problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
9.107. Prove that if $4 b+a=1$, then the inequality $a^{2}+4 b^{2} \geq \frac{1}{5}$ holds. | Solution.
We have
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 4 b + a = 1 , } \\
{ a ^ { 2 } + 4 b ^ { 2 } \geq \frac { 1 } { 5 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ a = 1 - 4 b , } \\
{ ( 1 - 4 b ) ^ { 2 } + 4 b ^ { 2 } \geq \frac { 1 } { 5 } }
\end{array} \Leftrightarrow \left\{\begin... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 48,694 |
9.108. Prove that the polynomial $m^{6}-m^{5}+m^{4}+m^{2}-m+1$ takes positive values for all real values of $m$. | Solution.
Rewrite the given polynomial as
$$
\begin{aligned}
& m^{6}+m^{2}-\left(m^{5}+m\right)+\left(m^{4}+1\right)=m^{2}\left(m^{4}+1\right)-m\left(m^{4}+1\right)+\left(m^{4}+1\right)= \\
& =\left(m^{4}+1\right)\left(m^{2}-m+1\right) . \text { Since } m^{4}+1>0 \text { and } m^{2}-m+1>0 \text { for } m \in R
\end{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,695 |
9.109. Find the domain of the function $f$, if
$f(x)=\sqrt[6]{4^{\frac{x+1}{x}}-17 \cdot 2^{\frac{1}{x}}+4}$ | ## Solution.
We have the inequality for $x \neq 0$:
$$
4^{\frac{x+1}{x}}-17 \cdot 2^{\frac{1}{x}}+4 \geq 0 \Leftrightarrow 4 \cdot\left(2^{\frac{1}{x}}\right)^{2}-17 \cdot 2^{\frac{1}{x}}+4 \geq 0
$$
We solve it as a quadratic inequality in terms of $2^{\frac{1}{x}}$. We get
$\left[\begin{array}{l}\frac{1}{2^{x}} \g... | x\in[-\frac{1}{2};0)\cup(0;\frac{1}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,696 |
9.110. Find the domain of the function $f$, if
$f(x)=\sqrt{9-\left(\frac{4 x-22}{x-5}\right)^{2}}$ | ## Solution.
We obtain the inequality $9-\left(\frac{4 x-22}{x-5}\right)^{2} \geq 0 \Leftrightarrow\left(\frac{4 x-22}{x-5}\right)^{2} \leq 9 \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow\left|\frac{4 x-22}{x-5}\right| \leq 3 \Leftrightarrow-3 \leq \frac{4 x-22}{x-5} \leq 3 \Leftrightarrow\left\{\begin{array}... | x\in[\frac{37}{7};7] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,697 |
9.111. Find the non-negative integer values of $x$ that satisfy the inequality $\frac{x+3}{x^{2}-4}-\frac{1}{x+2}<\frac{2 x}{2 x-x^{2}}$. | Solution.
Rewrite the inequality as $\frac{x+3}{x^{2}-4}-\frac{1}{x+2}+\frac{2 x}{x(x-2)}<0 \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow \frac{x+3}{(x-2)(x+2)}-\frac{1}{x+2}+\frac{2}{x-2}<0 \text { for } x \neq 0: \Leftrightarrow\left\{\begin{array}{l}
\frac{2 x+9}{(x-2)(x+2)}<0, \\
x \neq 0
\end{array} \Lef... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,698 |
9.112. For what values of $a$ is the inequality $\frac{a x}{x^{2}+4}<1$ true? If $x>0$, then we have $a x < x^{2} + 4$. This inequality holds for any values of $x \in R$ if $D=a^{2}-36<0$, $a^{2}<36$, $-6<a<6$.
Answer: $a \in(-6 ; 6)$.
9.113. Find the domain of the function $f$, if $f(x)=\sqrt{\log _{0.5}\left(x^{2}-... | ## Solution.
The domain of the given function will be found by solving the inequality
$$
\begin{aligned}
& \log _{0.5}\left(x^{2}-9\right)+4 \geq 0, \log _{0.5}\left(x^{2}-9\right) \geq-4 \Leftrightarrow \\
& \Leftrightarrow 09\end{aligned} \Leftrightarrow\left\{\begin{array}{l}-5 \leq x \leq 5, \\ {\left[\begin{arra... | x\in[-5;-3)\cup(3;5] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,699 |
9.114. For what values of $x$ is the following expression defined
$$
\log _{3}\left(1-\log _{0,5}\left(x^{2}-2 x-2.5\right)\right) ?
$$ | ## Solution.
We obtain the inequality
$$
\begin{aligned}
& 1-\log _{0.5}\left(x^{2}-2 x-2.5\right)>0, \log _{0.5}\left(x^{2}-2 x-2.5\right)0.5, x^{2}-2 x-3>0,(x+1)(x-3)>0 \Leftrightarrow \\
& \Leftrightarrow(x \in(-\infty ;-1) \cup(3 ;+\infty))
\end{aligned}
$$
Answer: $x \in(-\infty ;-1) \cup(3 ;+\infty)$. | x\in(-\infty;-1)\cup(3;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,700 |
9.115. Find the values of $m$ for which the inequality
$$
\frac{x^{2}-8 x+20}{m x^{2}+2(m+1) x+9 m+4}<0
$$
is satisfied for all real values of $x$. | ## Solution.
Since $x^{2}-8 x+20>0$ for $x \in R$, it is necessary that
$$
m x^{2}+2(m+1) x+9 m+40, \\
m<0
\end{array} \Leftrightarrow m<-\frac{1}{2}\right.\right.
$$
Answer: $m \in\left(-\infty ;-\frac{1}{2}\right)$. | \in(-\infty;-\frac{1}{2}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,701 |
9.116. For what values of $x$ does the difference $\frac{11 x^{2}-5 x+6}{x^{2}+5 x+6}-x$ take only negative values? | ## Solution.
We have
$$
\begin{aligned}
& \frac{11 x^{2}-5 x+6}{x^{2}+5 x+6}-x0 \Leftrightarrow \\
& \Leftrightarrow \frac{(x-1)(x-2)(x-3)}{(x+3)(x+2)}>0 \Leftrightarrow(x-1)(x-2)(x-3)(x+3)(x+2)>0
\end{aligned}
$$
Using the interval method, we get $x \in(-3 ;-2) \bigcup(1 ; 2) \cup(3 ; \infty)$.
\cup(1;2)\cup(3;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,702 |
9.117. For what values of $m$ is the inequality $\frac{x^{2}+m x-1}{2 x^{2}-2 x+3}<1$ satisfied for all $x$? | ## Solution.
From the condition we get
$$
\begin{aligned}
& \frac{x^{2}+m x-1}{2 x^{2}-2 x+3}-10
\end{aligned}
$$
Since $2 x^{2}-2 x+3>0$ for $x \in R$, then $x^{2}-(m+2) x+4>0$ for $x \in R \Leftrightarrow$ $\Leftrightarrow(m+2)^{2}-16<0,(m+2)^{2}<16,-4<m+2<4,-6<m<2$.
Answer: $m \in(-6 ; 2)$. | \in(-6;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,703 |
9.118. For what values of $m$ is the inequality $\frac{x^{2}-m x-2}{x^{2}-3 x+4}>-1$ satisfied for all $x$? | Solution.
From the condition, we have $\frac{2 x^{2}-(m+3) x+2}{x^{2}-3 x+4}>0$. Since $x^{2}-3 x+4>0$ for $x \in R$, then $2 x^{2}-(m+3) x+2>0$ for any $x$, from which $D=(m+3)^{2}-16<0$, $(m+3)^{2}<16, -4<m+3<4, -7<m<1$.
Answer: $m \in(-7 ; 1)$. | \in(-7;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,704 |
9.119. For what values of $a$ does the sum $a+\frac{-1+9 a+4 a^{2}}{a^{2}-3 a-10}$ take only positive values | Solution.
Let's write down the inequality
$$
\begin{aligned}
& a+\frac{-1+9 a+4 a^{2}}{a^{2}-3 a-10}>0, \frac{a^{3}-3 a^{2}-10 a-1+9 a+4 a^{2}}{a^{2}-3 a-10}>0 \\
& \frac{a^{3}+a^{2}-a-1}{a^{2}-3 a-10}>0, \frac{a^{2}(a+1)-(a+1)}{(a+2)(a-5)}>0, \frac{(a+1)\left(a^{2}-1\right)}{(a+2)(a-5)}>0 \\
& (a+1)^{2}(a-1)(a+2)(a-... | \in(-2;-1)\cup(-1;1)\cup(5;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,705 |
9.120. Find the integer values of $x$ that satisfy the inequality $\log _{4} x+\log _{2}(\sqrt{x}-1)<\log _{2} \log _{\sqrt{5}} 5$. | ## Solution.
