problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | 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
class |
|---|---|---|---|---|---|---|---|---|
1.009.
$$
\frac{0.4+8\left(5-0.8 \cdot \frac{5}{8}\right)-5: 2 \frac{1}{2}}{\left(1 \frac{7}{8} \cdot 8-\left(8.9-2.6: \frac{2}{3}\right)\right) \cdot 34 \frac{2}{5}} \cdot 90
$$ | Solution.
$$
\begin{aligned}
& \frac{0.4+8\left(5-0.8 \cdot \frac{5}{8}\right)-5: 2 \frac{1}{2}}{\left(1 \frac{7}{8} \cdot 8-\left(8.9-2.6: \frac{2}{3}\right)\right) \cdot 34 \frac{2}{5}} \cdot 90=\frac{\left(0.4+40-4-5 \cdot \frac{2}{5}\right) \cdot 90}{\left(\frac{15}{8} \cdot 8-\frac{89}{10}+\frac{13}{5} \cdot \fra... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.010. $\frac{\left(5 \frac{4}{45}-4 \frac{1}{6}\right): 5 \frac{8}{15}}{\left(4 \frac{2}{3}+0.75\right) \cdot 3 \frac{9}{13}} \cdot 34 \frac{2}{7}+\frac{0.3: 0.01}{70}+\frac{2}{7}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(5 \frac{4}{45}-4 \frac{1}{6}\right): 5 \frac{8}{15}}{\left(4 \frac{2}{3}+0.75\right) \cdot 3 \frac{9}{13}} \cdot 34 \frac{2}{7}+\frac{0.3: 0.01}{70}+\frac{2}{7}=\frac{\left(\frac{229}{45}-\frac{25}{6}\right): \frac{83}{15}}{\left(\frac{14}{3}+\frac{3}{4}\right) \cdot \frac... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.011. $\frac{\left(\frac{3}{5}+0.425-0.005\right): 0.1}{30.5+\frac{1}{6}+3 \frac{1}{3}}+\frac{6 \frac{3}{4}+5 \frac{1}{2}}{26: 3 \frac{5}{7}}-0.05$. | ## Solution.
$$
\frac{\left(\frac{3}{5}+0.425-0.005\right): 0.1}{30.5+\frac{1}{6}+3 \frac{1}{3}}+\frac{6 \frac{3}{4}+5 \frac{1}{2}}{26: 3 \frac{5}{7}}-0.05=
$$
$=\frac{(0.6+0.42) \cdot 10}{\frac{61}{2}+\frac{1}{6}+\frac{10}{3}}+\frac{12 \frac{1}{4} \cdot 26}{26 \cdot 7}-0.05=$
$=\frac{10.2}{34}+\frac{7}{4}-\frac{1}{... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.012. $\frac{3 \frac{1}{3} \cdot 1.9+19.5: 4 \frac{1}{2}}{\frac{62}{75}-0.16}: \frac{3.5+4 \frac{2}{3}+2 \frac{2}{15}}{0.5\left(1 \frac{1}{20}+4.1\right)}$. | ## Solution.
$\frac{3 \frac{1}{3} \cdot 1.9 + 19.5 : 4 \frac{1}{2}}{\frac{62}{75} - 0.16} : \frac{3.5 + 4 \frac{2}{3} + 2 \frac{2}{15}}{0.5 \left(1 \frac{1}{20} + 4.1\right)} = \frac{\frac{10}{3} \cdot \frac{19}{10} + \frac{39}{2} \cdot \frac{2}{9}}{\frac{62}{75} - \frac{4}{25}} \cdot \frac{\frac{1}{2} \left(\frac{21}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.014. $\frac{\left(4.5 \cdot 1 \frac{2}{3}-6.75\right) \cdot \frac{2}{3}}{\left(3 \frac{1}{3} \cdot 0.3+5 \frac{1}{3} \cdot \frac{1}{8}\right): 2 \frac{2}{3}}+\frac{1 \frac{4}{11} \cdot 0.22: 0.3-0.96}{\left(0.2-\frac{3}{40}\right) \cdot 1.6}$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(4.5 \cdot 1 \frac{2}{3}-6.75\right) \cdot \frac{2}{3}}{\left(3 \frac{1}{3} \cdot 0.3+5 \frac{1}{3} \cdot \frac{1}{8}\right): 2 \frac{2}{3}}+\frac{1 \frac{4}{11} \cdot 0.22: 0.3-0.96}{\left(0.2-\frac{3}{40}\right) \cdot 1.6}=\frac{\left(\frac{9}{2} \cdot \frac{5}{3}-\frac{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.016. $\left(16 \frac{1}{2}-13 \frac{7}{9}\right) \cdot \frac{18}{33}+2.2\left(\frac{8}{33}-\frac{1}{11}\right)+\frac{2}{11}$. | ## Solution.
