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\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
\(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{\frac{21}{44}+\frac{3}{13}}{\frac{20}{77}+\frac{5}{13}}+\frac{\frac{1}{6}+\frac{1}{4}}{\frac{5}{12}+\frac{5}{8}}\)
\(=\frac{\frac{405}{572}}{\frac{645}{1001}}+\frac{\frac{5}{12}}{\frac{25}{24}}\)
\(=\frac{1289}{860}\)
ta có \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\)
và \(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{2\left(\frac{1}{2.2}-\frac{1}{3.2}+\frac{1}{4.2}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2}{5}\)
Vậy \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
ĐS: 1
\(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3\times\frac{1}{7}-3\times\frac{1}{11}+3\times\frac{1}{13}}{5\times\frac{1}{7}-5\times\frac{1}{11}+5\times\frac{1}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\times\frac{1}{2}-\frac{5}{2}\times\frac{1}{3}+\frac{5}{2}\times\frac{1}{4}}\)
\(=\frac{3\times\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\times\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+1\div\frac{5}{2}\)
\(=\frac{3}{5}+1\times\frac{2}{5}\)
\(=\frac{3}{5}+\frac{2}{5}\)
\(=\frac{5}{5}\)
\(=1\)
\(\left(\frac{7}{8}-\frac{1}{4}\right).\left(\frac{5}{6}-\frac{3}{4}\right)\)
\(=\left(\frac{7}{8}-\frac{2}{8}\right).\left(\frac{10}{12}-\frac{9}{12}\right)\)
\(=\frac{5}{8}-\frac{1}{12}\)
\(=\frac{15}{24}-\frac{2}{24}\)
\(=\frac{13}{24}\)
Lộn :
\(\left(\frac{7}{8}-\frac{1}{4}\right).\left(\frac{5}{6}-\frac{3}{4}\right)\)
\(=\left(\frac{7}{8}-\frac{2}{8}\right).\left(\frac{10}{12}-\frac{9}{12}\right)\)
\(=\frac{5}{8}.\frac{1}{12}\)
\(=\frac{5}{96}\)