Let's switch to base 2. We have $\frac{1}{2} \log _{2} x+\log _{2}(\sqrt{x}-1) 0 , } \\
{ \sqrt { x - 1 } > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(\sqrt{x})^{2}-\sqrt{x}-20, \\
\sqrt{x}>1
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
-11
\end{array} \Leftrightarro... | x_{1}=2;x_{2}=3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,706 |
9.121. 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 $\frac{1}{2}$. | Solution.
Let, by contradiction, assume
1) $\frac{x^{2}+x+1}{x^{2}+1}>\frac{3}{2} \Leftrightarrow 2 x^{2}+2 x+2>3 x^{2}+3 \Leftrightarrow x^{2}-2 x+1<0 \Leftrightarrow$
$\Leftrightarrow(x-1)^{2}<0$, no solutions;
2) $\frac{x^{2}+x+1}{x^{2}+1}<\frac{1}{2} \Leftrightarrow 2 x^{2}+2 x+2<x^{2}+1 \Leftrightarrow x^{2}+2 ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 48,707 |
9.123. $y=\frac{\sqrt{4 x-x^{2}}}{\log _{3}|x-4|}$. | ## Solution.
We obtain
$$
\left\{\begin{array} { l }
{ 4 x - x ^ { 2 } \geq 0 , } \\
{ \operatorname { log } _ { 3 } | x - 4 | \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x ( x - 4 ) \leq 0 , } \\
{ | x - 4 | \neq 1 , } \\
{ x - 4 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
0 ... | x\in[0;3)\cup(3;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,708 |
9.124. $y=\log _{3}\left(0.64^{2-\log _{\sqrt{2}} x}-1.25^{8-\left(\log _{2} x\right)^{2}}\right)$. | ## Solution.
Consider the inequality $0.64^{2-\log _{\sqrt{2}} x}-1.25^{8-\left(\log _{2} x\right)^{2}}>0 \Leftrightarrow$
$$
\Leftrightarrow 0.64^{2-\log _{\sqrt{2}} x}>1.25^{8-\left(\log _{2} x\right)^{2}} \Leftrightarrow\left(\frac{4}{5}\right)^{4-2 \log _{\sqrt{2}} x}>\left(\frac{4}{5}\right)^{\left(\log _{2} x\r... | x\in(0;\frac{1}{64})\cup(4;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,709 |
9.125. $y=\sqrt{\log _{\frac{1}{3}} \log _{3}|x-3|}$. | ## Solution.
The domain of definition of the given function will be those values of $x$ for which the inequality
$$
\begin{aligned}
& \log _{\frac{1}{3}} \log _{3}|x-3| \geq 0 \Leftrightarrow 0 1 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ - 3 \leq x - 3 \leq 3 , } \\
{ [ \begin{array} { l }
{ x - 3 ... | x\in[0;2)\cup(4;6] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,710 |
9.126. $y=\sqrt{\log _{\frac{1}{2}}^{2}(x-3)-1}$. | ## Solution.
We obtain the inequality $\log _{\frac{1}{2}}^{2}(x-3)-1 \geq 0, \log _{\frac{1}{2}}^{2}(x-3) \geq 1 \Leftrightarrow$
$$
\Leftrightarrow\left[\begin{array} { l }
{ \log _ { \frac { 1 } { 2 } } ( x - 3 ) \geq 1 , } \\
{ \log _ { \frac { 1 } { 2 } } ( x - 3 ) \leq - 1 }
\end{array} \Leftrightarrow \left[\... | x\in(3;\frac{7}{2}]\cup[5;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,711 |
9.127. $y=\sqrt[4]{2-\lg |x-2|}$.
The translation is as follows:
9.127. $y=\sqrt[4]{2-\lg |x-2|}$.
This is the direct translation, preserving the original format and line breaks. | ## Solution.
We have the inequality $2-\lg |x-2| \geq 0, \lg |x-2| \leq 2 \Leftrightarrow 0 < |x-2| \leq 100$
\[
\begin{array}{l}
\Leftrightarrow \left\{\begin{array}{l}
0 < |x-2| \leq 100 \\
x - 2 \neq 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-100 \leq x - 2 \leq 100, \\
x - 2 \neq 0
\end{array} \Leftrig... | x\in[-98;2)\cup(2;102] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,712 |
9.128. $y=\log _{3}\left(2^{\log _{x-3} 0.5}-1\right)+\frac{1}{\log _{3}(2 x-6)}$. | ## Solution.
The domain of the given function will be found by solving the system of inequalities
$$
\left\{\begin{array} { l }
{ 2 ^ { \operatorname { log } _ { x - 3 } 0.5 } - 1 > 0 , } \\
{ \operatorname { log } _ { 3 } ( 2 x - 6 ) \neq 0 , } \\
{ 2 x - 6 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l... | x\in(3;\frac{7}{2})\cup(\frac{7}{2};4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,713 |
9.129. $y=\sqrt{\frac{x^{2}-1}{(x+3)(x-4)}-1}+\frac{1}{\log _{8}(x-4)}$. | ## Solution.
From the condition, we obtain the system of inequalities
$$
\left\{\begin{array} { l }
{ \frac { x ^ { 2 } - 1 } { ( x + 3 ) ( x - 4 ) } - 1 \geq 0 , } \\
{ \operatorname { log } _ { 8 } ( x - 4 ) \neq 0 , } \\
{ x - 4 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ \frac { x + 1 1 } { (... | x\in(4;5)\cup(5;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,714 |
9.130. $\left|\frac{3 x+1}{x-3}\right|<3$.
9.130. $\left|\frac{3 x+1}{x-3}\right|<3$.
The above text has been translated into English while retaining the original formatting and line breaks. However, the mathematical expression itself is unchanged as it is a universal notation. | ## Solution.
The inequality is equivalent to a system of two inequalities
$\left\{\begin{array}{l}\frac{3 x+1}{x-3}-3\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{3 x+1}{x-3}-30\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{10}{x-3}0\end{array} \Leftrightarrow\right.\right.\right.$
$\Leftrightarro... | x\in(-\infty;\frac{4}{3}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,715 |
9.131. $\log _{|x-1|} 0.5>0.5$. | ## Solution.
From the condition, we have two systems of inequalities:
1) $\left\{\begin{array}{l}01, \\ 0.5>\sqrt{|x-1|}\end{array} \Leftrightarrow\left\{\begin{array}{l}|x-1|>1, \\ |x-1|1.25, \\
x<0.75
\end{array}\right.}
\end{array}\right.
$
Using the interval method, we get $x \in(0 ; 0.75) \cup(1.25 ; 2)$.
\cup(1.25;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,716 |
9.132. $\log _{x} \frac{3 x-1}{x^{2}+1}>0$. | ## Solution.
The given inequality is equivalent to the following two systems of inequalities:
1) $\left\{\begin{array}{l}00\end{array} \Leftrightarrow\left\{\begin{array}{l}00 \\ 3 x-1>0\end{array}, \Leftrightarrow\left\{\begin{array}{l}00 \\ 3 x-1>0\end{array}, \Leftrightarrow \frac{1}{3}1, \\ \frac{3 x-1}{x^{2}+1}>... | x\in(\frac{1}{3};1)\cup(1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,717 |
$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$9.133 \frac{|x+2|-|x|}{\sqrt{4-x^{3}}}>0$. | Solution.
Domain of definition: $4-x^{3}>0, x^{3}<0$. Expanding the absolute values, we have the following three cases:
$$
\text { 1) }\left\{\begin{array} { l }
{ x 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x0
\end{array} \varnothing\right.\right.
$$
2) $\left\{\begin{array}{l}-2 \leq x0\end{array} ... | x\in(-1;\sqrt[3]{4}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,718 |
9.134. $0.5^{\sqrt{3}}<0.5^{\frac{\sin 2 x}{1-\cos 2 x}}<0.5$. | ## Solution.