$$
\begin{aligned}
& \left(16 \frac{1}{2}-13 \frac{7}{9}\right) \cdot \frac{18}{33}+2.2\left(\frac{8}{33}-\frac{1}{11}\right)+\frac{2}{11}=\left(\frac{33}{2}-\frac{124}{9}\right) \cdot \frac{6}{11}+ \\
& +\frac{22}{10}\left(\frac{8}{33}-\frac{3}{33}\right)+\frac{2}{11}=\frac{49}{18} \cdot \frac{6}{11}+\fr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.017. $\frac{0.128: 3.2+0.86}{\frac{5}{6} \cdot 1.2+0.8} \cdot \frac{\left(1 \frac{32}{63}-\frac{13}{21}\right) \cdot 3.6}{0.505 \cdot \frac{2}{5}-0.002}$. | ## Solution.
$$
\begin{aligned}
& \frac{0.128: 3.2+0.86}{\frac{5}{6} \cdot 1.2+0.8} \cdot \frac{\left(1 \frac{32}{63}-\frac{13}{21}\right) \cdot 3.6}{0.505 \cdot \frac{2}{5}-0.002}=\frac{0.04+0.86}{1+0.8} \cdot \frac{\left(\frac{95}{63}-\frac{39}{63}\right) \cdot \frac{18}{5}}{0.202-0.002}= \\
& =\frac{9}{18} \cdot \f... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.018. $\frac{3 \frac{1}{3}: 10+0.175: 0.35}{1.75-1 \frac{11}{17} \cdot \frac{51}{56}}-\frac{\left(\frac{11}{18}-\frac{1}{15}\right): 1.4}{\left(0.5-\frac{1}{9}\right) \cdot 3}$. | Solution.
$$
\begin{aligned}
& \frac{3 \frac{1}{3}: 10+0.175: 0.35}{1.75-1 \frac{11}{17} \cdot \frac{51}{56}}-\frac{\left(\frac{11}{18}-\frac{1}{15}\right): 1.4}{\left(0.5-\frac{1}{9}\right) \cdot 3}=\frac{\frac{1}{3}+\frac{1}{2}}{\frac{7}{4}-\frac{28}{17} \cdot \frac{51}{56}}-\frac{\frac{49}{90} \cdot \frac{5}{7}}{\f... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.019. $\frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3) \cdot 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5$. | ## Solution.
$$
\begin{aligned}
& \frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3): 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5=\frac{\frac{1}{2}+\frac{5}{8}}{\left(10-\frac{220}{23}\right) \cdot \frac{23}{50}+\frac{8}{5}}+\frac{17}{40}+\frac{19}{20}= \\
& =\frac{\frac{9}{8}}{\frac{1}{5}+\frac{8}{5}}+\frac{1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.020. $\left(\left(1 \frac{1}{7}-\frac{23}{49}\right): \frac{22}{147}-\left(0.6: 3 \frac{3}{4}\right) \cdot 2 \frac{1}{2}+3.75: 1 \frac{1}{2}\right): 2.2$. | ## Solution.
