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\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
\(A=\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\)
\(=\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{106.113}\right)\)
\(=\frac{3}{7}\left(\frac{8-1}{1.8}+\frac{15-8}{8.15}+\frac{22-15}{15.22}+...+\frac{113-106}{106.113}\right)\)
\(=\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+\frac{1}{106}-\frac{1}{113}\right)\)
\(=\frac{3}{7}\left(1-\frac{1}{113}\right)=\frac{48}{113}\)
\(B=\frac{25}{50.55}+\frac{25}{55.60}+...+\frac{25}{95.100}\)
\(=5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)
\(=5\left(\frac{55-50}{50.55}+\frac{60-55}{55.60}+...+\frac{100-95}{95.100}\right)\)
\(=5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=5\left(\frac{1}{50}-\frac{1}{100}\right)=\frac{1}{20}\)
Giá trị của biểu thức đã cho là:
\(A-B=\frac{48}{113}-\frac{1}{20}=\frac{847}{2260}\)