The given inequality is equivalent to the system of two inequalities
$\left\{\begin{array}{l}\frac{\sin 2 x}{1-\cos 2 x}>1, \\ \frac{\sin 2 x}{1-\cos 2 x}>0, \\ \frac{\sin 2 x}{1-\cos 2 x}-\sqrt{3}>0, \\ \frac{\sin 2 x-\sqrt{3}+\sqrt{3} \cos 2 x}{1-\cos 2 x}>0, \\ \frac{2 \sin x \cos x-\sqrt{3} \cos ^{2}... | x\in(\frac{\pi}{6}+\pik;\frac{\pi}{4}+\pik),k\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,719 |
9.135. a) $\frac{3 \log _{a} x+6}{\log _{a}^{2} x+2}>1$; б) $\log _{2} \log _{4} x+\log _{4} \log _{2} x \leq-4$.
9.135. a) $\frac{3 \log _{a} x+6}{\log _{a}^{2} x+2}>1$; b) $\log _{2} \log _{4} x+\log _{4} \log _{2} x \leq-4$. | ## Solution.
a) Rewrite the inequality as $\frac{\log _{a}^{2} x-3 \log _{a} x-4}{\log _{a}^{2} x+2} \leq 0$ for $0 < a < 1$, and $\frac{\log _{a}^{2} x-3 \log _{a} x-4}{\log _{a}^{2} x+2} \geq 0$ for $a > 1$. If $\log _{a}^{2} x+2 > 0$, then $\log _{a}^{2} x-3 \log _{a} x-4 \leq 0$ for $0 < a < 1$, and $\log _{a}^{2}... | x\in(1;\sqrt[4]{2}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,720 |
9.136. $\left(\frac{x^{2}}{8}+\frac{3 x}{4}+\frac{3}{2}+\frac{1}{x}\right) \cdot\left(1-x-\frac{(x-2)^{2} \cdot(1-x)}{(x+2)^{2}}\right)>0$. | ## Solution.
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+12 x+8}{8 x} \cdot \frac{(1-x)(x+2)^{2}-(x-2)^{2} \cdot(1-x)}{(x+2)^{2}}>0 \Leftrightarrow \\
& \Leftrightarrow \frac{\left(x^{3}+8\right)+\left(6 x^{2}+12 x\right)}{8 x} \cdot \frac{(1-x)\left((x+2)^{2}-(x-2)^{2}\right)}{(x+2)^{2}}>0 \Leftrightarrow
\end{aligned}
... | x\in(-2;0)\cup(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,721 |
9.137. $\left(\log _{2} x\right)^{4}-\left(\log _{\frac{1}{2}} \frac{x^{3}}{8}\right)^{2}+9 \cdot \log _{2} \frac{32}{x^{2}}<4 \cdot\left(\log _{\frac{1}{2}} x\right)^{2}$. | ## Solution.
Domain of definition: $x>0$.
Let's switch to base 2. We have
$$
\begin{aligned}
& \log _{2}^{4} x-\left(\log _{2} \frac{x^{3}}{8}\right)^{2}+9 \log _{2} \frac{32}{x^{2}}-4 \log _{2}^{\frac{2}{2}} x 2 } \\
{ \operatorname { log } _ { 2 } x 4 \\
0<x<\frac{1}{4}
\end{aligned}\right.} \\
\text { 0 }
\end{al... | x\in(\frac{1}{8};\frac{1}{4})\cup(4;8) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,722 |
9.140. $\frac{4}{\sqrt{2-x}}-\sqrt{2-x}<2$. | ## Solution.
Domain of definition: $x<2$, then we have $(\sqrt{2-x})^{2}+2(\sqrt{2-x})-4>0$. Solving this inequality as a quadratic equation in terms of $\sqrt{2-x}$, we get $\sqrt{2-x}-1+\sqrt{5} \Leftrightarrow 2-x>(\sqrt{5}-1)^{2}, x<-4+2 \sqrt{5}$.
Answer: $x \in(-\infty ;-4+2 \sqrt{5})$. | x\in(-\infty;-4+2\sqrt{5}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,725 |
9.141. $\sqrt{9^{x}-3^{x+2}}>3^{x}-9$.
9.141. $\sqrt{9^{x}-3^{x+2}}>3^{x}-9$. | ## Solution.
Let's rewrite the given inequality as $\sqrt{3^{2 x}-9 \cdot 3^{x}}>3^{x}-9$: It is equivalent to the following two systems of inequalities:
1) $\left\{\begin{array}{l}3^{x}-9\left(3^{x}-9\right)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}3^{x} \geq 3^{2}, \\ 3^{2 x}-9 \cdot 3^{x}>\left(3^{x}-9... | x\in(2;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,726 |
9.142. $\left|\frac{x^{2}-5 x+4}{x^{2}-4}\right| \leq 1$.
Solve the inequality $\left|\frac{x^{2}-5 x+4}{x^{2}-4}\right| \leq 1$. | ## Solution.
Domain of definition: $x \neq \pm 2$.
The given inequality is equivalent to the system of two inequalities
$$
\left\{\begin{array} { l }
{ \frac { x ^ { 2 } - 5 x + 4 } { x ^ { 2 } - 4 } \leq 1 , } \\
{ \frac { x ^ { 2 } - 5 x + 4 } { x ^ { 2 } - 4 } \geq - 1 }
\end{array} \Leftrightarrow \left\{\begin... | x\in[0;\frac{8}{5}]\cup[\frac{5}{2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,727 |
9.143. $\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+3 \geq 0, \\ x-1 \geq 0, \\ x-2 \geq 0\end{array} \Leftrightarrow x \geq 2\right.$.
Since both sides of the inequality are non-negative, squaring both sides, we get
$$
\begin{aligned}
& x+36-x . \text { The last inequality, taking into account the domain o... | x\in(\sqrt{\frac{28}{3}};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,728 |
9.144. $\frac{\left(x-\frac{1}{2}\right)(3-x)}{\log _{2}|x-1|}>0$. | Solution.
Domain of definition: $|x-1|>0, x \neq 1$.
The roots of the equations $\left(x-\frac{1}{2}\right)(3-x)=0$ and $\log _{2}|x-1|=0$ are the numbers $x_{1}=\frac{1}{2}, x_{2}=3, x_{3}=2, x_{4}=\dot{0}$, which are not solutions to the given inequality, so we mark them on the number line with hollow circles:
![]... | x\in(0;\frac{1}{2})\cup(2;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,729 |
9.145. $\sqrt{3} \cos^{-2} x < 4 \tan x$. | $$
\begin{aligned}
& \frac{\sqrt{3}}{\cos ^{2} x}-\frac{4 \sin x}{\cos x}\frac{\sqrt{3}}{2}, \Leftrightarrow \frac{\pi}{3}+2 \pi n<2 x<\pi-\frac{\pi}{3}+2 \pi n, n \in Z \\
\cos x \neq 0
\end{aligned}\right. \\
& \frac{\pi}{6}+\pi n<x<\frac{\pi}{3}+\pi n, n \in Z \\
& \text { Answer: } x \in\left(\frac{\pi}{6}+\pi n ; ... | x\in(\frac{\pi}{6}+\pin;\frac{\pi}{3}+\pin),n\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,730 |
9.146. $\sin 4 x+\cos 4 x \operatorname{ctg} 2 x>1$.
9.146. $\sin 4x + \cos 4x \cot 2x > 1$. | ## Solution.
Let's rewrite the inequality as
$$
\begin{aligned}
& \sin 4 x+\frac{\cos 4 x \cos 2 x}{\sin 2 x}>1 \Leftrightarrow \frac{\sin 4 x \sin 2 x+\cos 4 x \cos 2 x}{\sin 2 x}>1 \Leftrightarrow \\
& \Leftrightarrow \frac{\cos 2 x}{\sin 2 x}>1, \operatorname{ctg} 2 x>1, \pi n<2 x<\frac{\pi}{4}+\pi n, \frac{\pi n}... | x\in(\frac{\pin}{2};\frac{\pi}{8}(4n+1)),n\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,731 |
9.147. $2+\tan 2x+\cot 2x<0$. | ## Solution.
From the condition we have
$$
2+\operatorname{tg} 2 x+\frac{1}{\operatorname{tg} 2 x}<0 \Leftrightarrow \frac{\operatorname{tg}^{2} 2 x+2 \operatorname{tg} x+1}{\operatorname{tg} 2 x}<0, \frac{(\operatorname{tg} 2 x+1)^{2}}{\operatorname{tg} 2 x}<0 \Leftrightarrow
$$
$\Leftrightarrow\left\{\begin{array}... | x\in(\frac{\pi}{4}(2n-1);\frac{\pi}{8}(4n-1))\cup(\frac{\pi}{8}(4n-1);\frac{\pin}{2}),n\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,732 |
9.148. $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$.
