problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
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|---|---|---|---|---|---|---|---|---|
1.043. $\frac{0.125 X}{\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7}{0.675 \cdot 2.4-0.02}$. | Solution.
$X=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7 \cdot\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}{(0.675 \cdot 2.4-0.02) \cdot 0.125}=\frac{\left(\frac{91}{63}-\frac{17}{21}\right) \cdot \frac{7}{10} \cdot \frac{4}{15} \cdot \frac{135}{16}}{(1.62-0.02) \cdot 0.125}=$ $=\frac{\... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.046.
$$
\frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{(6.3+1.7)^{2}-4 \cdot 6.3 \cdot 1.7}}
$$ | Solution.
$$
\begin{aligned}
& \frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{(6.3+1.7)^{2}-4 \cdot 6.3 \cdot 1.7}}=\frac{\sqrt{6.3 \cdot 1.7} \cdot\left(\sqrt{\frac{6.3}{1.7}}-\sqrt{\frac{1.7}{6.3}}\right)}{\sqrt{6.3^{2}+2 \cdot 6.3 \cdot 1.7+1.7^{2}-4 \cdot 6.3 \cdo... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.049. $\frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+\left(\frac{2}{3}\right)^{-2}}+4.75$. | Solution.
$$
\begin{aligned}
& \frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+\left(\frac{2}{3}\right)^{-2}}+4.75=\frac{\frac{1}{2^{2}}+1}{-\frac{1}{(0.5)^{2}}-\frac{5}{(-2)^{-2}}+\left(\frac{3}{2}\right)^{2}}+4.75= \\
& =\frac{\frac{1}{4}+1}{\frac{1}{0.25}-\frac{5}{4}+\frac{9}{4}}+4.75=\frac{-\frac{4}{4}}{4+1}+4.75=\frac{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.025. $\frac{a^{3}-a-2 b-b^{2} / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^{3}+a^{2}+a b+a^{2} b}{a^{2}-b^{2}}+\frac{b}{a-b}\right) ;$
$a=23 ; b=22$. | Solution.
$$
\begin{aligned}
& \frac{a^{3}-a-2 b-b^{2} / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^{3}+a^{2}+a b+a^{2} b}{a^{2}-b^{2}}+\frac{b}{a-b}\right)= \\
& =\frac{\frac{a^{4}-a^{2}-2 a b-b^{2}}{a}}{\left(1-\sqrt{\frac{a+b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.
2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.
(Note: The original text and the translation are identical as the expression... | Solution.
Domain of definition: $\left\{\begin{array}{l}a>0, \\ a \neq \sqrt{2} .\end{array}\right.$
$$
\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a^{2} \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}=\sqrt{\frac{a^{2}+2 a \sqrt{2}+(\sqrt{2})^{2}}{a \sqrt{2}}}-
$$
$$
\begin{aligned}
& -\frac{a^{2} ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.039. $\frac{9 b^{4 / 3}-\frac{a^{3 / 2}}{b^{2}}}{\sqrt{a^{3 / 2} b^{-2}+6 a^{3 / 4} b^{-1 / 3}+9 b^{4 / 3}}} \cdot \frac{b^{2}}{a^{3 / 4}-3 b^{5 / 3}} ; \quad b=4$. | Solution.
$$
\begin{aligned}
& \frac{9 b^{4 / 3}-\frac{a^{3 / 2}}{b^{2}}}{\sqrt{a^{3 / 2} b^{-2}+6 a^{3 / 4} b^{-1 / 3}+9 b^{4 / 3}}} \cdot \frac{b^{2}}{a^{3 / 4}-3 b^{5 / 3}}=\frac{\frac{9 b^{4 / 3} \cdot b^{2}-a^{3 / 2}}{b^{2}}}{\sqrt{\frac{a^{3 / 2}}{b^{2}}+\frac{6 a^{3 / 4}}{b^{+1 / 3}}+9 b^{4 / 3}}} \times \\
& \... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.044. $\left(\frac{\sqrt{x-a}}{\sqrt{x+a}+\sqrt{x-a}}+\frac{x-a}{\sqrt{x^{2}-a^{2}}-x+a}\right): \sqrt{\frac{x^{2}}{a^{2}}-1} ; x>a>0$.
2.044. $\left(\frac{\sqrt{x-a}}{\sqrt{x+a}+\sqrt{x-a}}+\frac{x-a}{\sqrt{x^{2}-a^{2}}-x+a}\right): \sqrt{\frac{x^{2}}{a^{2}}-1} ; x>a>0$. | Solution.
$$
\begin{aligned}
& \left(\frac{\sqrt{x-a}}{\sqrt{x+a}+\sqrt{x-a}}+\frac{x-a}{\sqrt{x^{2}-a^{2}}-x+a}\right): \sqrt{\frac{x^{2}}{a^{2}}-1}= \\
& =\left(\frac{\sqrt{x-a}}{\sqrt{x+a}+\sqrt{x-a}}+\frac{(\sqrt{x-a})^{2}}{\sqrt{x-a}(\sqrt{x+a}-\sqrt{x-a})}\right): \sqrt{\frac{x^{2}-a^{2}}{a^{2}}}= \\
& =\left(\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.046. $\frac{\sqrt{1-x^{2}}-1}{x} \cdot\left(\frac{1-x}{\sqrt{1-x^{2}}+x-1}+\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}\right)$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ -1 \leq x<1 .\end{array}\right.$
$$
\begin{aligned}
& \frac{\sqrt{1-x^{2}}-1}{x} \cdot\left(\frac{1-x}{\sqrt{1-x^{2}}+x-1}+\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}\right)=\frac{\sqrt{1-x^{2}}-1}{x} \times \\
& \times\left(\frac{(\sqrt{1-x})^{2}}{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.051. $\frac{\left(a^{2}-b^{2}\right)\left(a^{2}+\sqrt[3]{b^{2}}+a \sqrt[3]{b}\right)}{a \sqrt[3]{b}+a \sqrt{a}-b \sqrt[3]{b}-\sqrt{a b^{2}}}: \frac{a^{3}-b}{a \sqrt[3]{b}-\sqrt[6]{a^{3} b^{2}}-\sqrt[3]{b^{2}}+a \sqrt{a}} ;$
$$
a=4.91 ; b=0.09
$$ | Solution.
$$
\begin{aligned}
& \frac{\left(a^{2}-b^{2}\right)\left(a^{2}+\sqrt[3]{b^{2}}+a \sqrt[3]{b}\right)}{a \sqrt[3]{b}+a \sqrt{a}-b \sqrt[3]{b}-\sqrt{a b^{2}}}: \frac{a^{3}-b}{a \sqrt[3]{b}-\sqrt[6]{a^{3} b^{2}}-\sqrt[3]{b^{2}}+a \sqrt{a}}= \\
& =\frac{(a-b)(a+b)\left(a^{2}+a \sqrt[3]{b}+\sqrt[3]{b^{2}}\right)}{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.054. $\frac{3 a^{2}+2 a x-x^{2}}{(3 x+a)(a+x)}-2+10 \cdot \frac{a x-3 x^{2}}{a^{2}-9 x^{2}}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq \pm \frac{a}{3}, \\ x \neq-a .\end{array}\right.$
$$
\begin{aligned}
& \frac{3 a^{2}+2 a x-x^{2}}{(3 x+a)(a+x)}-2+10 \cdot \frac{a x-3 x^{2}}{a^{2}-9 x^{2}}=\frac{-(x+a)(x-3 a)}{(3 x+a)(a+x)}-2+ \\
& +10 \cdot \frac{x(a-3 x)}{(a-3 x)(a+3 x)}=\frac{-x+3 a}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.058. $\left(\left(\frac{1}{a}+\frac{1}{b+c}\right):\left(\frac{1}{a}-\frac{1}{b+c}\right)\right):\left(1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$;
$$
a=1 \frac{33}{40} ; b=0.625 ; c=3.2
$$ | Solution.
