg ^{2}(-x)=4$.
Letting $\lg (-x)=y$, we find $y=2$, hence $\lg (-x)=2$, i.e. $-x=100, x=-100$.
Answer: $x=-100$. | -100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,789 |
$7.59 \quad \lg ^{4}(x-1)^{2}+\lg ^{2}(x-1)^{3}=25$. | 7.59 Here $x-1>0$, hence $x>1$. Using the fact that
$\lg (x-1)^{2}=2 \lg (x-1), \lg (x-1)^{3}=3 \lg (x-1)$, we can rewrite the equation as
$$
16 \lg ^{4}(x-1)+9 \lg ^{2}(x-1)=25
$$
Let $\lg ^{2}(x-1)=y \geq 0$, then we get the equation $16 y^{2}+9 y-25=0$, which has roots $y_{1}=1, y_{2}=-\frac{25}{16}$ (not suitable... | x_{1}=11,x_{2}=1.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,790 |
$7.60 \log _{2}(2-x)-\log _{2}(2-\sqrt{x})=\log _{2} \sqrt{2-x}-0.5$. | 7.60 Given the domain of the logarithmic function, we have
$$
\left\{\begin{array} { l }
{ 2 - x > 0 } \\
{ 2 - \sqrt { x } > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x < 2 \\
x < 4
\end{array}\right.
\right.
$$
From the above, we get \(0 < x < 2\) (since \(x < 2\) is the stricter condition). Given the eq... | x_{1}=0,x_{2}=\frac{16}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,791 |
7.61 $\log _{2} 3+2 \log _{4} x=x^{\frac{\log _{9} 16}{\log _{3} x}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.61 $\log _{2} 3+2 \log _{4} x=x^{\frac{\log _{9} 16}{\log _{3} x}}$. | 7.61 Here $x>0$. Simplify the right side of the equation
$$
x^{\frac{\log _{2} 16}{\log _{3} x}}=\left(x^{\log _{x} 3}\right)^{\log _{3} 4}=3^{\log _{3} 4}=4
$$
Now write the equation in the form
$\log _{4} 9+\log _{4} x^{2}=4 \log _{4} 4$, or $\log _{4}\left(9 x^{2}\right)=\log _{4} 4^{4}$,
i.e., $9 x^{2}=4^{4}$, ... | \frac{16}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,792 |
7.62 $\log _{\sqrt{3}} x+\log _{\sqrt[4]{3}} x+\log _{\sqrt[6]{3}} x+\ldots+\log _{\sqrt[6]{3}} x=36$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.62 $\log _{\sqrt{3}} x+\log _{\sqrt[4]{3}} x+\log _{\sqrt[6]{3}} x+\ldot... | 7.62 Applying formulas (7.8) and (7.6), we get
$$
\begin{aligned}
& \frac{1}{\log _{x} \sqrt{3}}+\frac{1}{\log _{x} \sqrt[4]{3}}+\frac{1}{\log _{x} \sqrt[6]{3}}+\ldots+\frac{1}{\log _{x} \sqrt[16]{3}}=36 \\
& \frac{2}{\log _{x} 3}+\frac{4}{\log _{x} 3}+\frac{6}{\log _{x} 3}+\ldots+\frac{16}{\log _{x} 3}=36 \\
& \frac{... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,793 |
$7.65 \quad 3^{\log _{3}^{2} x}+x^{\log _{3} x}=162$. | 7.65 We have $3^{\log _{3}^{2} x}=\left(3^{\log _{3} x}\right)^{\log _{3} x}=x^{\log _{3} x}$. Then the equation will be
$2 x^{\log _{3} x}=162$, or $x^{\log _{3} x}=81$.
Since $x>0$, we can take the logarithm of both sides with base 3:
$$
\log _{3}^{2} x=4 \Rightarrow \log _{3} x= \pm 2 \Rightarrow x_{1}=\frac{1}{9... | x_{1}=\frac{1}{9},x_{2}=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,795 |
$7.66 x^{2 \lg ^{2} x}=10 x^{3}$. | 7.66 Instruction. Take the logarithm of both sides of the equation to the base 10 and solve the resulting cubic equation by factoring.