9.148. $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$.
The inequality is $\frac{x^{4}+3 x^{3}+4 x^{2}-8}{x^{2}}<0$. | Solution.
The given inequality is equivalent to a system of two inequalities
$$
\begin{aligned}
& \left\{\begin{array}{l}
x^{4}+3 x^{3}+4 x^{2}-8<0, \\
x \neq 0
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\left(x^{4}-x^{3}\right)+\left(4 x^{3}-4\right)+\left(4 x^{2}-4\right)<0 \\
x \neq 0
\end{arra... | x\in(-2;0)\cup(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,733 |
9.149. $\frac{3}{6 x^{2}-x-12}<\frac{25 x-47}{10 x-15}-\frac{3}{3 x+4}$. | Solution.
Rewrite the given inequality as
$$
\begin{aligned}
& \frac{3}{(3 x+4)(2 x-3)}-\frac{25 x-47}{5(2 x-3)}+\frac{3}{3 x+4}0, \frac{75\left(x+\frac{79}{75}\right)(x-2)}{(3 x+4)(2 x-3)}>0 \Leftrightarrow \\
& \Leftrightarrow\left(x+\frac{79}{75}\right)(x-2)(3 x+4)(2 x-3)>0 .
\end{aligned}
$$
Using the interval m... | x\in(-\infty;-\frac{4}{3})\cup(-\frac{79}{75};\frac{3}{2})\cup(2;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,734 |
9.150. $\frac{\log _{0.3}|x-2|}{x^{2}-4 x}<0$. | ## Solution.
Domain of definition: $|x-2|>0, x \neq 2, x \neq 0, x \neq 4$.
The roots of the equations $\log _{0.3}|x-2|=0$ and $x^{2}-4 x=0$ are the numbers $x_{1}=1$, $x_{2}=3, x_{3}=0, x_{4}=4, x_{5}=2$ - does not fit the domain of definition.
These points divide the number line into 6 intervals. We select a valu... | x\in(-\infty;0)\cup(1;2)\cup(2;3)\cup(4;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,735 |
9.151. $\sqrt{x^{2}-4x}>x-3$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
9.151. $\sqrt{x^{2}-4x}>x-3$. | Solution.
The inequality is equivalent to the following two systems of inequalities:
1) $\left\{\begin{array}{l}x-3(x-3)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 3, \\ x>\frac{9}{2}\end{array} \Leftrightarrow x>\frac{9}{2}\right.\right.$.
Combining the obtained intervals, we have $x \in(-\infty ; ... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,736 | |
9.152. $\frac{1-\log _{4} x}{1+\log _{2} x} \leq \frac{1}{2}$. | Solution.
Let's switch to base 2.
$$
\begin{aligned}
& \text { We have } \frac{1-\frac{1}{2} \log _{2} x}{1+\log _{2} x}-\frac{1}{2} \leq 0, \frac{2-\log _{2} x-1-\log _{2} x}{2\left(1+\log _{2} x\right)} \leq 0, \\
& \frac{1-2 \log _{2} x}{1+\log _{2} x} \leq 0, \frac{\log _{2} x-\frac{1}{2}}{\log _{2} x+1} \geq 0 .... | x\in(0;\frac{1}{2})\cup[\sqrt{2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,737 |
9.154. $\frac{2-x}{x^{3}+x^{2}}>\frac{1-2 x}{x^{3}-3 x^{2}}$. | ## Solution.
Rewrite the given inequality as $\frac{2-x}{x^{2}(x+1)}-\frac{1-2 x}{x^{2}(x-3)}>0 \Leftrightarrow$ $\Leftrightarrow \frac{(2-x)(x-3)-(1-2 x)(x+1)}{x^{2}(x+1)(x-3)}>0, \frac{x^{2}+6 x-7}{x^{2}(x+1)(x-3)}>0, \frac{(x+7)(x-1)}{x^{2}(x+1)(x-3)}>0 \Leftrightarrow$ $\Leftrightarrow x^{2}(x+1)(x-3)(x+7)(x-1)>0$... | x\in(-\infty;-7)\cup(-1;0)\cup(0;1)\cup(3;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,739 |
9.155. $0.2^{\frac{6 \log _{4} x-3}{\log _{4} x}}>\sqrt[3]{0.008^{2 \log _{4} x-1}}$. | Solution.
Let's write the equation as $0.2^{\frac{6 \log _{4} x-3}{\log _{4} x}}>0.2^{2 \log _{4} x-1} \Leftrightarrow$
$$
\begin{aligned}
& \Leftrightarrow \frac{6 \log _{4} x-3}{\log _{4} x}0 \Leftrightarrow \\
& \Leftrightarrow \frac{2\left(\log _{4} x-\frac{1}{2}\right)\left(\log _{4} x-3\right)}{\log _{4} x}>0 \... | x\in(1;2)\cup(64;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,740 |
9.156. $(2,25)^{\log _{2}\left(x^{2}-3 x-10\right)}>\left(\frac{2}{3}\right)^{\log _{\frac{1}{2}}\left(x^{2}+4 x+4\right)}$.
9.156. $(2,25)^{\log _{2}\left(x^{2}-3 x-10\right)}>\left(\frac{2}{3}\right)^{\log _{\frac{1}{2}}\left(x^{2}+4 x+4\right)}$. | ## Solution.
Let's write the inequality as
$$
\begin{aligned}
& \left(\frac{3}{2}\right)^{2 \log _{2}\left(x^{2}-3 x-10\right)}>\left(\frac{3}{2}\right)^{-\log _{\frac{1}{2}}\left(x^{2}+4 x+4\right)} \Leftrightarrow \\
& \Leftrightarrow 2 \log _{2}\left(x^{2}-3 x-10\right)>-\log _{\frac{1}{2}}\left(x^{2}+4 x+4\right)... | x\in(-\infty;-2)\cup(6;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,741 |
9.157. $\log _{0.5}(x+3)<\log _{0.25}(x+15)$. | ## Solution.
Let's switch to base 2. We have $\log _{0.5}(x+3) 0 , } \\
{ x + 15 > 0 , } \\
{ x + 3 > \sqrt { x + 15 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x > -3, \\
x > -15, \\
x^{2} + 6x + 9 > x + 15
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ x > - 3 , } \\
{ x ^ { ... | x\in(1;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,742 |
9.158. $\log _{\frac{1}{3}}(x-1)+\log _{\frac{1}{3}}(x+1)+\log _{\sqrt{3}}(5-x)<1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x-1>0, \\ x+1>0, \\ 5-x>0\end{array} \Leftrightarrow 11 \Leftrightarrow \log _{3} \frac{(5-x)^{2}}{(x-1)(x+1)}>1 \Leftrightarrow \\
& \Leftrightarrow \frac{\cdot(5-x)^{2}}{(x-1)(x+1)}>3 \Leftrightarrow \frac{(5-x)^{2}}{(x-1)(x+1)}-3>0 \Leftrightarrow \frac{(5-... | x\in(2;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,743 |
9.159. $2 \log _{3} \log _{3} x+\log _{\frac{1}{3}} \log _{3}(9 \sqrt[3]{x}) \geq 1$. | ## Solution.
Domain: $\log _{3} x>0$.
Switch to base 3. $2 \log _{3} \log _{3} x-\log _{3} \log _{3} 9 \sqrt[3]{x} \geq 1 \Leftrightarrow$ $\Leftrightarrow \log _{3} \log _{3}^{2} x-\log _{3} \log _{3} 9 \sqrt[3]{x} \geq 1, \log _{3} \frac{\log _{3}^{2} x}{\log _{3} 9 \sqrt[3]{x}} \geq 1, \frac{\log _{3}^{2} x}{\log ... | x\in[27;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,744 |
9.160. $0.008^{x}+5^{1-3 x}+0.04^{\frac{3}{2}(x+1)}<30.04$. | ## Solution.
From the condition, we have $0.2^{3 x}+\frac{5}{5^{3 x}}+0.08 \cdot 0.2^{3 x}5^{-1} \Leftrightarrow 3 x>-1, x>-\frac{1}{3}$.