$$
\begin{aligned}
& \left(\left(\frac{1}{a}+\frac{1}{b+c}\right):\left(\frac{1}{a}-\frac{1}{b+c}\right):\left(1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)=\right. \\
& =\left(\frac{a+b+c}{a(b+c)}: \frac{-a+b+c}{a(b+c)}\right): \frac{2 b c+b^{2}+c^{2}-a^{2}}{2 b c}= \\
& =\left(\frac{a+b+c}{a(b+c)} \cdot \frac{a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.068. $\frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}} ; \quad a=7.4 ; b=\frac{5}{37}$. | Solution.
$$
\begin{aligned}
& \frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}}=\frac{\frac{a+b-2 c}{a b} \cdot(a+b+2 c)}{\frac{a^{2}+2 a b+b^{2}-4 c^{2}}{a^{2} b^{2}}}= \\
& =\frac{\frac{(a+b-2 c)(a+b+2 c)}{a b}}{\frac{(a+b... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.073. $\frac{\sqrt{5-2 \sqrt{6}}}{(\sqrt[4]{3}+\sqrt[4]{2})(\sqrt[4]{3}-\sqrt[4]{2})}$. | Solution.
$$
\begin{aligned}
& \frac{\sqrt{5-2 \sqrt{6}}}{(\sqrt[4]{3}+\sqrt[4]{2})(\sqrt[4]{3}-\sqrt[4]{2})}=\frac{\sqrt{3-2 \sqrt{3 \cdot 2}+2}}{(\sqrt[4]{3})^{2}-(\sqrt[4]{2})^{2}}= \\
& =\frac{\sqrt{(\sqrt{3})^{2}-2 \sqrt{3} \cdot \sqrt{2}+(\sqrt{2})^{2}}}{\sqrt{3}-\sqrt{2}}= \\
& =\frac{\sqrt{(\sqrt{3}-\sqrt{2})^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.078. $\left(\frac{1}{t^{2}+3 t+2}+\frac{2 t}{t^{2}+4 t+3}+\frac{1}{t^{2}+5 t+6}\right)^{2} \cdot \frac{(t-3)^{2}+12 t}{2}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}t \neq-3, \\ t \neq-2, \\ t \neq-1 .\end{array}\right.$
$$
\begin{aligned}
& \left(\frac{1}{t^{2}+3 t+2}+\frac{2 t}{t^{2}+4 t+3}+\frac{1}{t^{2}+5 t+6}\right)^{2} \cdot \frac{(t-3)^{2}+12 t}{2}= \\
& =\left(\frac{1}{(t+2)(t+1)}+\frac{2 t}{(t+3)(t+1)}+\frac{1}{(t+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.084. $\left(\frac{2-b}{b-1}+2 \cdot \frac{a-1}{a-2}\right):\left(b \cdot \frac{a-1}{b-1}+a \cdot \frac{2-b}{a-2}\right)$;
$a=\sqrt{2}+0.8 ; b=\sqrt{2}-0.2$. | Solution.
$$
\begin{aligned}
& \left(\frac{2-b}{b-1}+2 \cdot \frac{a-1}{a-2}\right):\left(b \cdot \frac{a-1}{b-1}+a \cdot \frac{2-b}{a-2}\right)=\frac{(2-b)(a-2)+2(a-1)(b-1)}{(b-1)(a-2)} \\
& \frac{b(a-1)(a-2)+a(2-b)(b-1)}{(b-1)(a-2)}=\frac{a b-2}{(b-1)(a-2)} \cdot \frac{(b-1)(a-2)}{a^{2} b-a b^{2}-2 a+2 b}= \\
& =\fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.105. $\left(\frac{1+\sqrt{1-x}}{1-x+\sqrt{1-x}}+\frac{1-\sqrt{1+x}}{1+x-\sqrt{1+x}}\right)^{2} \cdot \frac{x^{2}-1}{2}-\sqrt{1-x^{2}}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}-1<x<1, \\ x \neq 0 .\end{array}\right.$
$$
\begin{aligned}
& \left(\frac{1+\sqrt{1-x}}{1-x+\sqrt{1-x}}+\frac{1-\sqrt{1+x}}{1+x-\sqrt{1+x}}\right)^{2} \cdot \frac{x^{2}-1}{2}-\sqrt{1-x^{2}}= \\
& \left.\left.=\left(\frac{1+\sqrt{1-x}}{\sqrt{1-x}(\sqrt{1-x}+1}\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.123. $2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt[4]{75}-4 \sqrt{15 \sqrt{27}}$. | Solution.
$$
\begin{aligned}
& 2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt[4]{75}-4 \sqrt{15 \sqrt{27}}= \\
& =2 \sqrt{40 \sqrt{4 \cdot 3}}+3 \sqrt{5 \sqrt{16 \cdot 3}}-2 \sqrt[4]{25 \cdot 3}-4 \sqrt{15 \sqrt{9 \cdot 3}}= \\
& =2 \sqrt{40 \cdot 2 \sqrt{3}}+3 \sqrt{5 \cdot 4 \sqrt{3}}-2 \sqrt{\sqrt{25 \cdot 3}}-... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.126. $(4+\sqrt{15})(\sqrt{10}-\sqrt{6}) \cdot \sqrt{4-\sqrt{15}}=2$.
| ## Решение.
Возведем обе части равенства в квадрат. Тогда
$$
\begin{aligned}
& (4+\sqrt{15})^{2}(\sqrt{10}-\sqrt{6})^{2}(4-\sqrt{15})=4 \\
& (4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15})(10-2 \sqrt{60}+6)=4 \\
& \left(4^{2}-(\sqrt{15})^{2}\right)(4+\sqrt{15})(16-2 \sqrt{60})=4 \\
& (16-15)(4+\sqrt{15}) \cdot 2 \cdot(8-\sqr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.129. $\frac{25 \cdot \sqrt[4]{2}+2 \sqrt{5}}{\sqrt{250}+5 \sqrt[4]{8}}-\sqrt{\frac{\sqrt{2}}{5}+\frac{5}{\sqrt{2}}+2}=-1$. | ## Solution.
Let's set
$$
X=\frac{25 \cdot \sqrt[4]{2}+2 \sqrt{5}}{\sqrt{250}+5 \sqrt[4]{8}}=\frac{\sqrt[4]{5^{8} \cdot 2}+\sqrt[4]{5^{2} \cdot 2^{4}}}{\sqrt[4]{5^{6} \cdot 2^{2}}+\sqrt[4]{5^{4} \cdot 2^{3}}}=\frac{\sqrt[4]{5^{2} \cdot 2}\left(\sqrt[4]{5^{6}}+\sqrt[4]{2^{3}}\right)}{\sqrt[4]{5^{2} \cdot 2} \cdot \sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.130. $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2 \sqrt{3}}+1}}=\sqrt{2}$.
| ## Решение.