Answer: $x_{1}=0.1, x_{2,3}=10^{0.5(1 \pm \sqrt{3})}$. | x_{1}=0.1,x_{2,3}=10^{0.5(1\\sqrt{3})} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,796 |
$7.68 \log _{a} \sqrt{4+x}+3 \log _{a^{2}}(4-x)-\log _{a^{4}}\left(16-x^{2}\right)^{2}=2$.
For what values of $a$ does the equation have a solution? | 7.68 Given the domain of the logarithmic function, the square root, and the restrictions imposed on the base of the logarithm, we conclude that
$$
\left\{\begin{array} { l }
{ 4 + x > 0 , } \\
{ 4 - x > 0 , } \\
{ a > 0 , a \neq 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
-4 < x < 4, \\
a > 0, a \neq 1
\end{a... | 4-^{2}, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,797 |
$7.69 \log _{4 x+1} 7+\log _{9 x} 7=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$7.69 \log _{4 x+1} 7+\log _{9 x} 7=0$. | 7.69 Due to the restrictions imposed on the base of the logarithm, we have the system of inequalities $4 x+1>0$, $4 x+1 \neq 1$, $9 x>0$, $9 x \neq 1$, from which
$x>0, x \neq \frac{1}{9}$.
We will transform the given equation to the form
$\frac{1}{\log _{7}(4 x+1)}+\frac{1}{\log _{7}(9 x)}=0$.
Multiplying both sid... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,798 |
7.71 $\sqrt{\log _{x} \sqrt{5 x}}=-\log _{x} 5$. | 7.71 Since the left side of the equation is non-negative (arithmetic root), the inequality $-\log _{x} 5>0$ must be satisfied, i.e., $\log _{x} 5<0$, from which $0<x<1$. Squaring both sides of the equation:
$\log _{x} \sqrt{5 x}=\log _{x}^{2} 5$, or $\frac{1}{2} \log _{x} 5+\frac{1}{2}=\log _{x}^{2} 5$.
Let $\log _{x... | 0.04 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,799 |
7.72 $\log _{\sqrt{3}} x \cdot \sqrt{\log _{\sqrt{3}} 3-\log _{x} 9}+4=0$.
Chapter 7. Logarithms. Exponential and Logarithmic Equations 221 | 7.72 For the sum on the left side of the equation to be zero, it is necessary that the first term be negative, i.e., $\log _{\sqrt{3}} x<0$. From this, it follows that
$0<x<1$.
Under this condition, $\log _{x} 9<0$ and, therefore, the expression under the square root is positive. Let's move 4 to the right side of the... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,800 |
$7.73 \sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$.
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$7.73 \sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$. | 7.73 Given the domains of the logarithmic function and the square root, we obtain the system of inequalities
$$
\left\{\begin{array}{l}
x>0 \\
\log _{0.04} x+1 \geq 0 \\
\log _{0.2} x+3 \geq 0
\end{array}\right.
$$
We use the fact that $\log _{0.2} x=\log _{0.04} x^{2}=2 \log _{0.04} x$ (since $x>0$). Then, setting $... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,801 |
7.77 $\left\{\begin{array}{l}\left(x^{2}+y\right) 2^{y-x^{2}}=1, \\ 9\left(x^{2}+y\right)=6^{x^{2}-y} .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\left(x^{2}+y\right) 2^{y-x^{2}}=1, \\
9\left(x^{2}+y\right)=6^{x^{2}-y} .
\end{array}\right.