Answer: $x \in\left(-\frac{1}{3} ; \infty\right)$. | x\in(-\frac{1}{3};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,745 |
9.161. $0.4^{\log _{3} \frac{3}{x} \log _{3} 3 x}>6.25^{\log _{3} x^{2}+2}$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the inequality as $\left(\frac{2}{5}\right)^{\log _{3} \frac{3}{x} \log _{3} 3 x}>\left(\frac{2}{5}\right)^{-2 \log _{3} x^{2}-4} \Leftrightarrow$
$\Leftrightarrow \log _{3} \frac{3}{x} \log _{3} 3 x>-2 \log _{3} x^{2}-4 \Leftrightarrow$
$\Leftrightarrow\left(\log _... | x\in(0;\frac{1}{3})\cup(243;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,746 |
9.162. $0.3^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots}<\sqrt[3]{0.3^{3 x^{2}+5 x}}<1$. | Solution.
Let's write the inequality as $0.3^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots}\frac{3 x^{2}+5 x}{3}>0 \Leftrightarrow\left\{\begin{array}{l}
1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots>\frac{3 x^{2}+5 x}{3} \\
\frac{3 x^{2}+5 x}{3}>0
\end{array}\right.
$
The sum of the terms of an infinite decreasing geo... | x\in(-2;-\frac{5}{3})\cup(0;\frac{1}{3}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,747 |
9.163. $\frac{\lg 7-\lg \left(-8 x-x^{2}\right)}{\lg (x+3)}>0$.
9.163. $\frac{\log 7-\log \left(-8 x-x^{2}\right)}{\log (x+3)}>0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}-8 x-x^{2}>0, \\ x+3>0, \\ \lg (x+3) \neq 0\end{array} \Leftrightarrow\left[\begin{array}{l}-30, \log _{x+3} \frac{7}{-8 x-x^{2}}>0$. The obtained
inequality is equivalent to the following two systems of inequalities:
1) $\left\{\begin{array}{l}00\end{array}... | x\in(-3;-2)\cup(-1;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,748 |
9.164. $\log _{3} \log _{4} \frac{4 x-1}{x+1}-\log _{\frac{1}{3}} \log _{\frac{1}{4}} \frac{x+1}{4 x-1}<0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\log _{4} \frac{4 x-1}{x+1}>0, \\ \log _{\frac{1}{4}} \frac{x+1}{4 x-1}>0\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{4 x-1}{x+1}>1, \\ \frac{x+1}{4 x-1}>1\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{4 x-1}{x+1}-1>0, \\ \frac{x+1}{4 x-1}-... | x\in(\frac{2}{3};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,749 |
9.165. $2^{\log _{0.5}^{2} x}+x^{\log _{0.5} x}>2.5$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the inequality as $2^{\log _{0.5} x \cdot \log _{0.5} x}+x^{\log _{0.5} x}-2.5>0 \Leftrightarrow$ $\Leftrightarrow\left(2^{\log _{0.5} x}\right)^{\log _{0.5} x}+x^{\log _{0.5} x}-2.5>0 \Leftrightarrow\left(2^{\log _{2} \frac{1}{x}}\right)^{\log _{0.5} x}+x^{\log _{0.5... | x\in(0;\frac{1}{2})\cup(2;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,750 |
9.166. $3^{\lg x+2}<3^{\lg x^{2}+5}-2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.166. $3^{\lg x+2}<3^{\lg x^{2}+5}-2$. | ## Solution.
Domain of definition: $x>0$.
Let's write the given inequality as $243 \cdot\left(3^{\lg x}\right)^{2}-9 \cdot 3^{\lg x}-2>0$. Solving
it as a quadratic equation in terms of $3^{\lg x}$, we find $\left[\begin{array}{l}3^{\lg x}>3^{-2}, \\ 3^{\lg x}-2, x>0.01\end{array}\right.$.
Answer: $x \in(0.01 ; \i... | x\in(0.01;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,751 |
9.167. $\frac{1}{x+1}-\frac{2}{x^{2}-x+1} \leq \frac{1-2 x}{x^{3}+1}$. | ## Solution.
Domain of definition: $x \neq-1$.
From the condition we have $\frac{x^{2}-x+1-2 x-2}{x^{3}+1} \leq \frac{1-2 x}{x^{3}+1}$,
$$
\begin{aligned}
& \frac{x^{2}-3 x-1}{x^{3}+1}-\frac{1-2 x}{x^{3}+1} \leq 0, \frac{x^{2}-3 x-1-1+2 x}{x^{3}+1} \leq 0, \frac{x^{2}-x-2}{x^{3}+1} \leq 0 \\
& \frac{(x+1)(x-2)}{(x+1... | x\in(-\infty;-1)\cup(-1;2] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,752 |
9.169. $\frac{1}{x^{2}-4}+\frac{4}{2 x^{2}+7 x+6} \leq \frac{1}{2 x+3}+\frac{4}{2 x^{3}+3 x^{2}-8 x-12}$. | ## Solution.
Rewrite the inequality as
$\frac{1}{(x-2)(x+2)}+\frac{4}{(x+2)(2 x+3)}-\frac{1}{2 x+3}-\frac{4}{(2 x+3)(x-2)(x+2)} \leq 0 \Leftrightarrow$
$\Leftrightarrow \frac{2 x+3+4 x-8-x^{2}+4-4}{(2 x+3)(x-2)(x+2)} \leq 0, \frac{x^{2}-6 x+5}{(2 x+3)(x-2)(x+2)} \geq 0$,
$\frac{(x-1)(x-5)}{(2 x+3)(x-2)(x+2)} \geq 0... | x\in(-2;-\frac{3}{2})\cup[1;2)\cup[5;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,754 |
9.170. $\frac{10(5-x)}{3(x-4)}-\frac{11}{3} \cdot \frac{6-x}{x-4} \geq \frac{5(6-x)}{x-2}$. | ## Solution.
Domain of definition: $x \neq 2, x \neq 4$.
We have $\frac{x-16}{3(x-4)}-\frac{5(6-x)}{x-2} \geq 0 \Leftrightarrow \frac{(x-16)(x-2)-15(6-x)(x-4)}{3(x-4)(x-2)} \geq 0$,
$\frac{x^{2}-34 x+76}{(x-4)(x-2)} \geq 0, \frac{\left(x-\frac{7}{2}\right)(x-7)}{(x-4)(x-2)} \geq 0 \Leftrightarrow$
$$
\Leftrightarro... | x\in(-\infty;2)\cup[\frac{7}{2};4)\cup[7;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,755 |
9.171. $0.6^{\lg ^{2}(-x)+3} \leq\left(\frac{5}{3}\right)^{2 \lg x^{2}}$. | ## Solution.
Domain of definition: $x<0$.
Since $\lg x^{2 k}=2 k \lg |x|$, taking into account the domain of definition, we can rewrite the given inequality as
$\left(\frac{3}{5}\right)^{\lg ^{2}(-x)+3} \leq\left(\frac{3}{5}\right)^{-4 \lg (-x)} \Leftrightarrow \lg ^{2}(-x)+3 \geq-4 \lg (-x), \lg ^{2}(-x)+4 \lg (-x)... | x\in(-\infty;-0.1]\cup[-0.001;0) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,756 |
9.173. $\left(\frac{3}{5}\right)^{13 x^{2}} \leq\left(\frac{3}{5}\right)^{x^{4}+36}<\left(\frac{3}{5}\right)^{12 x^{2}}$. | Solution.
The given inequality is equivalent to the inequality $13 x^{2} \geq x^{4}+36>12 x^{2} \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}x^{4}+36>12 x^{2}, \\ x^{4}+36 \leq 13 x^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{4}-12 x^{2}+36>0, \\ x^{4}-13 x^{2}+36 \leq 0\end{array} \Leftrightarr... | x\in[-3;-\sqrt{6})\cup(-\sqrt{6};-2]\cup[2;\sqrt{6})\cup(\sqrt{6};3] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,758 |
9.174. $|x-3|^{2 x^{2}-7 x}>1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.174. $|x-3|^{2 x^{2}-7 x}>1$. | ## Solution.
Rewrite the inequality as $|x-3|^{2 x^{2}-7 x}>|x-3|^{0}$. It is equivalent to the following conjunction of two systems of inequalities:
The solution to the first system of ... | x\in(-\infty;0)\cup(2;3)\cup(3;\frac{7}{2})\cup(4;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,759 |
9.175. $\log _{\frac{1}{5}} x+\log _{4} x>1$. | ## Solution.