Возведем обе части равенства в квадрат. Тогда
$$
\begin{aligned}
& \frac{2 \sqrt[4]{27}-2 \sqrt{(\sqrt[4]{27})^{2}-(\sqrt{\sqrt{3}-1})^{2}}}{\sqrt[4]{27}-\sqrt{2 \sqrt{3}+1}}=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.135. $\frac{x^{3}-a^{-2 / 3} \cdot b^{-1}\left(a^{2}+b^{2}\right) x+b^{1 / 2}}{b^{3 / 2} \cdot x^{2}} ; x=a^{2 / 3} b^{-1 / 2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}a \neq 0, \\ b \neq 0 .\end{array}\right.$
$$
\begin{aligned}
& \frac{\left(a^{2 / 3} b^{-1 / 2}\right)^{3}-a^{2 / 3} \cdot b^{-1}\left(a^{2}+b^{2}\right)^{2 / 3} b^{-1 / 2}+b^{1 / 2}}{b^{3 / 2} \cdot\left(a^{2 / 3} b^{-1 / 2}\right)^{2}}= \\
& =\frac{a^{2} b... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.136. $\frac{1-b}{\sqrt{b}} \cdot x^{2}-2 x+\sqrt{b} ; \quad x=\frac{\sqrt{b}}{1-\sqrt{b}}$.
2.136. $\frac{1-b}{\sqrt{b}} \cdot x^{2}-2 x+\sqrt{b} ; \quad x=\frac{\sqrt{b}}{1-\sqrt{b}}$. | ## Solution.
Domain of definition: $0<b \neq 1$.
$$
\begin{aligned}
& \frac{1-b}{\sqrt{b}} \cdot\left(\frac{\sqrt{b}}{1-\sqrt{b}}\right)^{2}-2 \cdot \frac{\sqrt{b}}{1-\sqrt{b}}+\sqrt{b}=\frac{(1-\sqrt{b})(1+\sqrt{b})}{\sqrt{b}} \cdot \frac{b}{(1-\sqrt{b})^{2}}-\frac{2 \sqrt{b}}{1-\sqrt{b}}+ \\
& +\sqrt{b}=\frac{(1+\s... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.141. $\frac{(1-y)(y+2)}{y^{2}(y+1)^{2}} ; \quad y=\frac{\sqrt{3}-1}{2}$. | Solution.
$$
\begin{aligned}
& \frac{\left(1-\frac{\sqrt{3}-1}{2}\right) \cdot\left(\frac{\sqrt{3}-1}{2}+2\right)}{\left(\frac{\sqrt{3}-1}{2}\right)^{2} \cdot\left(\frac{\sqrt{3}-1}{2}+1\right)^{2}}=\frac{-\left(\frac{\sqrt{3}-1}{2}-1\right) \cdot\left(\frac{\sqrt{3}-1}{2}+2\right)}{\left(\frac{\sqrt{3}-1}{2} \cdot\le... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.145. $\frac{1-a x}{1+a x} \cdot \sqrt{\frac{1+b x}{1-b x}} ; \quad x=\frac{1}{a} \cdot \sqrt{\frac{2 a-b}{b}} ; \quad 0<\frac{b}{2}<a<b$. | Solution.
$\frac{1-a \cdot \frac{1}{a} \cdot \sqrt{\frac{2 a-b}{b}}}{1+a \cdot \frac{1}{a} \cdot \sqrt{\frac{2 a-b}{b}}} \cdot \sqrt{\frac{1+b \cdot \frac{1}{a} \cdot \sqrt{\frac{2 a-b}{b}}}{1-b \cdot \frac{1}{a} \cdot \sqrt{\frac{2 a-b}{b}}}}=\frac{1-\sqrt{\frac{2 a-b}{b}}}{1+\sqrt{\frac{2 a-b}{b}}} \times$
$$
\begi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.154. What is the value of $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}$, given that the difference $\sqrt{25-x^{2}}-\sqrt{15-x^{2}}=2$ (the value of $x$ does not need to be found)? | Solution.
Domain of definition: $\left\{\begin{array}{l}25-x^{2} \geq 0, \\ 15-x^{2} \geq 0\end{array} \Leftrightarrow-\sqrt{15} \leq x \leq \sqrt{15}\right.$.
Multiplying both sides of the equation by $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}$, we have
$$
\begin{aligned}
& \left(\sqrt{25-x^{2}}-\sqrt{15-x^{2}}\right)\left(\... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.040. $\frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=2$. | Solution.
$$
\begin{aligned}
& \frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=\frac{1-\cos 4 \alpha}{\frac{1}{\cos ^{2} 2 \alpha}-1}+\frac{1+\cos 4 \alpha}{\frac{1}{\sin ^{2} 2 \alpha}-1}= \\
& =\frac{(1-\cos 4 \alpha) \cos ^{2} 2 \alpha}{1-\cos ^{2} 2 \alpha}+\frac{(1+\cos... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.081. $\sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)$. | Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)= \\
& =\left(\sin \left(\frac{3}{2} \pi-\alpha\right)\right)^{2}\left(1-\operatorname{tg}^{2} \alpha\righ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.088. $\frac{\operatorname{ctg}\left(270^{\circ}-\alpha\right)}{1-\operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)} \cdot \frac{\operatorname{ctg}^{2}\left(360^{\circ}-\alpha\right)-1}{\operatorname{ctg}\left(180^{\circ}+\alpha\right)}$. | ## Solution.
$\frac{\operatorname{ctg}\left(270^{\circ}-\alpha\right)}{1-\operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)} \cdot \frac{\operatorname{ctg}^{2}\left(360^{\circ}-\alpha\right)-1}{\operatorname{ctg}\left(180^{\circ}+\alpha\right)}=\frac{\operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha} \cdot \f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.098. $\sin \left(2 \alpha-\frac{3}{2} \pi\right)+\cos \left(2 \alpha-\frac{8}{3} \pi\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right)$. | ## Solution.
Let
$$
\begin{aligned}
& X=\sin \left(2 \alpha-\frac{3}{2} \pi\right)+\cos \left(2 \alpha-\frac{8}{3} \pi\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right)= \\
& =-\sin \left(\frac{3}{2} \pi-2 \alpha\right)+\cos \left(\frac{8}{3} \pi-2 \alpha\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right) \\
& -\sin \... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.105. $\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}$. | ## Solution.
$\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (\pi-6 \alpha)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos 6 \alpha}{\cos 2 \alpha}=$
$=\frac{\sin 6 \alpha \cos 2 \alpha-\cos 6 \alpha \sin 2 \alpha}{\sin 2 \al... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.148. $\left(\sin 160^{\circ}+\sin 40^{\circ}\right)\left(\sin 140^{\circ}+\sin 20^{\circ}\right)+\left(\sin 50^{\circ}-\sin 70^{\circ}\right) \times$
$$
\times\left(\sin 130^{\circ}-\sin 110^{\circ}\right)=1
$$ | Solution.
