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a ) \(5\frac{3}{4}:3+2\frac{1}{4}.\frac{1}{3}-\frac{3}{8}=\frac{23}{4}:\frac{3}{1}+\frac{9}{4}.\frac{1}{3}=\frac{23}{12}+\frac{3}{4}=\frac{8}{3}\)
b ) \(\frac{3}{5}:\frac{5}{6}:\frac{6}{7}:\frac{7}{8}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{3.6.7.8}{5.5.6.7}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{24}{25}+\frac{7}{8}+\frac{2}{5}+\frac{23}{35}=\frac{4049}{1400}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
\(A=5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+...+100}\)
A = \(5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+..+100}\)
\(=5x\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=5x\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=2x5x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=10x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{100x101}\right)\)
\(=10x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=10x\left(1-\frac{1}{101}\right)\)
\(=10x\frac{100}{101}\)
\(=\frac{1000}{101}\)