Tính :
\(\dfrac{1}{100\cdot99}-\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)
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B2
13 tháng 2 2019
\(C=\dfrac{1}{100}-\dfrac{1}{100\cdot99}-\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)
\(C=\dfrac{1}{100}-\left(\dfrac{1}{2\cdot1}+\dfrac{1}{3\cdot2}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)
\(C=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(C=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(C=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-98}{100}=-\dfrac{49}{50}\)
13 tháng 2 2019
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=-\frac{98}{100}=-\frac{49}{50}\)
NT
0
5 tháng 9 2017
bài này dễ lắm,mình giải đây:
C = \(\frac{1}{100}\)- \(\frac{1}{100.99}\)-\(\frac{1}{99.98}\)\(\frac{1}{98.97}\)- ... - \(\frac{1}{3.2}\)- \(\frac{1}{2.1}\)
C = \(\frac{-1}{1.2}\)+ \(\frac{-1}{2.3}\) + ... +\(\frac{-1}{98.99}\)+ \(\frac{1}{99.100}\)+ \(\frac{1}{100}\)
C = \(\frac{-1}{1}\)- \(\frac{-1}{2}\)
Mình bận rồi , phần sau tự làm nha.
17 tháng 5 2017
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
NT
18 tháng 5 2017
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)
17 tháng 6 2021
Giải:
\(\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+...+\dfrac{5}{98.99}+\dfrac{5}{99.100}\)
\(=5.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5.\left(1-\dfrac{1}{100}\right)\)
\(=5.\dfrac{99}{100}\)
\(=\dfrac{99}{20}\)
Chúc em học tốt!
XT
17 tháng 6 2021
Giải:
51.2+52.3+53.4+...+598.99+599.10051.2+52.3+53.4+...+598.99+599.100
=5.(11.2+12.3+13.4+...+198.99+199.100)=5.(11.2+12.3+13.4+...+198.99+199.100)
=5.(1−12+12−13+13−14+...+198−199+199−1100)=5.(1−12+12−13+13−14+...+198−199+199−1100)
=5.(1−1100)=5.(1−1100)
=5.99100=5.99100
=9920=9920
\(A=\dfrac{1}{99.100}-\dfrac{1}{98.99}-....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ =-\left(-\dfrac{1}{99.100}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}\right)\\ =-\left(-\dfrac{1}{99.100}+\dfrac{98}{99}\right)\\ =\dfrac{1}{99.100}-\dfrac{98}{99}\\ =\dfrac{1}{99}\left(\dfrac{1}{100}-98\right)=\dfrac{-9799}{9900}\)