$\left(\sin 160^{\circ}+\sin 40^{\circ}\right)\left(\sin 140^{\circ}+\sin 20^{\circ}\right)+\left(\sin 50^{\circ}-\sin 70^{\circ}\right)\left(\sin 130^{\circ}-\sin 110^{\circ}\right)=$
$$
\begin{aligned}
& =\left(\sin \left(180^{\circ}-20^{\circ}\right)+\sin 40^{\circ}\right)\left(\sin \left(180^{\circ}-40^... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3.153. $\sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}$. | Solution.
$$
\begin{aligned}
& \sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}= \\
& =\frac{1-\cos \frac{\pi}{4}}{2}+\frac{1+\cos \frac{3 \pi}{4}}{2}+\frac{1-\cos \frac{5 \pi}{4}}{2}+\frac{1+\cos \frac{7 \pi}{4}}{2}=
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.154. $\operatorname{tg} 435^{\circ}+\operatorname{tg} 375^{\circ}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.154. $\operatorname{tan} 435^{\circ}+\operatorname{tan} 375^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \tan 435^{\circ}+\tan 375^{\circ}=\tan\left(450^{\circ}-15^{\circ}\right)+\tan\left(360^{\circ}+15^{\circ}\right)= \\
& =\cot 15^{\circ}+\tan 15^{\circ}=\frac{\cos 15^{\circ}}{\sin 15^{\circ}}+\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=\frac{\cos ^{2} 15^{\circ}+\sin ^{2} 15^{\circ}}{\si... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.177. Calculate $(1+\operatorname{ctg} \alpha)(1+\operatorname{ctg} \beta)$, if $\alpha+\beta=\frac{3 \pi}{4}$. | Solution.
$$
\begin{aligned}
& (1+\operatorname{ctg} \alpha)(1+\operatorname{ctg} \beta)=\left(1+\frac{\cos \alpha}{\sin \alpha}\right)\left(1+\frac{\cos \beta}{\sin \beta}\right)=\frac{\sin \alpha+\cos \alpha}{\sin \alpha} \times \\
& \times \frac{\sin \beta+\cos \beta}{\sin \beta}=\frac{\cos \alpha \cos \beta+\sin \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.178. Calculate $(1+\operatorname{tg} \alpha)(1+\operatorname{tg} \beta)$, if $\alpha+\beta=\frac{\pi}{4}$. | Solution.
$$
(1+\operatorname{tg} \alpha)(1+\operatorname{tg} \beta)=\left(1+\frac{\sin \alpha}{\cos \alpha}\right)\left(1+\frac{\sin \beta}{\cos \beta}\right)=\frac{\cos \alpha+\sin \alpha}{\cos \alpha} \times
$$
$$
\begin{aligned}
& \times \frac{\cos \beta+\sin \beta}{\cos \beta}=\frac{\cos \alpha \cos \beta+\sin \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.018. The denominator of the geometric progression is $1 / 3$, the fourth term of this progression is $1 / 54$, and the sum of all its terms is 121/162. Find the number of terms in the progression. | Solution.
From the condition we have $\left\{\begin{array}{l}b_{4}=\frac{1}{54}, \\ S_{n}=\frac{121}{162}\end{array}\right.$.
Using formulas (4.6) and (4.11), we get
$$
\begin{aligned}
& b_{4}=b_{1} q^{3}=b_{1}\left(\frac{1}{3}\right)^{3} ; \frac{b_{1}}{27}=\frac{1}{54}, b_{1}=\frac{1}{2} ; \\
& S_{n}=\frac{b_{1}\le... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.028. Find the number of terms in a finite geometric progression, where the first, second, and last terms are 3, 12, and 3072, respectively. | Solution.
From the condition we have $b_{1}=3, b_{2}=12, \ldots, b_{n}=3072$.
By formula (4.6) we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } = 3 , } \\
{ b _ { 1 } q = 1 2 , } \\
{ b _ { 1 } q ^ { n - 1 } = 3 0 7 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=3, \\
q=4, \\
4^{n-1}=1024
\end{array}... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.031. It is known that the interior angles of a certain convex polygon, the smallest of which is $120^{\circ}$, form an arithmetic progression with a difference of $5^{\circ}$. Determine the number of sides of the polygon. | Solution.
From the condition, we have $a_{1}=120^{\circ}, d=5^{\circ}$. Using the formulas for the sum of terms of an arithmetic progression (4.5) and the sum of the interior angles of an $n$-sided polygon $S_{n}=180^{\circ}(n-2)$, we get
$$
\frac{240^{\circ}+(n-1) 5^{\circ}}{2} \cdot n=180^{\circ}(n-2), n^{2}-25 n+1... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.012. $(x-1)\left(x^{2}-3\right)+(2 x-1)\left(x^{2}+2\right)=3$. | Solution.
Domain of definition: $x \in R$.
## We have
$$
\begin{aligned}
& x^{3}-x^{2}-3 x+3+2 x^{3}-x^{2}+4 x-2=3 \Leftrightarrow \\
& \Leftrightarrow 3 x^{3}-2 x^{2}+x-2=0 \Leftrightarrow 3 x^{3}-3 x^{2}+x^{2}-x+2 x-2=0 \Leftrightarrow \\
& \Leftrightarrow 3 x^{2}(x-1)+x(x-1)+2(x-1)=0 \Leftrightarrow(x-1)\left(3 x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.014. $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$. | ## Solution.
Domain: $x \in R$.
$\frac{2 x^{4}+9 x^{2}}{\left(x^{2}+4\right)\left(x^{2}+5\right)}=0 \Leftrightarrow 2 x^{4}+9 x^{2}=0 \Leftrightarrow x^{2}\left(2 x^{2}+9\right)=0$,
$x^{2}=0, x_{1}=0$ or $2 x^{2}+9=0, x_{2,3} \in \varnothing$.
Answer: $x=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.031. $\sqrt{3 x+4}+\sqrt{x-4}=2 \sqrt{x}$. | ## Solution.
Domain of definition: $3 x+4 \geq 0, x-4 \geq 0, x \geq 0 \Rightarrow x \geq 4$.
Squaring both sides of the equation, we get
$$
\begin{aligned}
& 3 x+4+2 \sqrt{(3 x+4)(x-4)}+x-4=4 x \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{(3 x+4)(x-4)}=0
\end{aligned}
$$
Squaring again, we get: $(3 x+4)(x-4)=0$. F... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$.
6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$. | Solution.
Let $\sqrt{x+11}=y \geq 0$ or $x+11=y^{2}$, i.e., $x=y^{2}-11$. Then
$$
\sqrt{y^{2}+y-11}+\sqrt{y^{2}-y-11}=4 \text { or } \sqrt{y^{2}+y-11}=4-\sqrt{y^{2}-y-11}
$$
Squaring both sides of the equation, we get
$$
y^{2}+y-11=16-8 \sqrt{y^{2}-y-11}+y^{2}-y-11, 8 \sqrt{y^{2}-y-11}=16-2 y
$$
or $4 \sqrt{y^{2}-... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$.
6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$. | Solution.
Write the equation as $\sqrt{1+x \sqrt{x^{2}-24}}=x-1$. Squaring both sides of the equation, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 1 + x \sqrt { x ^ { 2 } - 2 4 } = x ^ { 2 } - 2 x + 1 , } \\
{ x - 1 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \sqrt{x^{2}-24}=x^{2}-2 x... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.037. $\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=2$. | Solution.
Domain of definition: $x \geq 0$.
Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 1+\sqrt{x}+3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})^{2}}+1-\sqrt{x}=8 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$. | ## Solution.
Domain of definition: $x \geq 0$.
Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 24+\sqrt{x}-3 \sqrt[3]{(24+\sqrt{x})^{2}(5+\sqrt{x})}+3 \sqrt[3]{(24+\sqrt{x})(5+\sqrt{x})^{2}}-5-\sqrt{x}=1 \Leftrightarrow \\
& \Leftrightarrow-3 \sqrt[3]{(24+\sqrt{x})}(5+\sqrt{x})(\sq... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$.
6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$. | ## Solution.
Domain of definition: $x^{3}+8 \geq 0 \Leftrightarrow x^{3} \geq-8 \Leftrightarrow x \geq-2$.
Let $\sqrt[4]{x^{3}+8}=y, y>0$, and the equation becomes $y^{2}+y=6 \Leftrightarrow$ $\Leftrightarrow y^{2}+y-6=0$, from which $y_{1}=-3, y_{2}=2 ; y_{1}=-3$ is not suitable. Then $\sqrt[4]{x^{3}+8}=2, x^{3}+8=1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.052.
$$
\frac{1}{x-\sqrt{x^{2}-x}}-\frac{1}{x+\sqrt{x^{2}-x}}=\sqrt{3}
$$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-x \geq 0, \\ x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x-1) \geq 0, \\ x \neq 0\end{array} \Leftrightarrow x \in(-\infty ; 0) \cup[1 ;+\infty)\right.\right.$.
From the condition we get
$$
\begin{aligned}
& \frac{x+\sqrt{x^{2}-x}-x+\sq... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.053. $\frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$ | ## Solution.
Domain of definition: $x \neq \pm 1$.
Let $\sqrt[3]{x}=y, y \neq \pm 1$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& \frac{y^{4}-1}{y^{2}-1}-\frac{y^{2}-1}{y+1}=4 \Leftrightarrow \frac{\left(y^{2}-1\right)\left(y^{2}+1\right)}{y^{2}-1}-\frac{(y-1)(y+1)}{y+1}=4 \Leftrightarrow \\
& \Leftrig... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.058. $\sqrt[3]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-3, \\ x \neq 5\end{array}\right.$
Let $\sqrt[7]{\frac{5-x}{x+3}}=z, z \neq 0$. The equation in terms of $z$ becomes $z+\frac{1}{z}=2 \Leftrightarrow z^{2}-2 z+1=0 \Leftrightarrow(z-1)^{2}=0 \Leftrightarrow z-1=0, z=1$.
Then $\sqrt[7]{\frac{5-x}{x+3}}=1 \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$.
6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+2 \geq 0, \\ 3 x+8 \geq 0, \\ 2 x+6 \geq 0\end{array} \Leftrightarrow x \geq-2\right.$.
Write the equation in the form $\sqrt{x+2}-\sqrt{2 x+6}=-\sqrt{3 x+8}$ and square both sides:
$$
\begin{aligned}
& x+2-2 \sqrt{(x+2)(2 x+6)}+2 x+6=3 x+8 \Leftrightarrow \\... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.065. $\sqrt{2 x+5}+\sqrt{5 x+6}=\sqrt{12 x+25}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}2 x+5 \geq 0, \\ 5 x+6 \geq 0, \\ 12 x+25 \geq 0\end{array} \Leftrightarrow x \geq-\frac{6}{5}\right.$.
By squaring both sides of the equation, we have
$$
\begin{aligned}
& 2 x+5+2 \sqrt{(2 x+5)(5 x+6)}+5 x+6=12 x+25 \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.126. For what integer value of $k$ is one of the roots of the equation $4 x^{2}-(3 k+2) x+\left(k^{2}-1\right)=0$ three times smaller than the other? | Solution.
From the condition, by Vieta's theorem, we have
where $k \in \mathbb{Z}$. From this, $37 k^{2}-36 k-76=0, k_{1}=2, k_{2}=-\frac{38}{37} \notin \mathbb{Z}$ (does not fit).
Answer... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.129. For what value of $a$ do the equations $x^{2}+a x+8=0$ and $x^{2}+x+a=0$ have a common root? | ## Solution.
Let $x_{1}$ be the common root, then
$$
\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+8=0, \\
x_{1}^{2}+x_{1}+a=0
\end{array} \Rightarrow a x_{1}-x_{1}+8-a=0, x_{1}=\frac{a-8}{a-1}\right.
$$
From the second equation of the system, we have
$$
\begin{aligned}
& \left(\frac{a-8}{a-1}\right)^{2}+\left(\frac{a-... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.131. Without solving the equation $x^{2}-(2 a+1) x+a^{2}+2=0$, find the value of $a$ for which one of the roots is twice the other. | Solution.
From the condition, by Vieta's theorem, we have
$$
\left\{\begin{array} { l }
{ x _ { 1 } + x _ { 2 } = 2 a + 1 , } \\
{ x _ { 1 } \cdot x _ { 2 } = a ^ { 2 } + 2 , } \\
{ x _ { 2 } = 2 x _ { 1 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 3 x _ { 1 } = 2 a + 1 , } \\
{ 2 x _ { 1 } ^ { 2 } ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.134. For what integer value of $b$ do the equations $2 x^{2}+(3 b-1) x-3=0$ and $6 x^{2}-(2 b-3) x-1=0$ have a common root? | Solution.
Let $x_{1}$ be the common root. Then
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 x _ { 1 } ^ { 2 } + ( 3 b - 1 ) x _ { 1 } - 3 = 0 , } \\
{ 6 x _ { 1 } ^ { 2 } - ( 2 b - 3 ) x _ { 1 } - 1 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
6 x^{2}+(9 b-3) x-9=0, \\
6 x^{2}-(2 b-3) x_{1}-1=0
\... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.001. $\sqrt{25^{\frac{1}{\log _{6} 5}}+49^{\frac{1}{\log _{8} 7}}}$ | ## Solution.
$$
\begin{aligned}
& \sqrt{\frac{1}{25^{\log _{6} 5}}+49^{\frac{1}{\log _{8} 7}}}=\sqrt{5^{2 \log _{5} 6}+7^{2 \log _{7} 8}}=\sqrt{5^{\log _{5} 6^{2}}+7^{\log _{7} 8^{2}}}= \\
& =\sqrt{6^{2}+8^{2}}=10
\end{aligned}
$$
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.003. $-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.003. $-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}$. | Solution.
$$
-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}=-\log _{2} \log _{2} 2^{\frac{1}{8}}=-\log _{2} \frac{1}{8} \log _{2} 2=-\log _{2} 2^{-3}=3
$$
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.004. $-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}$.
7.004. $-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}$. | ## Solution.
$-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}=-\log _{3} \log _{3} 3^{\frac{1}{9}}=-\log _{3} \frac{1}{9} \log _{3} 3=-\log _{3} 3^{-2}=2$.
Answer: 2.
## Solution.
^{\frac{2}{\log _{25} 7}}-125^{\log _{25} 6}\right)$
| ## Решение.
$$
\frac{81^{\frac{1}{\log _{5} 9}}+3^{\frac{3}{\log _{\sqrt{6}}}}}{409} \cdot\left((\sqrt{7})^{\frac{2}{\log _{25} 7}}-125^{\log _{25} 6}\right)=
$$
$=\frac{9^{2 \log _{9} 5}+3^{3 \log _{3} \sqrt{6}}}{409} \cdot\left(\left(7^{\frac{1}{2}}\right)^{2 \log _{7} 25}-5^{3 \log _{5} 26}\right)=\frac{9^{\log _{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.009. $\left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{64} N}}\right)^{\frac{1}{15}}$ (the bases of the logarithms are consecutive natural powers of the number 2). | ## Solution.
$$
\begin{aligned}
& \left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{512} N}}\right)^{\frac{1}{15}}= \\
& =\left(N^{\log _{N} 2} \cdot N^{\log _{N} 4} \cdot N^{\log _{N} 8} \cdots N^{\log _{N} 512}\right)^{\frac{1}{15}}= \\
& =(2 \c... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.021. $3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$.