\] | 7.77 Multiplying both sides of the second equation by $6^{y-x^{2}}>0$, we get the system
$$
\left\{\begin{array}{l}
\left(x^{2}+y\right) \cdot 2^{y-x^{2}}=1 \\
9\left(x^{2}+y\right) \cdot 2^{y-x^{2}} \cdot 3^{y-x^{2}}=1
\end{array}\right.
$$
Dividing the second equation by the first term by term:
$9 \cdot 3^{y-x^{2}... | (\sqrt{3};1)(-\sqrt{3};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,803 |
$7.78 \quad\left\{\begin{array}{l}y-\log _{3} x=1 \\ x^{y}=3^{12}\end{array}\right.$
$7.79\left\{\begin{array}{l}\log _{x}(3 x+2 y)=2, \\ \log _{y}(2 x+3 y)=2 .\end{array}\right.$ | 7.78 Instruction. Take the logarithm of the second equation to the base 3.
Answer: $(27 ; 4),\left(\frac{1}{81} ;-3\right)$. | (27;4),(\frac{1}{81};-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,804 |
$7.80 \quad\left\{\begin{array}{l}x^{x-2 y}=36, \\ 4(x-2 y)+\log _{6} x=9\end{array}\right.$
(find only integer solutions). | 7.80 Here $x>0$. By taking the logarithm of the first equation to the base 6, we arrive at the system
$$
\left\{\begin{array} { l }
{ ( x - 2 y ) \log _ { 6 } x = 2 , } \\
{ 4 ( x - 2 y ) + \log _ { 6 } x = 9 , }
\end{array} \text { or } \left\{\begin{array}{l}
4(x-2 y) \log _{6} x=8 \\
4(x-2 y)+\log _{6} x=9
\end{ar... | (6;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,805 |
8.2 Prove that if $a>0$ and $b>0$, then
$$
\frac{2 \sqrt{a b}}{\sqrt{a}+\sqrt{b}} \leq \sqrt[4]{a b}
$$ | 8.2 Let's consider the difference between the left and right sides of the inequality:
$$
\begin{aligned}
& \frac{2 \sqrt{a b}}{\sqrt{a}+\sqrt{b}}-\sqrt[4]{a b}=\frac{2 \sqrt{a b}-\sqrt[4]{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}= \\
= & \frac{\sqrt[4]{a b}(2 \sqrt[4]{a b}-\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}=-\fr... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 49,806 |
$8.4 \quad \frac{(x+3)(5-x)}{2 x-5}>0$.
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$8.4 \quad \frac{(x+3)(5-x)}{2 x-5}>0$. | 8.4 The roots of the equations $(x+3)(5-x)=0$ and $2 x-5=0$ are the numbers $x_{1}=-3, x_{2}=2.5, x_{3}=5$, which are not solutions to the given inequality. Therefore, on the number line, we mark them with light circles (Fig. 8.4). These points divide the number line into four intervals. It is easy to determine that fo... | (-\infty,-3)\cup(2.5,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,808 |
8.5 $\quad \frac{x^{2}(2 x-9)(x-1)^{3}}{(x+4)^{5}(2 x-6)^{4}} \leq 0$. | 8.5 Assuming $x \neq 0$ and $x \neq 3$, divide both sides of the inequality by the positive fraction $\frac{x^{2}}{(2 x-6)^{4}}$ and immediately note that $x=0$ satisfies the given inequality, while $x=3$ does not. Moreover, replace the factors with odd exponents with the corresponding first-degree factors (it is clear... | (-\infty,-4)\cup[1,3)\cup(3,4.5]\cup{0} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,809 |
8.6 $\quad \frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$. | 8.6 Let's move 1 to the left side of the inequality and perform the transformations:
$$
\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1 \Rightarrow \frac{x^{2}-3 x+2}{x^{2}+3 x+2}-1 \geq 0 \Rightarrow \frac{6 x}{(x+1)(x+2)} \leq 0
$$
We apply the method of intervals and, using Fig. 8.6, we establish that \( x < -2 \) or \( -1... | (-\infty,-2)\cup(-1,0] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,810 |
$8.8 \quad \frac{1}{x^{2}-5 x+6} \leq \frac{1}{2}$. | 8.8 Transform the given inequality into an equivalent one:
$$
\frac{1}{x^{2}-5 x+6}-\frac{1}{2} \leq 0 ; \frac{x^{2}-5 x+4}{2\left(x^{2}-5 x+6\right)} \geq 0
$$
The roots \( x_{1}=1, x_{2}=4 \) of the equation \( x^{2}-5 x+4=0 \) are solutions to the inequality (on Fig. 8.7, we mark them with filled circles); the roo... | (-\infty,1]\cup(2,3)\cup[4,\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,812 |
$8.9 \quad \frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$.