Domain of definition: $x>0$.
Switch to base 4. We have $\frac{\log _{4} x}{\log _{4} \frac{1}{5}}+\log _{4} x>1 \Leftrightarrow$
$\Leftrightarrow \frac{\log _{4} x}{-\log _{4} 5}+\log _{4} x>1 \Leftrightarrow \log _{4} x \cdot\left(\frac{1-\log _{4} 5}{-\log _{4} 5}\right)>1 \Leftrightarrow$
$\Leftrigh... | x\in(4^{\log_{0.8}0.2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,760 |
9.176. $-9<x^{4}-10 x^{2}<56$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
9.176. $-9<x^{4}-10 x^{2}<56$. | Solution.
From the condition, we get
Using the interval method, we find \( x \in (-\sqrt{14} ; -3) \cup (-1 ; 1) \cup (3 ; \sqrt{14}) \).
$$
\text { }
$$
Answer: \( x \in (-\sqrt{14} ; ... | x\in(-\sqrt{14};-3)\cup(-1;1)\cup(3;\sqrt{14}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,761 |
9.177. $216 x^{6}+19 x^{3}<1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
9.177. $216 x^{6}+19 x^{3}<1$. | ## Solution.
$216 x^{6}+19 x^{3}-1<0$. Solve the inequality as a quadratic equation with respect to
$$
x^{3},-\frac{1}{8}<x^{3}<\frac{1}{27} \Leftrightarrow-\frac{1}{2}<x<\frac{1}{3}
$$
Answer: $x \in\left(-\frac{1}{2} ; \frac{1}{3}\right)$. | x\in(-\frac{1}{2};\frac{1}{3}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,762 |
9.178. $x^{0.5 \log _{0.5} x-3} \geq 0.5^{3-2.5 \log _{0.5} x}$. | ## Solution.
Domain of definition: $x>0$.
By logarithmizing both sides of the inequality with base 0.5, we get
$$
\begin{aligned}
& \log _{0.5} x^{0.5 \log _{0.5} x-3} \leq \log _{0.5} 0.5^{3-2.5 \log _{0.5} x} \Leftrightarrow \\
& \Leftrightarrow\left(0.5 \log _{0.5} x-3\right) \log _{0.5} x \leq\left(3-2.5 \log _{... | x\in[0.125;4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,763 |
9.179. $|x-6|>\left|x^{2}-5 x+9\right|$. | ## Solution.
Since $x^{2}-5 x+9>0$ for $x \in R$, the given inequality has the form $|x-6|>x^{2}-5 x+9$. It is equivalent to two inequalities
1) $x-6>x^{2}-5 x+9 \Leftrightarrow x^{2}-6 x+15<0, \varnothing$;
2) $x-6<-x^{2}+5 x-9 \Leftrightarrow x^{2}-4 x+3<0 \Leftrightarrow 1<x<3$.
Answer: $x \in(1 ; 3)$. | x\in(1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,764 |
9.180. $\frac{6 x}{x-2}-\sqrt{\frac{12 x}{x-2}}-2 \sqrt[4]{\frac{12 x}{x-2}}>0$. | ## Solution.
Domain of definition: $\frac{x}{x-2} \geq 0 ; x \neq 2$.
$$
\begin{aligned}
& \text { Let } \sqrt[4]{\frac{12 x}{x-2}}=y \geq 0 . \text { In terms of } y \text {, the inequality is } \\
& \frac{y^{4}}{2}-y^{2}-2 y>0 \Leftrightarrow y\left(y^{3}-2 y-4\right)>0 \Leftrightarrow y(y-2)\left(y^{2}+2 y+2\right... | x\in(2;8) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,765 |
9.181. $\log _{0.3} \log _{6} \frac{x^{2}+x}{x+4}<0$. | ## Solution.
The original inequality is equivalent to the inequality $\log _{6} \frac{x^{2}+x}{x+4}>1 \Leftrightarrow$
$$
\Leftrightarrow \frac{x^{2}+x}{x+4}>6 \Leftrightarrow \frac{x^{2}+x}{x+4}-6>0 \Leftrightarrow \frac{x^{2}+x-6 x-24}{x+4}>0 \Leftrightarrow
$$
$$
\Leftrightarrow \frac{x^{2}-5 x-24}{x+4}>0 \Leftri... | x\in(-4;-3)\cup(8;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,766 |
9.182. $\log _{2 x}\left(x^{2}-5 x+6\right)<1$. | Solution.
The original inequality is equivalent to the following systems of inequalities:
1) $\left\{\begin{array}{l}0<2 x\end{array} \Leftrightarrow\left\{\begin{array}{l}0<0\end{array} \Leftrightarrow\left\{\begin{array}{l}0<6, \\ x>1, \\ x^{2}-5 x+6<0\end{array}, \Leftrightarrow\left\{\begin{array}{l}x>\frac{1}{2}... | x\in(0;\frac{1}{2})\cup(1;2)\cup(3;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,767 |
9.183. $\log _{\frac{1}{2}} \log _{2} \log _{x-1} 9>0$. | Solution.
The original inequality is equivalent to the inequality $0<\log _{2} \log _{x-1} 9<1 \Leftrightarrow$ $\Leftrightarrow 1<\log _{x-1} 9<2$. We obtain two systems of inequalities:
1) $\left\{\begin{array}{l}0<x-1<1, \\ (x-1)^{2}<9<x-1,\end{array} \varnothing\right.$;
 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,768 |
9.184. $\log _{0.25}\left|\frac{2 x+1}{x+3}+\frac{1}{2}\right|>\frac{1}{2}$. | ## Solution.
The given inequality is equivalent to the inequality
$$
0 - \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x \neq - 1 , } \\
{ x \neq - 3 , } \\
{ \frac { 2 x + 1 } { x + 3 } 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \neq -1, \\
x \neq -3, \\
-3 - \frac{4}{... | x\in(-\frac{4}{3};-1)\cup(-1;-\frac{1}{2}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,769 |
9.185. $x^{2}\left(x^{4}+36\right)-6 \sqrt{3}\left(x^{4}+4\right)<0$. | ## Solution.
We have $x^{6}-6 \sqrt{3} x^{4}+36 x^{2}-24 \sqrt{3}<0 \Leftrightarrow\left(x^{2}-2 \sqrt{3}\right)^{3}<0 \Leftrightarrow$ $\Leftrightarrow x^{2}-2 \sqrt{3}<0 \Leftrightarrow x^{2}<2 \sqrt{3}$. From this, $-\sqrt{2 \sqrt{3}}<x<\sqrt{2 \sqrt{3}}$, $-\sqrt[4]{12}<x<\sqrt[4]{12}$.
Answer: $x \in(-\sqrt[4]{1... | x\in(-\sqrt[4]{12};\sqrt[4]{12}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,770 |
9.186. $\frac{x^{3}+3 x^{2}-x-3}{x^{2}+3 x-10}<0$.
9.186. $\frac{x^{3}+3 x^{2}-x-3}{x^{2}+3 x-10}<0$. | ## Solution.
Let's rewrite the given inequality as
$$
\begin{aligned}
& \frac{\left(x^{3}-x\right)+\left(3 x^{2}-3\right)}{(x+5)(x-2)}<0 \Leftrightarrow \frac{x\left(x^{2}-1\right)+3\left(x^{2}-1\right)}{(x+5)(x-2)}<0, \frac{\left(x^{2}-1\right)(x+3)}{(x+5)(x-2)}<0 \\
& (x-1)(x+1)(x+3)(x+5)(x-2)<0 . x \in(-\infty ;-5... | x\in(-\infty;-5)\cup(-3;-1)\cup(1;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,771 |
9.188. $\sqrt{x+3}+\sqrt{x-2}-\sqrt{2 x+4}>0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+3 \geq 0, \\ x-2 \geq 0, \\ 2 x+4 \geq 0\end{array} \Leftrightarrow x \geq 2\right.$.
Let's rewrite the given inequality as $\sqrt{x+3}+\sqrt{x-2}>\sqrt{2 x+4}$ and square both sides. We have
$$
\begin{aligned}
& x+3+2 \sqrt{(x+3)(x-2)}+x-2>2 x+4 \Leftrigh... | x\in(\frac{-1+\sqrt{34}}{2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,773 |
9.189. $\log _{5} \sqrt{3 x+4} \cdot \log _{x} 5>1$. | ## Solution.
