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Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)
A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)
A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
A=\(\dfrac{98}{99}-\dfrac{49}{100}\)
A=\(\dfrac{4949}{9900}\)
Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)
Vậy A<\(\dfrac{3}{4}\)
Đặt \(U=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
\(\Rightarrow U=\dfrac{1.2}{1.2.3}+\dfrac{1.2}{3.2.5}+\dfrac{1.2}{5.2.7}+...+\dfrac{1.2}{2003.2.2005}\)
\(\Rightarrow U=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{2005}\right)\Rightarrow U=\dfrac{1}{2}.\dfrac{2004}{2005}\Rightarrow U=\dfrac{1002}{2005}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\)
\(2M=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\)
\(=1-\dfrac{1}{999}=\dfrac{998}{999}\)
\(\Rightarrow M=\dfrac{998}{999}.\dfrac{1}{2}=\dfrac{499}{999}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{995.997}+\dfrac{1}{997.999}\\ =\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{995.997}+\dfrac{2}{997.999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{997}-\dfrac{1}{999}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{999}\right)=\dfrac{1}{2}.\dfrac{998}{999}=\dfrac{499}{999}\)
\(B=\dfrac{1}{1.3}\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2003.2005}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\\ =\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}.\dfrac{2004}{2005}\\ =\dfrac{1002}{2005}\)
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
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+....+\(\dfrac{1}{x}\)- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= 1- \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= \(\dfrac{1006}{2011}\)
=> x +1= 2011
=> x= 2011-1
=> x=2010
Bài này mk lm đại nha bn 
! Cs j sai mong bn bỏ qua 
.
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{33}{99}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{95}-\dfrac{2}{97}+\dfrac{2}{97}-\dfrac{2}{99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{99}=\dfrac{64}{99}\)
chúc bạn học tốt
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{99}\)
\(B=\dfrac{99}{99}-\dfrac{1}{99}\)
\(B=\dfrac{98}{99}\)
#YVA
B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)
B=\(\dfrac{98}{99}:2\)
B=\(\dfrac{49}{99}\)