$$
\begin{aligned}
& \left(\left(1 \frac{1}{7}-\frac{23}{49}\right): \frac{22}{147}-\left(0.6: 3 \frac{3}{4}\right) \cdot 2 \frac{1}{2}+3.75: 1 \frac{1}{2}\right): 2.2= \\
& =\left(\left(\frac{8}{7}-\frac{23}{49}\right) \frac{147}{22}-0.16 \cdot 2.5+2.5\right): 2.2=\left(\frac{33}{49} \cdot \frac{147}{22}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.022. $\frac{0.5+\frac{1}{4}+\frac{1}{6}+0.125}{\frac{1}{3}+0.4+\frac{14}{15}}+\frac{(3.75-0.625) \frac{48}{125}}{12.8 \cdot 0.25}$. | Solution.
$\frac{0.5+\frac{1}{4}+\frac{1}{6}+0.125}{\frac{1}{3}+0.4+\frac{14}{15}}+\frac{(3.75-0.625) \frac{48}{125}}{12.8 \cdot 0.25}=\frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}}{\frac{1}{3}+\frac{2}{5}+\frac{14}{15}}+\frac{3.125 \cdot 48}{3.2 \cdot 125}=$
$=\frac{25}{24} \cdot \frac{3}{5}+\frac{1.2}{3.2}=... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.024. $\frac{0.725+0.6+\frac{7}{40}+\frac{11}{20}}{0.128 \cdot 6 \frac{1}{4}-0.0345: \frac{3}{25}} \cdot 0.25$. | ## Solution.
$$
\begin{aligned}
& \frac{0.725+0.6+\frac{7}{40}+\frac{11}{20}}{0.128 \cdot 6 \frac{1}{4}-0.0345: \frac{3}{25}} \cdot 0.25=\frac{1.325+\frac{29}{40}}{0.128 \cdot 6.25-0.0345: 0.12} \cdot 0.25= \\
& =\frac{1.325+0.725}{0.8-0.2875} \cdot 0.25=\frac{2.05}{0.5125} \cdot 0.25=1
\end{aligned}
$$
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.025. $\left((520 \cdot 0.43): 0.26-217 \cdot 2 \frac{3}{7}\right)-\left(31.5: 12 \frac{3}{5}+114 \cdot 2 \frac{1}{3}+61 \frac{1}{2}\right)$. | Solution.
$$
\begin{aligned}
& \left((520 \cdot 0.43): 0.26-217 \cdot 2 \frac{3}{7}\right)-\left(31.5: 12 \frac{3}{5}+114 \cdot 2 \frac{1}{3}+61 \frac{1}{2}\right)= \\
& =\left(223.6: 0.26-217 \cdot \frac{17}{7}\right)-\left(\frac{63}{2} \cdot \frac{5}{63}+114 \cdot \frac{7}{3}+\frac{123}{2}\right)= \\
& =(860-527)-\l... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.026. $\frac{(3.4-1.275) \cdot \frac{16}{17}}{\frac{5}{18} \cdot\left(1 \frac{7}{85}+6 \frac{2}{17}\right)}+0.5\left(2+\frac{12.5}{5.75+\frac{1}{2}}\right)$. | Solution.
$$
\begin{aligned}
& \frac{(3.4-1.275) \cdot \frac{16}{17}}{\frac{5}{18} \cdot\left(1 \frac{7}{85}+6 \frac{2}{17}\right)}+0.5\left(2+\frac{12.5}{5.75+\frac{1}{2}}\right)=\frac{2.125 \cdot \frac{16}{17}}{\frac{5}{18}\left(\frac{92}{85}+\frac{104}{17}\right)}+\frac{1}{2}\left(2+\frac{12.5}{6.25}\right)= \\
& =... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.027. $\left(\frac{3.75+2 \frac{1}{2}}{2 \frac{1}{2}-1.875}-\frac{2 \frac{3}{4}+1.5}{2.75-1 \frac{1}{2}}\right) \cdot \frac{10}{11}$. | Solution.