Domain of definition: $01$. From this, we have $\log _{x} \sqrt{3 x+4}>1$. The last inequality is equivalent to two systems of inequalities:
1) $\left\{\begin{array}{l}00,\end{array}\right.\right.\right.$
2); $\left\{\begin{array}{l}x>1, \\ \sqrt{3 x+4}>x\end{array} \Leftrightarrow\left\{\begin{array}{l}... | x\in(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,774 |
9.190. $\frac{x^{3}-2 x^{2}-5 x+6}{x-2}>0$.
9.190. $\frac{x^{3}-2 x^{2}-5 x+6}{x-2}>0$. | Solution.
Rewrite the given inequality as $\frac{(x-1)(x+2)(x-3)}{x-2}>0 \Leftrightarrow$ $\Leftrightarrow(x-1)(x+2)(x-3)(x-2)>0$. Using the method of intervals, we get $x \in(-\infty ;-2) \cup(1 ; 2) \cup(3 ; \infty)$.
\cup(1;2)\cup(3;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,775 |
9.191. $2 \cos x(\cos x-\sqrt{8} \tan x)<5$. | Solution.
Domain of definition: $\cos x \neq 0$.
Rewrite the inequality as $2 \cos x\left(\cos x-\sqrt{8} \frac{\sin x}{\cos x}\right)-50, \\ \cos x \neq 0\end{array} \Leftrightarrow\right.\right.$
\cup(\frac{\pi}{2}+2\pin;\frac{5\pi}{4}+2\pin),n\inZ | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,776 |
9.192. $\sqrt{x^{3}+3 x+4}>-2$, | ## Solution.
The solutions to this inequality are the values of the unknown that belong to the domain of definition: $x^{3}+3 x+4 \geq 0,(x+1)\left(x^{2}-x+4\right) \geq 0$. Here $x^{2}-x+4>0$ for $x \in R$. Then $x+1 \geq 0, x \geq-1$.
Answer: $x \in[-1 ; \infty)$. | x\in[-1;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,777 |
9.194. $25 \cdot 2^{x}-10^{x}+5^{x}>25$. | Solution.
Let's write the given inequality as $\left(25 \cdot 2^{x}-25\right)-\left(10^{x}-5^{x}\right)>0$, $25\left(2^{x}-1\right)-5^{x}\left(2^{x}-1\right)>0,\left(2^{x}-1\right)\left(25-5^{x}\right)>0 \Leftrightarrow$ $\Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}2^{x}-1>0, \\ 25-5^{x}>0 ;\end{array}... | x\in(0;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,779 |
9.195. $\log _{3} \log _{x^{2}} \log _{x^{2}} x^{4}>0$. | ## Solution.
The given inequality is equivalent to the inequality $\log _{x^{2}} \log _{x^{2}}\left(x^{2}\right)^{2}>1 \Leftrightarrow$ $\Leftrightarrow \log _{x^{2}} 2 \log _{x^{2}} x^{2}>1, \log _{x^{2}} 2>1 \Leftrightarrow$
\cup(1;\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,780 |
9.196. $0.5^{2 \sqrt{x}}+2>3 \cdot 0.5^{\sqrt{x}}$. | ## Solution.
Domain of definition: $x \geq 0$.
Let's rewrite the given inequality as $0.5^{2 \sqrt{x}} - 3 \cdot 0.5^{\sqrt{x}} + 2 > 0$ and solve it as a quadratic inequality in terms of $0.5^{\sqrt{x}}$. We get $\left[\begin{array}{l}0.5^{\sqrt{x}} > 2, \\ 0.5^{\sqrt{x}} < 1\end{array} \Leftrightarrow\right.$
 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,781 |
9.197. $x^{2}(x+3 \sqrt{5})+5(3 x+\sqrt{5})>0$. | ## Solution.
From the condition
$$
x^{3}+3 \sqrt{5} x^{2}+15 x+5 \sqrt{5}>0 \Leftrightarrow(x+\sqrt{5})^{3}>0, x+\sqrt{5}>0, x>-\sqrt{5}
$$
Answer: $x \in(-\sqrt{5} ; \infty)$. | x\in(-\sqrt{5};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,782 |
9.198. $9^{\log _{2}(x-1)-1}-8 \cdot 5^{\log _{2}(x-1)-2}>9^{\log _{2}(x-1)}-16 \cdot 5^{\log _{2}(x-1)-1}$. | ## Solution.
Domain of definition: $x>1$.
$$
\begin{aligned}
& \text { We have } \frac{9^{\log _{2}(x-1)}}{9}-\frac{8 \cdot 5^{\log _{2}(x-1)}}{25}>9^{\log _{2}(x-1)}-\frac{16 \cdot 5^{\log _{2}(x-1)}}{5}, \\
& \frac{9^{\log _{2}(x-1)}}{9}-9^{\log _{2}(x-1)}>\frac{8 \cdot 5^{\log _{2}(x-1)}}{25}-\frac{16 \cdot 5^{\lo... | x\in(1;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,783 |
9.199. $\frac{\log _{2}(\sqrt{4 x+5}-1)}{\log _{2}(\sqrt{4 x+5}+11)}>\frac{1}{2}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\sqrt{4 x+5}-1>0, \\ \sqrt{4 x+5}+11>0, \\ \log _{2}(\sqrt{4 x+5}+11) \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{4 x+5}>1, \\ \sqrt{4 x+5}>-11, \\ \sqrt{4 x+5}+11 \neq 1\end{array} \Leftrightarrow\right.\right.$
$$
\Leftrightarrow \sqrt{4 x+5}... | x\in(5;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,784 |
9.200. $\frac{\log _{0.5}(\sqrt{x+3}-1)}{\log _{0.5}(\sqrt{x+3}+5)}<\frac{1}{2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+3 \geq 0, \\ \sqrt{x+3}>1,\end{array} \Leftrightarrow x>-2\right.$.
For $x>-2, \sqrt{x+3}+5>1$, therefore, $\log _{0.5}(\sqrt{x+3}+5)\log _{0.5}(\sqrt{x+3}+5) \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}x>-2, \\ (\sqrt{x+3}-1)^{2}-2, \\ (\sqrt{x... | x\in(-2;13) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,785 |
9.202. $x^{\log _{2} x}+16 x^{-\log _{2} x}<17$. | Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the given inequality as $x^{\log _{2} x}+\frac{16}{x^{\log _{2} x}}-17<0 \Leftrightarrow$ $\Leftrightarrow x^{2 \log _{2} x}-17 x^{\log _{2} x}+16<0$. Solving this inequality as a quadratic in terms of $x^{\log _{2} x}$, we get $1<x^{\log _{2} x}<16$. Taking the l... | x\in(\frac{1}{4};1)\cup(1;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,787 |
9.203. $5^{\log _{5}^{2} x}+x^{\log _{5} x}<10$. | Solution.
Domain of definition: $x>0$.
Rewrite the inequality as
$5^{\log _{5} x \log _{5} x}+x^{\log _{5} x}<10 \Leftrightarrow x^{\log _{5} x}+x^{\log _{5} x}<10 \Leftrightarrow$
$\Leftrightarrow 2 x^{\log _{5} x}<10, x^{\log _{5} x}<5$. Taking the logarithm of both sides of this inequality to the base 5, we get ... | x\in(\frac{1}{5};5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,788 |
9.204. $\log _{3}\left(\log _{2}\left(2-\log _{4} x\right)-1\right)<1$. | ## Solution.
The given inequality is equivalent to the double inequality
$$
\begin{aligned}
& 0<\log _{2}\left(2-\log _{4} x\right)-1<3 \Leftrightarrow 1<\log _{2}\left(2-\log _{4} x\right)<4 \Leftrightarrow \\
& \Leftrightarrow 2<2-\log _{4} x<16,0<-\log _{4} x<14 \Leftrightarrow-14<\log _{4} x<0, \\
& \frac{1}{4^{1... | x\in(2^{-28};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,789 |
9.205. $\left(x^{2}+4 x+10\right)^{2}-7\left(x^{2}+4 x+11\right)+7<0$. | ## Solution.
Rewrite the given inequality as
$$
\begin{aligned}
& \left(x^{2}+4 x+10\right)^{2}-7\left(x^{2}+4 x+10\right)0 \text { for } x \in R,
\end{aligned}
$$
then the obtained inequality is equivalent to the inequality $x^{2}+4 x+3<0$. Hence, $-3<x<-1$.