$8.9 \quad \frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$. | 8.9 Factoring the numerator and denominator, we get $\frac{\left(x^{2}-4\right)\left(x^{2}+2\right)}{(x+1)^{2}}0$ for $x \neq-1$, and $x^{2}+2>0$ for any $x$. Solving the inequality $x^{2}-4<0$, we find $-2<x<2$. For the final answer, we need to consider that $x \neq-1$.
Answer: $(-2,-1) \cup(-1,2)$. | (-2,-1)\cup(-1,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,813 |
8.10 $x^{6}-9 x^{3}+8>0$.
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8.10 $x^{6}-9 x^{3}+8>0$. | 8.10 Hint. Introduce an auxiliary variable $x^{3}=y$.
Answer: $(-\infty, 1) \cup(2, \infty)$. | (-\infty,1)\cup(2,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,814 |
$8.13 \frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$.
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$8.13 \frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$. | 8.13 The given inequality can only be considered under the condition $4 x^{2}-19 x+12>0$, from which we find that $x<0.75$ or $x>4$.
For these values of $x$, we have $\sqrt{4 x^{2}-19 x+12}>0$ and it is sufficient to solve the inequality $x-7<0$, i.e.:
$$
x<7
$$
From the system of inequalities (1), (2) we obtain tha... | (-\infty;0.75)\cup(4,7) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,816 |
$8.14 \frac{\sqrt{x}-3}{x-2}>0$.
Solve the inequality:
$8.14 \frac{\sqrt{x}-3}{x-2}>0$. | 8.14 Given that $x$ cannot take negative values, we will divide the points $x_{1}=2$ (root of the equation $x-2=0$) and $x_{2}=9$ (root of the equation $\sqrt{x}-3=0$) into intervals, but not the entire number line, only its part $[0, \infty)$. Using the sign curve (Fig. 8.9), we find the solutions to the given inequal... | [0,2)\cup(9,\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,817 |
$8.16 x-3<\sqrt{x-2}$.
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$8.16 x-3<\sqrt{x-2}$. | 8.16 According to the indication $6^{0}$, this irrational inequality is equivalent to the following combination of two systems of inequalities:
$$
\left\{\begin{array} { l }
{ x - 2 \geq 0 } \\
{ x - 3 \geq 0 } \\
{ x ^ { 2 } - 6 x + 9 < x - 2 }
\end{array} \quad \left\{\begin{array}{l}
x-2 \geq 0 \\
x-3<0
\end{array... | [2,\frac{7+\sqrt{5}}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,818 |
$8.26 \quad \log _{0.2}^{2}(x-1)^{2}>4$.
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$8.26 \quad \log _{0.2}^{2}(x-1)^{2}>4$. | 8.26 The given inequality can be rewritten as $\left|\log _{0.2}(x-1)\right|>2$, from which we have either.
$\log _{0.2}(x-1) > 2$ or $\log _{0.2}(x-1) < -2$.
Solving inequality (1). Since $-2=\log _{0.2} 0.2^{-2}$, taking into account that $0.2^2 = 0.04$, we get $x-1 < 0.04$, i.e., $x < 1.04$. Solving inequality (2)... | (1;1.04)\cup(26,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,824 |
$8.27 \quad \log _{2}\left(1+\log _{\frac{1}{9}} x-\log _{9} x\right)<1$.