$\left(\frac{3.75+2 \frac{1}{2}}{2 \frac{1}{2}-1.875}-\frac{2 \frac{3}{4}+1.5}{2.75-1 \frac{1}{2}}\right) \cdot \frac{10}{11}=\left(\frac{3.75+2.5}{2.5-1.875}-\frac{2.75+1.5}{2.75-1.5}\right) \cdot \frac{10}{11}=$
$=\left(\frac{6.25}{0.625}-\frac{4.25}{1.25}\right) \cdot \frac{10}{11}=\left(10-\frac{17}{5}\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.028. $((21.85: 43.7+8.5: 3.4): 4.5): 1 \frac{2}{5}+1 \frac{11}{21}$. | ## Solution.
$$
\begin{aligned}
& ((21.85: 43.7+8.5: 3.4): 4.5): 1 \frac{2}{5}+1 \frac{11}{21}=\left((0.5+2.5): 4 \frac{1}{2}\right): \frac{7}{5}+\frac{32}{21}= \\
& =\left(3 \cdot \frac{2}{9}\right) \cdot \frac{5}{7}+\frac{32}{21}=\frac{10}{21}+\frac{32}{21}=\frac{42}{21}=2
\end{aligned}
$$
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.029. $\left(1 \frac{2}{5}+3.5: 1 \frac{1}{4}\right): 2 \frac{2}{5}+3.4: 2 \frac{1}{8}-0.35$. | ## Solution.
$$
\begin{aligned}
& \left(1 \frac{2}{5}+3.5: 1 \frac{1}{4}\right): 2 \frac{2}{5}+3.4: 2 \frac{1}{8}-0.35= \\
& =(1.4+3.5: 1.25): 2.4+3.4: 2.125-0.35=(1.4+2.8): 2.4+ \\
& +1.6-0.35=4.2: 2.4+1.25=1.75+1.25=3
\end{aligned}
$$
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.032. $\frac{\left(3^{-1}-\sqrt{1 \frac{7}{9}}\right)^{-2}: 0.25}{\frac{37}{300}: 0.0925}+12.5 \cdot 0.64$. | ## Solution.
$$
\begin{aligned}
& \frac{\left(3^{-1}-\sqrt{1 \frac{7}{9}}\right)^{-2}: 0.25}{\frac{37}{300}: 0.0925}+12.5 \cdot 0.64=\frac{\left(\frac{1}{3}-\sqrt{\frac{16}{9}}\right)^{-2} \cdot 4}{\frac{37}{300} \cdot \frac{400}{37}}+8= \\
& =\frac{\left(\frac{1}{3}-\frac{4}{3}\right)^{-2} \cdot 4}{\frac{4}{3}}+8=3(-... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.033. $\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5$. | ## Solution.
$\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2.5}{\left(1.3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5=\frac{\left(\frac{5}{8}+\frac{65}{24}\right) \cdot \frac{2}{5} \cdot \frac{1}{2}}{\left(\frac{13}{10}+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}}=\frac{\frac{10}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.039. $\left(\frac{(3.2-1.7): 0.003}{\left(\frac{29}{35}-\frac{3}{7}\right) \cdot 4: 0.2}-\frac{\left(1 \frac{13}{20}-1.5\right) \cdot 1.5}{\left(2.44+1 \frac{14}{25}\right) \cdot \frac{1}{8}}\right): 62 \frac{1}{20}+1.364: 0.124$. | Solution.
$$
\begin{aligned}
& \left(\frac{(3.2-1.7): 0.003}{\left(\frac{29}{35}-\frac{3}{7}\right) \cdot 4: 0.2}-\frac{\left(1 \frac{13}{20}-1.5\right) \cdot 1.5}{\left(2.44+1 \frac{14}{25}\right) \cdot \frac{1}{8}}\right): 62 \frac{1}{20}+1.364: 0.124= \\
& =\left(\frac{1.5: 0.003}{\frac{14}{35} \cdot 4 \cdot 5}-\fr... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.041. $\frac{\left(4-3.5 \cdot\left(2 \frac{1}{7}-1 \frac{1}{5}\right)\right): 0.16}{X}=\frac{3 \frac{2}{7}-\frac{3}{14}: \frac{1}{6}}{41 \frac{23}{84}-40 \frac{49}{60}}$. | ## Solution.