Answer: $x \in(-3 ;-1)$. | x\in(-3;-1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,790 |
9.206. Arrange the three numbers $a_{1}=\log _{1} \sin 2 x$,
$$
a_{2}=-1-\log _{2} \sin x, a_{3}=\log _{\frac{1}{2}}(1-\cos 2 x), \text { if } 0<x<\frac{\pi}{4}
$$ | ## Solution.
Let's rewrite $a_{1}, a_{2}, a_{3}$ as follows:
$a_{1}=-\log _{2} 2 \sin x \cos x=-1-\log _{2} \sin x-\log _{2} \cos x$,
$a_{2}=-1-\log _{2} \sin x$,
$a_{3}=-\log _{2} 2 \sin ^{2} x=-1-2 \log _{2} \sin x$.
For $0<x<\frac{\pi}{4}, \log _{2} \sin x<0, \log _{2} \cos x<0$, therefore, the arrangement of t... | a_{2},a_{1},a_{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,791 |
9.207. $\left\{\begin{array}{l}0,2^{\cos x} \leq 1, \\ \frac{x-1}{2-x}+\frac{1}{2}>0 .\end{array}\right.$ | Solution.
Domain of definition: $x \neq 2$.
Rewrite the given system of inequalities as
$\left\{\begin{array}{l}0.2^{\cos x} \leq 0.2^{0}, \\ \frac{2 x-2+2-x}{2(2-x)}>0\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x \geq 0, \\ x(x-2)<0\end{array} \Leftrightarrow\right.\right.$
$\Leftrightarrow\left\{\begin... | x\in(0;\frac{\pi}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,792 |
9.208. $\sqrt{x^{2}-9 x+20} \leq \sqrt{x-1} \leq \sqrt{x^{2}-13}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.208. $\sqrt{x^{2}-9 x+20} \leq \sqrt{x-1} \leq \sqrt{x^{2}-13}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-9 x+20 \geq 0, \\ x-1 \geq 0, \\ x^{2}-13 \geq 0\end{array} \Leftrightarrow\left[\begin{array}{l}\sqrt{13} \leq x \leq 4, \\ x \geq 5 .\end{array}\right.\right.$
The given inequality is equivalent to the system of inequalities
$\left\{\begin{array}{l}\sqr... | x\in[5;7]\cup{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,793 |
9.209. $\left\{\begin{array}{l}\frac{x^{2}+4}{x^{2}-16 x+64}>0 \\ \lg \sqrt{x+7}>\lg (x-5)-2 \lg 2\end{array}\right.$
The system of inequalities is:
9.209. $\left\{\begin{array}{l}\frac{x^{2}+4}{x^{2}-16 x+64}>0 \\ \lg \sqrt{x+7}>\lg (x-5)-2 \lg 2\end{array}\right.$ | Solution.
Domain of definition: $x>5, x \neq 8$.
The given system has the form:
$$
\begin{aligned}
& \Leftrightarrow\left\{\begin{array}{l}
x \neq 8, \\
x^{2}-26 x-875
\end{array}, \Left... | x\in(5;8)\cup(8;29) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,794 |
9.210. $\frac{5 x-7}{x-5}<4-\frac{x}{5-x}+\frac{3 x}{x^{2}-25}<4$.
9.210. $\frac{5 x-7}{x-5}<4-\frac{x}{5-x}+\frac{3 x}{x^{2}-25}<4$. | Solution.
Domain of definition: $x \neq \pm 5$.
Rewrite the inequality as a system of inequalities
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \frac { 5 x - 7 } { x - 5 } < 4 - \frac { x } { 5 - x } + \frac { 3 x } { x ^ { 2 } - 2 5 } } \\
{ 4 - \frac { x } { 5 - x } + \frac { 3 x } { x ^ { 2 } - 2 5 } < 4 }
... | x\in(-8;-\frac{13}{2})\cup(0;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,795 |
9.212. $\left\{\begin{array}{l}\left(\frac{2}{3}\right)^{x} \cdot\left(\frac{8}{9}\right)^{-x}>\frac{27}{64} \\ 2^{x^{2}-6 x-3.5}<8 \sqrt{2}\end{array}\right.$ | ## Solution.
Let's rewrite the given system of inequalities as
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \frac { 2 ^ { x - 3 x } } { 3 ^ { x - 2 x } } > \frac { 27 } { 64 } , } \\
{ 2 ^ { x ^ { 2 } - 6 x - 3.5 } \left(\frac{3}{4}\right)^{3}, \\
x^{2}-6 x-3.5<3.5
\end{array} \Leftrightarrow\right.\right. \\
&... | x\in(-1;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,797 |
9.214. $\left\{\begin{array}{l}\left|x^{2}-4 x\right|<5 \\ |x+1|<3\end{array}\right.$
Solve the system of inequalities:
9.214. $\left\{\begin{array}{l}\left|x^{2}-4 x\right|<5 \\ |x+1|<3\end{array}\right.$ | ## Solution.
The system of inequalities is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 4 x - 5 , } \\
{ - 3 0 } \\
{ - 4 < x < 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-1<x<5 \\
x \in R, \\
-4<x<2
\end{array} \Leftrightarrow-1<x<2\right.\right.\right.
$$
Ans... | x\in(-1;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 48,799 |
9.215. Find the domain of the function
$$
y=\sqrt[4]{\frac{x^{2}-6 x-16}{x^{2}-12 x+11}}+\frac{2}{x^{2}-49}
$$ | ## Solution.
The domain of the given function consists of all values of $x$ that satisfy the system of inequalities
$\left\{\begin{array}{l}\frac{x^{2}-6 x-16}{x^{2}-12 x+11} \geq 0, \\ x^{2}-49 \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{(x-8)(x+2)}{(x-1)(x-11)} \geq 0, \\ x^{2} \neq 49\end{array} ... | x\in(-\infty;-7)\cup(-7;-2]\cup(1;7)\cup(7;8]\cup(11;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,800 |
10.191. The center of the circle inscribed in a right trapezoid is at distances of 3 and 9 cm from the ends of its lateral side. Find the sides of the trapezoid. | Solution.
Given $O C=3$ cm, $O D=9$ cm (see Fig. 10.1). Let $N, P, M, E$ be the points of tangency of the circle with the sides of the trapezoid $B C, C D, A D$, $A B$ respectively. By the property of tangents:
1) $O N \perp B C, O P \perp C D, O M \perp A D$.
2) $\angle N C O=\angle P C O, \angle M D O=\angle P D O$... | \frac{9\sqrt{10}}{5},\frac{6\sqrt{10}}{5},3\sqrt{10},\frac{18\sqrt{10}}{5}\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,801 |
10.192. Two circles touch each other externally. Their radii are in the ratio 3:1, and the length of their common external tangent is $6 \sqrt{3}$. Determine the perimeter of the figure formed by the external tangents and the external parts of the circles. | Solution.
Let $O_{2} B=r$; then $O_{1} A=R=3 r$ (Fig. 10.2). Draw $O_{2} C \| A B$; we have $A C=r, O_{1} C=2 r, O_{1} O_{2}=4 r$, i.e., $O_{1} O_{2}=2 O_{1} C$ and, therefore, $\angle C O_{2} O_{1}=30^{\circ}$. Since $O_{2} C=A B=6 \sqrt{3}$, from $\triangle O_{1} C O_{2}$ we find $O_{1} O_{2}=O_{2} C: \cos 30^{\circ... | 14\pi+12\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,802 |
10.193. Inside a right angle, there is a point $M$, the distances from which to the sides of the angle are 4 and $8 \mathrm{~cm}$. A line passing through point $M$ cuts off a triangle from the right angle with an area of $100 \mathrm{~cm}^{2}$. Find the legs of the triangle. | Solution.
Given $\angle C=90^{\circ}, M P=4 \mathrm{~cm}, M Q=8 \mathrm{~cm}, S_{\triangle A B C}=100 \mathrm{~cm}^{2}$ (Fig. 10.3); we need to find $B C$ and $A C$. Let $B C=x, A C=y$; then $0.5 x y=100$, i.e., $x y=200$. Since $\triangle B P M \sim \triangle M Q A$, we have $\frac{M P}{A Q}=\frac{B P}{M Q}$ or $\fra... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,803 |
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