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Ta có : \(\frac{15}{59}:\frac{24}{97}=\frac{485}{472}\)mà \(\frac{485}{472}>1\)
\(\Rightarrow\frac{15}{59}>\frac{24}{97}\)
a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)
b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
\(A=\frac{18}{11}+\frac{11}{13}+\frac{15}{17}+\frac{17}{15}\)
\(=1+\frac{7}{11}+1-\frac{2}{13}+1-\frac{2}{17}+1+\frac{2}{15}\)
\(=4+\frac{7}{11}+\frac{2}{15}-\frac{2}{13}-\frac{2}{17}\)
\(\frac{7}{11}>\frac{2}{11}>\frac{2}{13};\frac{2}{15}>\frac{2}{17}\Rightarrow\frac{7}{11}+\frac{2}{15}>\frac{2}{11}+\frac{2}{17}\)
\(\Rightarrow\frac{7}{11}+\frac{2}{15}-\frac{2}{13}-\frac{2}{17}>0\)
\(\Rightarrow A>4+0=4\)
Vậy A>4
\(\frac{22}{7}\)> \(\frac{11}{5}\)vì 22 : 7 = 3,14 ; 11: 5 = 2,2
\(\frac{15}{59}\)< \(\frac{24}{97}\)vì 15 : 59 = 0,21 ; 24 : 97 = 0,24
\(\frac{11}{19}\)< \(\frac{13}{18}\)vì 11 : 19 = 0,57 ; 13 : 18 = 0,72
\(\frac{7}{10}\)> \(\frac{4}{9}\)vì 7 : 10 = 0,7 ; 4 : 9 = 0,44
Ủa 15:59 ko ra0,21