$$
\begin{aligned}
& X=\frac{\left(4-3.5 \cdot\left(2 \frac{1}{7}-1 \frac{1}{5}\right)\right): 0.16 \cdot\left(41 \frac{23}{84}-40 \frac{49}{60}\right)}{3 \frac{2}{7}-\frac{3}{14}: \frac{1}{6}}= \\
& =\frac{\left(4-3.5 \cdot\left(\frac{15}{7}-\frac{6}{5}\right)\right): 0.16 \cdot \frac{16}{35}}{\frac{23}{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.043. $\frac{0.125 X}{\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7}{0.675 \cdot 2.4-0.02}$. | Solution.
$X=\frac{\left(1 \frac{28}{63}-\frac{17}{21}\right) \cdot 0.7 \cdot\left(\frac{19}{24}-\frac{21}{40}\right) \cdot 8 \frac{7}{16}}{(0.675 \cdot 2.4-0.02) \cdot 0.125}=\frac{\left(\frac{91}{63}-\frac{17}{21}\right) \cdot \frac{7}{10} \cdot \frac{4}{15} \cdot \frac{135}{16}}{(1.62-0.02) \cdot 0.125}=$ $=\frac{\... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.046.
$$
\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 |
1.047. $\left(\frac{\sqrt{561^{2}-459^{2}}}{4 \frac{2}{7} \cdot 0.15+4 \frac{2}{7}: \frac{20}{3}}+4 \sqrt{10}\right): \frac{1}{3} \sqrt{40}$. | Solution.
$$
\begin{aligned}
& \left(\frac{\sqrt{561^{2}-459^{2}}}{4 \frac{2}{7} \cdot 0.15+4 \frac{2}{7} : \frac{20}{3}}+4 \sqrt{10}\right): \frac{1}{3} \sqrt{40}=\left(\frac{\sqrt{(561+459)(561-459)}}{\frac{30}{7} \cdot \frac{3}{20}+\frac{30}{7} \cdot \frac{3}{20}}+4 \sqrt{10}\right) \times \\
& \times \frac{3}{2 \s... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.049. $\frac{2^{-2}+5^{0}}{(0.5)^{-2}-5(-2)^{-2}+\left(\frac{2}{3}\right)^{-2}}+4.75$. | Solution.
$$
\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 |
2.019. $\frac{\left(x^{2}-y^{2}\right)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}}}-\left(\sqrt[3]{x y}+\sqrt[3]{y^{2}}\right) x=64$. | Solution.
$$
\begin{aligned}
& \text { Domain: } z=\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}} \neq 0 \\
& \frac{\left(x^{2}-y^{2}\right)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^{5}}+\sqrt[3]{x^{2} y^{3}}-\sqrt[3]{x^{3} y^{2}}-\sqrt[3]{y^{5}}}-\left(\sqrt[3]{x y}+\sqrt[3]{y^{2}}\right)= \\... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.025. $\frac{a^{3}-a-2 b-b^{2} / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^{2}}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^{3}+a^{2}+a b+a^{2} b}{a^{2}-b^{2}}+\frac{b}{a-b}\right) ;$
$a=23 ; b=22$. | Solution.
$$
\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 |
2.109. $\left(-4 a^{3} \sqrt{\frac{\sqrt{a x}}{a^{2}}}\right)^{3}+\left(-10 a \sqrt{x} \cdot \sqrt{(a x)^{-1}}\right)^{2}+\left(-2\left(\sqrt[3]{a^{4} \sqrt{\frac{x}{a}}}\right)^{2}\right)^{3} ;$
$$
a=3 \frac{4}{7} ; x=0.28
$$ | ## Solution.