7.021. $3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$. | Solution.
Domain of definition: $3^{x}-5^{2-x}>0$.
$\log _{5} 8+2 \log _{5} 5-\log _{5}\left(3^{x}-25 \cdot 5^{-x}\right)=x \Leftrightarrow \log _{5} \frac{8 \cdot 25}{3^{x}-25 \cdot 5^{-x}}=x$,
from which $\frac{200}{3^{x}-25 \cdot 5^{-x}}=5^{x} \Leftrightarrow 15^{x}=15^{2}$. Therefore, $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.028. $5^{2\left(\log _{5} 2+x\right)}-2=5^{x+\log _{5} 2}$. | ## Solution.
$\left(5^{x+\log _{5} 2}\right)^{2}-5^{x+\log _{5} 2}-2=0$; solving this equation as a quadratic equation in terms of $5^{x+\log _{5} 2}$, we find $5^{x+\log _{5} 2}=-1$ and $5^{x+\log _{5} 2}=2 ; 5^{x+\log _{5} 2}=-1$ has no solutions.
Thus,
$$
5^{x+\log _{5} 2}=2 \Rightarrow \log _{5} 5^{x+\log _{5} 2... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.029. $0.25^{\log _{2} \sqrt{x+3}-0.5 \log _{2}\left(x^{2}-9\right)}=\sqrt{2(7-x)}$. | ## Solution.
Domain of definition: $\quad\left\{\begin{array}{l}x+3>0, \\ x^{2}-9>0.33 . \Rightarrow x_{1}=5, x_{2}=-1 ; x_{2}=-1 \text{ does not fit the domain of definition.}\end{array}\right.$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.031. $\log _{5}(x-2)+\log _{\sqrt{5}}\left(x^{3}-2\right)+\log _{0.2}(x-2)=4$. | ## Solution.
Domain of definition: $\quad x-2>0, x>2$.
From the condition we have
$$
\log _{5}(x-2)+2 \log _{5}\left(x^{3}-2\right)-\log _{5}(x-2)=4, \log _{5}\left(x^{3}-2\right)=2
$$
from which $x^{3}-2=25, x^{3}=27$. Then $x=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.034. $\lg \left(3^{x}-2^{4-x}\right)=2+0.25 \lg 16-0.5 x \lg 4$. | Solution.
Domain of definition: $3^{x}-2^{4-x}>0$.
From the condition
$$
\begin{aligned}
& \lg \left(3^{x}-2^{4-x}\right)=\lg 100+\lg 2-\lg 2^{x} \Rightarrow \lg \left(3^{x}-2^{4-x}\right)=\lg \frac{100 \cdot 2}{2^{x}} \\
& 3^{x}-2^{4-x}=\frac{200}{2^{x}}
\end{aligned}
$$
From here $6^{x}=216$, hence $x=3$.
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.035. $\log _{3}\left(81^{x}+3^{2 x}\right)=3 \log _{27} 90$. | ## Solution.
From the condition $\log _{3}\left(81^{x}+3^{2 x}\right)=\log _{3} 90, 9^{2 x}+9^{x}-90=0$, from which we find $9^{x}=-10$, which is not suitable, or $9^{x}=9$, from which we have $x=1$.
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.039. $\lg \left(10^{\lg \left(x^{2}-21\right)}\right)-2=\lg x-\lg 25$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-21>0, \\ x>0,\end{array} x>\sqrt{21}\right.$.
From the condition we have
$$
\lg \left(x^{2}-21\right)-\lg 100=\lg x-\lg 25, \lg \frac{x^{2}-21}{100}=\lg \frac{x}{25}, \quad \frac{x^{2}-21}{100}=\frac{x}{25}
$$
We obtain the quadratic equation $x^{2}-4... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$. | Solution.
$$
\begin{aligned}
& x(\lg 5-\lg 10)=\lg \left(2^{x}+1\right)-\lg 6, \quad x \lg \frac{5}{10}=\lg \frac{2^{x}+1}{6} \\
& \lg 2^{-x}=\lg \frac{2^{x}+1}{6}, 2^{-x}=\frac{2^{x}+1}{6}, 2^{2 x}+2^{x}-6=0
\end{aligned}
$$
Solving this equation as a quadratic in terms of $2^{x}$, we find $2^{x}=-3$ (not valid), $2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.047. Find the natural number $n$ from the equation
$$
3^{2} \cdot 3^{5} \cdot 3^{8} \cdots 3^{3 n-1}=27^{5}
$$ | ## Solution.
$3^{2+5+8+\ldots+3 n-1}=3^{15}, 2+5+8+\ldots+3 n-1=15$.
On the left side of the equation, we have the sum of the terms of an arithmetic progression $S_{k}$, where $a_{1}=2, d=3, a_{k}=3 n-1, k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-1-2}{3}+1=n$.
Then $S_{k}=\frac{a_{1}+a_{k}}{2} \cdot k=\frac{2+3 n-1}{2} \cd... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.050. $\log _{2} \frac{x-5}{x+5}+\log _{2}\left(x^{2}-25\right)=0$. | ## Solution.
Domain of definition: $\frac{x-5}{x+5}>0$ or $x \in(-\infty ;-5) \cup(5 ; \infty)$.
We have $\log _{2} \frac{(x-5)\left(x^{2}-25\right)}{x+5}=0,(x-5)^{2}=1$, from which $x-5=-1$ or $x-5=1$. Then $x_{1}=4, x_{2}=6 ; x_{1}=4$ does not fit the domain of definition.
Answer: 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.053. $\lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.053. $\lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\lg x>0, \\ \lg x^{3}-2>0,\end{array} \quad x>\sqrt[3]{100}\right.$.
From the condition we have
$\lg \left(\lg x \cdot\left(\lg x^{3}-2\right)\right)=0, \quad \lg x(3 \lg x-2)=1$,
$3 \lg ^{2} x-2 \lg x-1=0$.
Solving this equation as a quadratic equation in te... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.055. $\log _{3}\left(3^{x}-8\right)=2-x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.055. $\log _{3}\left(3^{x}-8\right)=2-x$. | Solution.
Domain of definition: $3^{x}-8>0$.
By the definition of logarithm, we have $3^{x}-8=3^{2-x}, 3^{x}-8=\frac{9}{3^{x}}$, $3^{2 x}-8 \cdot 3^{x}-9=0$, from which, solving this equation as a quadratic equation in terms of $3^{x}$, we find $3^{x}=-1, \varnothing$; or $3^{x}=9$, from which $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.057. $5^{x+6}-3^{x+7}=43 \cdot 5^{x+4}-19 \cdot 3^{x+5}$. | Solution.
We have $5^{6} \cdot 5^{x}-43 \cdot 5^{4} \cdot 5^{x}=3^{7} \cdot 3^{x}-19 \cdot 3^{5} \cdot 3^{x},\left(\frac{5}{3}\right)^{x}=\left(\frac{5}{3}\right)^{-3}$, from which $x=-3$.
