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Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)
`5/(1.3)+5/(3.5)+....+5/(99.101)`
`=5/2(2/(1.3)+2/(3.5)+....+2/(99.101))`
`=5/2(1-1/3+1/3-1/5+...+1/99-1/101)`
`=5/2(1-1/101)`
`=5/2*100/101`
`=250/101`
Ta có: \(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+...+\dfrac{5}{99\cdot101}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{100}{101}\)
\(=\dfrac{250}{101}\)
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{99^2}{98.100}\)
\(=\dfrac{2.2.3.3.4.4.....99.99}{1.3.2.4.3.5.....98.100}\)
\(=\dfrac{2.3.4.....99}{1.2.3.4.....98}.\dfrac{2.3.4.....99}{3.4.5.....100}\)
\(=\dfrac{99}{98}\cdot\dfrac{2}{100}\)
\(=\dfrac{99}{4900}\)
A = \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ...........+ \(\dfrac{2}{99.101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +............+ \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{100}{101}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
\(\dfrac{x}{1\cdot3}+\dfrac{x}{3\cdot5}+...+\dfrac{x}{99\cdot101}=50\)
=>\(x\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\right)=50\)
=>\(\dfrac{x}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)=50\)
=>\(\dfrac{x}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=50\)
=>\(\dfrac{x}{2}\cdot\left(1-\dfrac{1}{101}\right)=50\)
=>\(x\cdot\dfrac{50}{101}=50\)
=>x=101
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