$$
\begin{aligned}
& \left(-4 a^{3} \sqrt{\frac{\sqrt{a x}}{a^{2}}}\right)^{3}+\left(-10 a \sqrt{x} \cdot \sqrt{(a x)^{-1}}\right)^{2}+\left(-2\left(\sqrt[3]{a \sqrt{\frac{x}{a}}}\right)^{2}\right)^{3}= \\
& =\frac{-64 a^{3} \sqrt{a x}}{a^{2}}+\frac{100 a^{2} x}{a x}-\frac{8 a^{2} \sqrt{x}}{\sqrt{a}}=-64 ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.123. $2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt[4]{75}-4 \sqrt{15 \sqrt{27}}$. | Solution.
$$
\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
$$
\begin{aligned}
& (4+\sqrt{15})^{2}(\sqrt{10}-\sqrt{6})^{2}(4-\sqrt{15})=4 \\
& (4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15})(10-2 \sqrt{60}+6)=4 \\
& \left(4^{2}-(\sqrt{15})^{2}\right)(4+\sqrt{15})(16-2 \sqrt{60})=4 \\
& (16-15)(4+\sqrt{15}) \cdot 2 \cdot(8-\sqr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.127. $\sqrt{3-\sqrt{5}} \cdot(3+\sqrt{5}) \cdot(\sqrt{10}-\sqrt{2})=8$. | ## Solution.
Let's square both sides of the equation. Then
$$
(\sqrt{3-\sqrt{5}})^{2}(3+\sqrt{5})^{2}(\sqrt{2}(\sqrt{5}-1))^{2}=64
$$
$(3-\sqrt{5})(3+\sqrt{5})^{2} \cdot 2(\sqrt{5}-1)^{2}=64$,
$(3-\sqrt{5})(3+\sqrt{5})(3+\sqrt{5})(5-2 \sqrt{5}+1)=32$,
$\left(3^{2}-(\sqrt{5})^{2}\right)(3+\sqrt{5})(6-2 \sqrt{5})=32... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.129. $\frac{25 \cdot \sqrt[4]{2}+2 \sqrt{5}}{\sqrt{250}+5 \sqrt[4]{8}}-\sqrt{\frac{\sqrt{2}}{5}+\frac{5}{\sqrt{2}}+2}=-1$. | ## Solution.
Let's set
$$
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 |
2.156. Calculate the sum of the cubes of two numbers if their sum and product are 11 and 21, respectively. | Solution.
Let $a+b=11$ and $a b=21$. Then
$$
\begin{aligned}
& a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)=(a+b)\left((a+b)^{2}-3 a b\right)=11\left(11^{2}-3 \cdot 21\right)= \\
& =11(121-63)=638
\end{aligned}
$$
Answer: 638. | 638 | 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 |
4.003. In a shooting competition, for each miss in a series of 25 shots, the shooter received penalty points: for the first miss - one penalty point, and for each subsequent miss - half a point more than for the previous one. How many times did the shooter hit the target, having received 7 penalty points? | Solution.
Let $a_{1}=1$ be the first term of the arithmetic progression, $d=\frac{1}{2}$ be its common difference, and $S_{n}=7$ be the sum of the first $n$ terms of this progression, where $n$ is the number of terms. Using formula (4.5), we have
$$
\frac{2+(n-1) \cdot \frac{1}{2}}{2} \cdot n=7, n^{2}+3 n-28=0
$$
fr... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.012. The sum of the third and ninth terms of an arithmetic progression is 8. Find the sum of the first 11 terms of this progression. | Solution.
From the condition, we have $a_{3}+a_{9}=8$. Using formula (4.1), we get $a_{1}+2 d+a_{1}+8 d=8, 2 a_{1}+10 d=8$, and using formula (4.5), we find $S_{11}=\frac{2 a_{1}+10 d}{2} \cdot 11=44$.
Answer: 44. | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.013. The sum of the first three terms of an increasing arithmetic progression is 15. If 1 is subtracted from the first two terms of this progression, and 1 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first 10 terms of the arithmetic progression. | ## Solution.