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$8.27 \quad \log _{2}\left(1+\log _{\frac{1}{9}} x-\log _{9} x\right)<1$. | 8.27 Noticing that $\log _{\frac{1}{9}} x=-\log _{9} x$, we can rewrite the given inequality as
$\log _{2}\left(1-2 \log _{9} x\right)0 \quad$ (the inequality $\log _{2}\left(1-2 \log _{9} x\right)0$ is automatically satisfied).
Answer: $\left(\frac{1}{3}, 3\right)$. | (\frac{1}{3},3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,825 |
$8.28 \log _{\pi}(x+27)-\log _{\pi}(16-2 x)<\log _{\pi} x$. | 8.28 The given inequality is equivalent to the system
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x + 27 > 0 , } \\
{ 16 - 2 x > 0 , } \\
{ x > 0 , } \\
{ \operatorname { log } _ { \pi } \frac { x + 27 } { 16 - 2 x } - 27 \\
x > 0, \\
\frac{x + 27}{16 - 2 x} > 0
\end{array}\right.\right.
\end{aligned}
$$
Solvi... | (3;4.5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,826 |
$8.29 \frac{1}{2}+\log _{9} x-\log _{3}(5 x)>\log _{\frac{1}{3}}(x+3)$.
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$8.29 \frac{1}{2}+\log _{9} x-\log _{3}(5 x)>\log _{\frac{1}{3}}(x+3)$. | 8.29 We will convert all terms to logarithms with base 9:
$$
\begin{aligned}
& \frac{1}{2}=\log _{9} 9^{\frac{1}{2}}=\log _{9} 3 ; \log _{3}(5 x)=\log _{9}\left(25 x^{2}\right) \\
& \log _{\frac{1}{3}}(x+3)=-\log _{3}(x+3)=-\log _{9}(x+3)^{2}=\log _{9}(x+3)^{-2}
\end{aligned}
$$
$$
\left\{\begin{array}{l}
x>0 \\
x+3>... | (0,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,827 |
$8.30 \log _{\frac{x-1}{x+5}} 0.3>0$.
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However, the provided text is already in a mathematical form which is universally understood and does not require translation. The text in... | 8.30 According to the condition, the logarithm of the number 0.3, which is less than 1, is positive. This can only be the case if the base of the logarithm is a positive number less than 1 (Fig. 8.13). Thus, we have the system of inequalities $0 < a < 1$.
Answer: $(1, \infty)$.
 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,828 |
$8.34 \log _{x} \log _{9}\left(3^{x}-9\right)<1$.
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$8.34 \log _{x} \log _{9}\left(3^{x}-9\right)<1$. | 8.34 This inequality can be considered under the conditions
$$
\left\{\begin{array} { l }
{ 3 ^ { x } - 9 > 0 , } \\
{ \operatorname { log } _ { 9 } ( 3 ^ { x } - 9 ) > 0 , }
\end{array} \text { or } \left\{\begin{array} { l }
{ 3 ^ { x } > 3 ^ { 2 } , } \\
{ 3 ^ { x } - 9 > 1 , }
\end{array} \text { or } \left\{\be... | (\log_{3}10,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,830 |
$8.37 y=\sqrt{-\frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}}}$.
## Group 5
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 8.37 The following conditions must be satisfied
$$
\left\{\begin{array}{l}
x-1>0 \\
-x^{2}+2 x+8>0 \\
-\frac{\log _{0.3}(x-1)}{\sqrt{-x^{2}+2 x+8}} \geq 0
\end{array}\right.