Answer: -3. | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.066. $2^{\frac{1}{\sqrt{x}-1}} \cdot 0.5^{\frac{1}{\sqrt{x}+1}}=4^{\frac{\sqrt{x}}{x+\sqrt{x}}}$ | Solution.
Domain of definition: $0<x \neq 1$.
We have: $2^{\frac{1}{\sqrt{x}-1}} \cdot 2^{-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}, 2^{\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}$.
Then $\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}=\frac{2 \sqrt{x}}{x+\sqrt{x}}, x-\sqrt{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.069. $\log _{\sqrt{5}}\left(4^{x}-6\right)-\log _{\sqrt{5}}\left(2^{x}-2\right)=2$. | Solution.
Domain of Definition (DOD): $\left\{\begin{array}{l}4^{x}-6>0 \\ 2^{x}-2>0 .\end{array}\right.$
We have $\log _{\sqrt{5}} \frac{4^{x}-6}{2^{x}-2}=2, \frac{2^{2 x}-6}{2^{2}-2}=5,2^{2 x}-5 \cdot 2^{x}+4=0$. Solving this equation as a quadratic in terms of $2^{x}$, we find $\left(2^{x}\right)=1$, from which we... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.071. $3 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}=0.2$. | ## Solution.
From the condition $3 \cdot 5^{2 x}-2 \cdot 5^{x}=1,3 \cdot 5^{2 x}-2 \cdot 5^{x}-1=0$. Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{1}{3}, \varnothing$; or $5^{x}=1$, from which $x=0$.
Answer: 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$.
7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$. | Solution.
Domain of definition: $x-5 \geq 0, x \geq 5$.
$$
3^{2 \sqrt{x-5}}-6 \cdot 3^{\sqrt{x-5}}-27=0
$$
We solve the equation as a quadratic equation in terms of $3^{\sqrt{x-5}}$. We have $3^{\sqrt{x-5}}=-3$. (not suitable) or $3^{\sqrt{x-5}}=9$, from which $\sqrt{x-5}=2$, or $x-5=4$. Then $x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.080. $\lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}6+x \geq 0, \\ x>0, \\ x \neq 1,\end{array} \quad 0<x \neq 1\right.$.
We will switch to base 10. We have
$\lg (\sqrt{6+x}+6)=2 \lg \sqrt{x}, \quad \lg (\sqrt{6+x}+6)=\lg x$.
Then $\sqrt{6+x}+6=x, \sqrt{6+x}=x-6 \Rightarrow\left\{\begin{array}{l}x^{2}-13 x+30=0... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.084. $\frac{2^{x}+10}{4}=\frac{9}{2^{x-2}}$.
7.084. $\frac{2^{x}+10}{4}=\frac{9}{2^{x-2}}$. | ## Solution.
From the condition
$$
\frac{2^{x}+10}{4}=\frac{9}{2^{x} \cdot 2^{-2}}, \frac{2^{x}+10}{4}=\frac{36}{2^{x}}, 2^{2 x}+10 \cdot 2^{x}-144=0 .
$$
Solving this equation as a quadratic in terms of $2^{x}$, we find $2^{x}=-18, \varnothing$, or $2^{x}=8$, from which $x=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.091. $\log _{2}\left(4 \cdot 3^{x}-6\right)-\log _{2}\left(9^{x}-6\right)=1$. | ## Solution.
oDZ: $\left\{\begin{array}{l}4 \cdot 3^{x}-6>0 \\ 9^{x}-6>0\end{array}\right.$
We have $\log _{2} \frac{4 \cdot 3^{x}-6}{3^{2 x}-6}=1, \frac{4 \cdot 3^{x}-6}{3^{2 x}-6}=2 \Rightarrow 3^{2 x}-2 \cdot 3^{x}-3=0$. Solving it as a quadratic equation in terms of $3^{x}$, we find $3^{x}=-1, \varnothing$; or $3... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.098. $\lg \sqrt{x-3}+\lg \sqrt{x+3}=2-0.5 \lg 625$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x-3>0, \\ x+3>0\end{array} x>3\right.$.
## We have
$\lg \sqrt{x-3}+\lg \sqrt{x+3}=\lg 100-\lg 25, \lg \sqrt{x^{2}-9}=\lg 4, \sqrt{x^{2}-9}=4$, from which $x^{2}=25, x_{1}=-5, x_{2}=5, x_{1}=-5$ does not satisfy the domain of definition.
Answer: 5 . | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.099. $\lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$. | ## Solution.
Domain of definition: $3-x>0, x<3$.
## Rewrite the equation as
$$
3 \lg (3-x)=\lg \left(27-x^{3}\right), \lg (3-x)^{3}=\lg \left(27-x^{3}\right)
$$
Then $(3-x)^{3}=27-x^{3} \Rightarrow x^{2}-9 x=0$, from which $x_{1}=0, x_{2}=9 ; x_{2}=9$ does not satisfy the domain of definition.
Answer: 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.101. $\lg 8-\lg \sqrt{x+6}=\lg 16-\lg (x-2)$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x+6>0, \\ x-2>0,\end{array} \quad x>2\right.$.
We have
$$
\lg \frac{8}{\sqrt{x+6}}=\lg \frac{16}{x-2}, \quad \frac{8}{\sqrt{x+6}}=\frac{16}{x-2}, \quad 2 \sqrt{x+6}=x-2, x^{2}-8 x-20=0
$$
from which $x_{1}=10, x_{2}=-2 ; x_{2}=-2$ does not satisfy the domain o... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.103. $\frac{\lg (2 x-19)-\lg (3 x-20)}{\lg x}=-1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}2 x-19>0, \\ 3 x-20>0,\end{array} \quad x>\frac{19}{2}\right.$.
## From the condition
$$
\begin{aligned}
& \lg (2 x-19)-\lg (3 x-20)=-\lg x, \lg (2 x-19)+\lg x=\lg (3 x-20) \\
& x(2 x-19)=3 x-20, x^{2}-11 x+10=0 .
\end{aligned}
$$
From here $x_{1}=10, x_{2}... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.104. $\frac{\lg x^{2}}{\lg (6 x-5)}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ 6 x-5>0,\end{array}, \frac{5}{6}<x \neq 1\right.$.
We have $\lg x^{2}=\lg (6 x-5)$, from which $x^{2}=6 x-5, x^{2}-6 x+5=0$, hence $x_{1}=5$ and $x_{2}=1 ; x_{2}=1$ does not satisfy the domain of definition.
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.109. $5^{2 x-1}+2^{2 x}-5^{2 x}+2^{2 x+2}=0$. | Solution.
Write the equation as
$$
\begin{aligned}
& \frac{5^{2 x}}{5}-5^{2 x}=-2^{2 x}-4 \cdot 2^{2 x},-\frac{4}{5} \cdot 5^{2 x}=-5 \cdot 2^{2 x} \\
& \left(\frac{5}{2}\right)^{2 x}=\left(\frac{5}{2}\right)^{2}, x=1
\end{aligned}
$$
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$.
7.110. $\log _{2}\left(9-2^{x}\right)=10^{\lg (3-x)}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}9-2^{x}>0, \\ 3-x>0,\end{array} \quad x<3\right.$.
We have $\log _{2}\left(9-2^{x}\right)=3-x, 9-2^{x}=2^{3-x}, 2^{2 x}-9 \cdot 2^{x}+8=0$. Solving this equation as a quadratic in terms of $2^{x}$, we get $\left(2^{x}\right)=1$ or $\left(2^{x}\right)_{2}=8$, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.111. $\frac{1}{3} \lg \left(271+3^{2 \sqrt{x}}\right)+\lg 10=2$. | ## Solution.