From the condition, we have: $a_{1}-1, a_{1}+d-1, a_{1}+2 d+1 \cdots$ - three consecutive terms of a geometric progression. Using formula (4.5), we find $S_{3}=\frac{2 a_{1}+2 d}{2} \cdot 3=15$ or $a_{1}+d=5$. Using formula (4.7), we get $\left(a_{1}+d-1\right)^{2}=\left(a_{1}-1\right)\left(a_{1}+2 d+1\ri... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.015. Calculate
$$
\left(1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}\right)-\left(2^{2}+4^{2}+6^{2}+(2 n)^{2}+\ldots+200^{2}\right)
$$ | ## Solution.
From the condition we have
$$
\begin{aligned}
& 1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}-2^{2}-4^{2}-6^{2}-(2 n)^{2}-\ldots-200^{2}= \\
& =(1-2)^{2}+\left(3^{2}-4^{2}\right)+\left(5^{2}-6^{2}\right)+\ldots+\left((2 n-1)^{2}-(2 n)^{2}\right)+\ldots+\left(199^{2}-200^{2}\right)= \\
& =(1-2)(1+2)+(3-... | -20100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.018. The denominator of the geometric progression is $1 / 3$, the fourth term of this progression is $1 / 54$, and the sum of all its terms is 121/162. Find the number of terms in the progression. | Solution.
From the condition we have $\left\{\begin{array}{l}b_{4}=\frac{1}{54}, \\ S_{n}=\frac{121}{162}\end{array}\right.$.
Using formulas (4.6) and (4.11), we get
$$
\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
$$
\left\{\begin{array} { l }
{ b _ { 1 } = 3 , } \\
{ b _ { 1 } q = 1 2 , } \\
{ b _ { 1 } q ^ { n - 1 } = 3 0 7 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=3, \\
q=4, \\
4^{n-1}=1024
\end{array}... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.029. Find the sum of all positive even two-digit numbers that are divisible by 3. | ## Solution.
From the condition, we have $a_{1}=12, a_{n}=96, d=12$.
Using formulas (4.4) and (4.5), we get
$$
n=\frac{a_{n}-a_{1}}{d}+1 ; n=\frac{96-12}{6}+1=15, S_{n}=\frac{12+96}{2} \cdot 15=810 \text {. }
$$
Answer: 810. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.031. It is known that the interior angles of a certain convex polygon, the smallest of which is $120^{\circ}$, form an arithmetic progression with a difference of $5^{\circ}$. Determine the number of sides of the polygon. | Solution.
From the condition, we have $a_{1}=120^{\circ}, d=5^{\circ}$. Using the formulas for the sum of terms of an arithmetic progression (4.5) and the sum of the interior angles of an $n$-sided polygon $S_{n}=180^{\circ}(n-2)$, we get
$$
\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 |
6.054. $\sqrt{5+\sqrt[3]{x}}+\sqrt{5-\sqrt[3]{x}}=\sqrt[3]{x}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}5+\sqrt[3]{x} \geq 0 \\ 5-\sqrt[3]{x} \geq 0\end{array} \Leftrightarrow -125 \leq x \leq 125\right.$.
By squaring both sides of the equation, we get the equation
$$
\begin{aligned}
& 5+\sqrt[3]{x}+2 \sqrt{(5+\sqrt[3]{x})(5-\sqrt[3]{x})}+5-\sqrt[3]{x}=\sqrt[3]{x... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.055. $\sqrt{x \sqrt[5]{x}}+\sqrt[5]{x \sqrt{x}}=56$.
6.055. $\sqrt{x \sqrt[5]{x}}+\sqrt[5]{x \sqrt{x}}=56$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition we have
$$
x^{\frac{6}{10}}-x^{\frac{3}{10}}=56 \Leftrightarrow\left(x^{\frac{3}{10}}\right)^{2}-x^{\frac{3}{10}}-56=0
$$
Let $x^{\frac{3}{10}}=y \geq 0$. The equation in terms of $y$ becomes $y^{2}-y-56=0$, from which $y_{1}=-7$ or $y_{2}=8 ; y_{1}=... | 1024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.058. $\sqrt[3]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-3, \\ x \neq 5\end{array}\right.$
Let $\sqrt[7]{\frac{5-x}{x+3}}=z, z \neq 0$. The equation in terms of $z$ becomes $z+\frac{1}{z}=2 \Leftrightarrow z^{2}-2 z+1=0 \Leftrightarrow(z-1)^{2}=0 \Leftrightarrow z-1=0, z=1$.