$$
From the second inequality, it follows that $\sqrt{-x^{2}+2 x+8}>0$, and thus, in the third inequality, $\log _{0.3}(x-1) \leq 0$. Therefore,... | [2,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,833 |
8.38 Prove that the product of the sum of three positive numbers and the sum of their reciprocals is not less than 9. | 8.38 Here it is required to prove that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9$ under the condition $a>0, b>0, c>0$. Let's consider the difference between the left and right sides of the inequality:
$$
\begin{aligned}
& (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9= \\
& =1+\frac{b}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 49,834 |
8.39 Prove that if $a$ is any real number, then the inequality $\frac{a^{2}+a+2}{\sqrt{a^{2}+a+1}} \geq 2$ holds. | 8.39 Here $a^{2}+a+1>0$ for any $a$. Let's transform the left part of the inequality:
$$
\begin{aligned}
& \frac{a^{2}+a+2}{\sqrt{a^{2}+a+1}}=\frac{a^{2}+a+1}{\sqrt{a^{2}+a+1}}+\frac{1}{\sqrt{a^{2}+a+1}}= \\
& =\sqrt{a^{2}+a+1}+\frac{1}{\sqrt{a^{2}+a+1}}
\end{aligned}
$$
this is the sum of two positive reciprocal num... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 49,835 |
8.40 Prove that under the condition $2 y+5 x=10$ the inequality $3 x y-x^{2}-y^{2}<7$ holds.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 8.40 Instruction. Express $y=0.5(10-5 x)$, substitute this expression into the inequality and consider the difference between
the left and right sides of the inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 49,836 |
8.41 Show that for any real values of $x$ the function $y=\frac{x^{2}+x+1}{x^{2}+1}$ cannot take values greater than $\frac{3}{2}$ and less than $\frac{1}{2}$. | 8.41 We need to prove that
$$
\frac{1}{2} \leq \frac{x^{2}+x+1}{x^{2}+1} \leq \frac{3}{2}, x \in R
$$
If this inequality is true, then multiplying all its parts by $2\left(x^{2}+1\right)>0$, we get the true inequality
$$
x^{2}+1 \leq 2\left(x^{2}+x+1\right) \leq 3\left(x^{2}+1\right)
$$
Subtracting $x^{2}+1$ from a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 49,837 |
8.42 Find the integers $x$, satisfying the inequality $\left|\frac{2}{x-13}\right|>\frac{8}{9}$ | 8.42 Here, the condition $x-13 \neq 0$ must be satisfied, i.e., $x \neq 13$. At the same time, $\left|\frac{2}{x-13}\right|>0$ and this inequality is equivalent to the following: $\left|\frac{x-13}{2}\right|<\frac{9}{8}$. Solving it, we get
$$
-\frac{9}{8}<\frac{x-13}{2}<\frac{9}{8} ;-9<4 x-52<9 ; 43<4 x<61
$$
Thus, ... | 11;12;14;15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,838 |
8.43 For what values of $p$ are both roots of the quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ negative? | 8.43 The quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ has roots if $D=4(p+1)^{2}-4(9 p-5) \geq 0$. Let $x_{1}$ and $x_{2}$ be the roots of the trinomial; then by the condition $x_{1}0$, $x_{1}+x_{2}=-2(p+1) 0 , } \\
{ p + 1 > 0 , }
\end{array} \text { or } \left\{\begin{array}{l}
p^{2}-7 p+6 \geq 0 \\
p>\frac{5}{9} \\
p>... | \frac{5}{9}<p\leq1orp\geq6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,839 |
8.44 Find all values of $a$ for which the expression
$$
\sqrt{(a+1) x^{2}-2(a-1) x+3 a-3}
$$
makes sense for any $x \in \boldsymbol{R}$. | 8.44 The expression is defined under the condition $(a+1) x^{2}-2(a-1) x+3 a-3 \geq 0$.
1) Let $a+1=0$, i.e., $a=-1$. Then the inequality
will take the form $4 x-6 \geq 0$, from which $x \geq \frac{3}{2}$, i.e., it is not satisfied for all $x$.
2) Let $a+1 \neq 0$. Then the inequality (1) is satisfied for any $x$ if... | 1\leq\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,840 |
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