Domain: $x \geq 0$.
From the condition $\frac{1}{3} \lg \left(271+3^{2 \sqrt{x}}\right)+1=2, \lg \left(271+3^{2 \sqrt{x}}\right)=3$. Then $271+3^{2 \sqrt{x}}=$ $=1000,3^{2 \sqrt{x}}=3^{6}$, from which $\sqrt{x}=3, x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.112. $\left.(\sqrt[5]{27})^{\frac{x}{4}-\sqrt{\frac{x}{3}}}\right)^{\frac{x}{4}+\sqrt{\frac{x}{3}}}=\sqrt[4]{3^{7}}$. | ## Solution.
Domain of definition: $x \geq 0$.
Rewrite the equation as $3^{\frac{3}{5}\left(\frac{x}{4}-\sqrt{\frac{x}{3}}\right)\left(\frac{x}{4}+\sqrt{\frac{x}{3}}\right)}=3^{\frac{7}{4}}$. Then
$$
\frac{3}{5}\left(\frac{x}{4}-\sqrt{\frac{x}{3}}\right)\left(\frac{x}{4}+\sqrt{\frac{x}{3}}\right)=\frac{7}{4}, 3 x^{2... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.114. $\lg (x(x+9))+\lg \frac{x+9}{x}=0$. | ## Solution.
Domain of definition: $x(x+9)>0, x \in(-\infty ;-9) \cup(0 ; \infty)$.
We have $\lg \frac{x(x+9)(x+9)}{x}=0$, from which $(x+9)^{2}=1$. Then $(x+9)_{1}=-1$, $x_{1}=-10$ or $(x+9)_{2}=1, x_{2}=-8 ; x_{2}=-8$ does not fit the domain of definition.
Answer: -10. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.117. $2^{\log _{3} x^{2}} \cdot 5^{\log _{3} x}=400$. | Solution.
Domain: $x>0$.
From the condition $4^{\log _{3} x} \cdot 5^{\log _{3} x}=400, 20^{\log _{3} x}=20^{2}$, hence $\log _{3} x=2$, $x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.121. $\log _{2}\left(4^{x}+4\right)=x+\log _{2}\left(2^{x+1}-3\right)$ | ## Solution.
Domain of definition: $2^{x+1}-3>0$.
Rewrite the equation as
$$
\begin{aligned}
& \log _{2}\left(2^{2 x}+4\right)-\log _{2}\left(2 \cdot 2^{x}-3\right)=x, \quad \log _{2} \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=x \\
& \frac{2^{2 x}+4}{2 \cdot 2^{x}-3}=2^{x}, 2^{2 x}-3 \cdot 2^{x}-4=0 .
\end{aligned}
$$
Solvi... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.127. $\log _{2}\left(25^{x+3}-1\right)=2+\log _{2}\left(5^{x+3}+1\right)$ | Solution.
Domain of definition: $25^{x+3}-1>0, 25^{x+3}>25^{0}, x>-3$.
From the condition
$\log _{2}\left(25^{3} \cdot 25^{x}-1\right)=\log _{2} 4\left(5^{3} \cdot 5^{x}+1\right) 25^{3} \cdot 5^{2 x}-1=4 \cdot 5^{3} \cdot 5^{x}+4$, $3125 \cdot 5^{2 x}-100 \cdot 5^{x}-1=0$,
from which, solving this equation as a qua... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.010. Find the positive integer values of $x$ that satisfy the inequality $\frac{5 x+1}{x-1}>2 x+2$. | Solution.
We have
$$
\begin{aligned}
& \frac{5 x+1}{x-1}-2 x-2>0 \Leftrightarrow \frac{5 x+1-2(x+1)(x-1)}{x-1}>0 \Leftrightarrow \frac{-2 x^{2}+5 x+3}{x-1}>0 \Leftrightarrow \\
& \Leftrightarrow\left(2 x^{2}-5 x-3\right)(x-1)<0 \Leftrightarrow 2\left(x+\frac{1}{2}\right)(x-3)(x-1)<0
\end{aligned}
$$
Using the number... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.012. Find the natural values of $x$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
\log _{\sqrt{2}}(x-1)<4 \\
\frac{x}{x-3}+\frac{x-5}{x}<\frac{2 x}{3-x}
\end{array}\right.
$$ | Solution.
From the condition
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 0 0 \text { for } x \in R, \\
x(x-3)<0 .
\end{array}\right.
\end{aligned}
$$
Using the number line, we find the solution to the system $x=2$.
 We have $D E=\sqrt{289-225}=8 \mathrm{~cm}$, $A D=2 E D+B C=16+B C, B C+16+B C=34, B C=9 \mathrm{~cm}, A D=25 \mathrm{~cm}$.
Answer: 9 cm $; 25$ c... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.013. Given a triangle with sides 12, 15, and \(18 \, \text{cm}\). A circle is drawn that touches both smaller sides and has its center on the larger side. Find the segments into which the center of the circle divides the larger side of the triangle. | ## Solution.
Using the cosine theorem, from Fig. 10.13 we have:
$$
\begin{aligned}
& A B^{2}=B C^{2}+A C^{2}-2 \cdot B C \cdot A C \cdot \cos \alpha \Rightarrow \\
& \Rightarrow 12^{2}=15^{2}+18^{2}-2 \cdot 15 \cdot 18 \cos \alpha, \cos \alpha=\frac{3}{4}, \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{7}}{4} \\
&... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.017. Given a point $P$, which is 7 cm away from the center of a circle with a radius of $11 \mathrm{~cm}$. A chord of length 18 cm is drawn through this point. What are the lengths of the segments into which the chord is divided by point $P$? | ## Solution.
Draw the diameter $C D$ through point $P$ (Fig. 10.17), which will divide it into segments $P D$ and $C P$ of lengths $11-7=4$ and $11+7=18$ (cm). Let $A P=x$; then $P B=18-x$. Since $A P \cdot P B=C P \cdot P D=4 \cdot 18$, we have $x(18-x)=72$ or $x^{2}-18 x+72=0$, from which $x_{1}=12, x_{2}=6$, i.e., ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.024. The lengths of the parallel sides of the trapezoid are 25 and 4 cm, and the lengths of the non-parallel sides are 20 and 13 cm. Find the height of the trapezoid. | ## Solution.
Given $B C=4 \text{ cm}, A D=25 \text{ cm}, A B=20 \text{ cm}, C D=13 \text{ cm}$ (Fig. 10.24). Draw $B E \perp A D$ and $C F \perp A D$. Let $B E=C F=h, A E=x$, $F D=y$. Then from $\triangle A B E$ and $\triangle C F D$ we find $h^{2}=20^{2}-x^{2}=13^{2}-y^{2}$. Considering that $y=25-4-x=21-x$, we have ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.033. A tangent, parallel to the base, is drawn to the circle inscribed in an isosceles triangle with a base of 12 cm and a height of 8 cm. Find the length of the segment of this tangent, enclosed between the sides of the triangle. | ## Solution.
Let's find the length of the side $B C$ (Fig. 10.33): $B C=\sqrt{B M^{2}+M C^{2}}=$ $=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $M O / O B=A M / A B$ or $r /(8-r)=6 / 10$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \sim \triangl... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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