Then $\sqrt[7]{\frac{5-x}{x+3}}=1 \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.061. $2 \sqrt[3]{x}+5 \sqrt[6]{x}-18=0$. | ## Solution.
Domain of definition: $x \geq 0$.
Let $\sqrt[6]{x}=y \geq 0$. The equation in terms of $y$ becomes $2 y^{2}+5 y-18=0$, from which we find $y_{1}=-\frac{9}{2}, y_{2}=2 ; y_{1}=-\frac{9}{2}<0$ is not valid.
Then $\sqrt[6]{x}=2, x=2^{6}=64$.
Answer: $x=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.063. $\frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the equation as
$$
\frac{\sqrt[6]{x^{3}}+\sqrt[6]{x^{2}}}{\sqrt[6]{x^{3}}-\sqrt[6]{x^{2}}}=3 \Leftrightarrow \frac{\sqrt[6]{x^{3}}(\sqrt[6]{x}+1)}{\sqrt[6]{x^{3}}(\sqrt[6]{x}-1)}=3 \Leftrightarrow \frac{\sqrt[6]{x}+1}{\sqrt[6]{x}-1}=3 \Leftrightarrow
$$
$$
\b... | 64 | 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
$$
\begin{aligned}
& \left(\frac{a-8}{a-1}\right)^{2}+\left(\frac{a-... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.130. In the equation $x^{2}-2 x+c=0$, determine the value of $c$ for which its roots $x_{1}$ and $x_{2}$ satisfy the condition $7 x_{2}-4 x_{1}=47$. | Solution.
From the condition by Vieta's theorem we have $\left\{\begin{array}{l}x_{1}+x_{2}=2, \\ x_{1} \cdot x_{2}=c, \\ 7 x_{2}-4 x_{1}=47 .\end{array}\right.$ From here, $x_{2}=2-x_{1}$ and we obtain
$$
\left\{\begin{array} { l }
{ x _ { 1 } ( 2 - x _ { 1 } ) = c } \\
{ 7 ( 2 - x _ { 1 } ) - 4 x _ { 1 } = 4 7 }
\... | -15 | 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 |
7.002. $81^{\frac{1}{\log _{5} 3}}+27^{\log _{9} 36}+3^{\frac{4}{\log _{7} 9}}$. | Solution.
$81^{\frac{1}{\log _{5} 3}}+27^{\log _{9} 36}+3^{\frac{4}{\log _{7} 9}}=3^{4 \log _{3} 5}+3^{\frac{3}{2^{2} \log _{3} 36}}+3^{\frac{4}{2} \log _{3} 7}=5^{4}+36^{\frac{3}{2}}+49=$ $=625+216+49=890$.
Answer: 890. | 890 | 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.
 \cdot 49^{\log _{7} 2}$. | Solution.
$$
\begin{aligned}
& \left(81^{\frac{1}{4}-\frac{1}{2} \log _{9} 4}+25^{\log _{125} 8}\right) \cdot 49^{\log _{7} 2}=\left(\frac{81^{\frac{1}{4}}}{\left(9^{2}\right)^{\frac{1}{2} \log _{9} 4}}+5^{2 \log _{5} 32^{3}}\right) \cdot 7^{2 \log _{7} 2}= \\
& =\left(\frac{3}{4}+4\right) \cdot 4=19
\end{aligned}
$$
... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.008. $\frac{81^{\frac{1}{\log _{5} 9}}+3^{\frac{3}{\log _{\sqrt{6}} 3}}}{409} \cdot\left((\sqrt{7})^{\frac{2}{\log _{25} 7}}-125^{\log _{25} 6}\right)$
| ## 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 |
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