2/1*3+2/3*5+2/5*7+...+2/101*99
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\(\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{8}{9}\right)=1+1+1+1=4\)
\(=\dfrac{2}{5}\cdot\dfrac{7}{3}\cdot\dfrac{3}{7}\cdot\dfrac{5}{2}+2020=1+2020=2021\)
\(5\dfrac{3}{4}:3+2\dfrac{1}{4}\times\dfrac{1}{3}-\dfrac{1}{8}=\left(5\dfrac{3}{4}+2\dfrac{1}{4}\right)\times\dfrac{1}{3}-\dfrac{1}{8}\\ =8\times\dfrac{1}{3}-\dfrac{1}{8}=\dfrac{8}{3}-\dfrac{1}{8}=\dfrac{61}{24}\\ ----\\ \dfrac{3}{5}:\dfrac{5}{6}:\dfrac{6}{7}:\dfrac{7}{8}+\dfrac{2}{5}+\dfrac{23}{35}\\ =\dfrac{3}{5}\times\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{18}{25}\times\dfrac{4}{3}+\dfrac{2}{5}+\dfrac{23}{35}=\dfrac{353}{175}\)
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{101-99}{99x101}=\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)