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\(=\dfrac{2}{5}\cdot\dfrac{7}{3}\cdot\dfrac{3}{7}\cdot\dfrac{5}{2}+2020=1+2020=2021\)
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{101-99}{99x101}=\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(2\times3\times5\times7\)
\(=\left(2\times5\right)\times\left(3\times7\right)\)
\(=10\times21\)
\(=210\)
\(\frac{9}{10}\)+\(\frac{7}{9}\)+\(\frac{5}{8}\)+\(\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)\)\(+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=1+1+1+1+1\)
\(=5\)
9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
= ( 9/10 + 1/10 ) + ( 7/9 + 2/9 ) + ( 5/8 + 3/8 ) + ( 3/7 + 4/7 ) + ( 3/5 + 2/5 )
= 1 + 1 + 1 + 1 + 1
= 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)
= \(1+1+1+1+1\)
= \(1\times5\)
= \(5\)
Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :
A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)
A = 1 + 1 + 1 + 1 + 1
A = 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)\(=\frac{9}{10}+\frac{1}{10}+\frac{7}{9}+\frac{2}{9}+\frac{5}{8}+\frac{3}{8}+\frac{3}{7}+\frac{4}{7}+\frac{3}{5}+\frac{2}{5}\)
\(=\left[\frac{9}{10}+\frac{1}{10}\right]+\left[\frac{7}{9}+\frac{2}{9}\right]+\left[\frac{5}{8}+\frac{3}{8}\right]+\left[\frac{3}{7}+\frac{4}{7}\right]+\left[\frac{3}{5}+\frac{2}{5}\right]\)
\(=1+1+1+1+1\)
\(=5\)
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= 1 + 1 + 1 + 1 + 1
= 5
\(a)\frac{4}{9}+\frac{3}{8}+\frac{5}{8}=\frac{4}{9}+(\frac{3}{8}+\frac{5}{8})=\frac{4}{9}+1=\frac{4+9}{9}=\frac{13}{9}\)
\(b)\frac{3}{7}+\frac{2}{3}+\frac{4}{7}=(\frac{3}{7}+\frac{4}{7})+\frac{2}{3}=1+\frac{2}{3}=\frac{3+2}{3}=\frac{5}{3}\)
`2/5 xx 4/7 + 3/5 xx 4/7`
`= 4/7 xx (2/5 + 3/5)`
`= 4/7 xx 1`
`= 4/7`
\(3-\dfrac{3}{5}\times\dfrac{7}{2}:\dfrac{7}{5}\)
\(=3-\dfrac{3}{5}\times\dfrac{7}{2}\times\dfrac{5}{7}\)
\(=\dfrac{6}{2}-\dfrac{3}{2}=\dfrac{3}{2}\)