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3 tháng 3 2022
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{-4751}{9603}\)
HQ
Hà Quang Minh
Giáo viên
4 tháng 8 2023
a, \(A=-\dfrac{1}{20}-\left(\dfrac{1}{20\cdot19}+\dfrac{1}{19\cdot18}+...+\dfrac{1}{2\cdot1}\right)\\ \Rightarrow A=-\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\\ \Rightarrow A=-\dfrac{1}{20}-1+\dfrac{1}{20}=-1\)
b, \(B=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{2\cdot99}=-\dfrac{16}{33}\)
9 tháng 10 2021
\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}+\dfrac{1}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{1}{2}\cdot\dfrac{98}{99}\\ =\dfrac{1}{99}-\dfrac{49}{99}=-\dfrac{48}{99}=-\dfrac{16}{33}\)
22 tháng 1 2017
\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{97.99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)
\(=\frac{1}{97.99}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)
\(=\frac{1}{97.99}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{9603}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)
\(=\frac{1}{9603}-\frac{1}{2}.\frac{96}{97}\)
\(=\frac{-4751}{9603}\)
22 tháng 1 2017
A=-(1/99.97+1/97.95+...+1/5.3+1/3.1)
2B=2/99.97+2/97.95+...+2/5.3+2/3.1
2B=1-1/3+1/3-1/5+...+1/97-1/99
2B=1-1/99
2B=98/99
B=49/99
Suy ra A=-1/49/49
Mình giải đúng rồi bạn cứ yên tâm
NH
2
26 tháng 8 2018
\(\frac{1}{99}-\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\right)\)
\(=\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{99}-\frac{1}{2}\cdot\frac{98}{99}=\frac{1}{99}-\frac{49}{99}=\frac{-48}{99}=\frac{-16}{33}\)
26 tháng 8 2018
cảm on bạn két quả của mình cũng thế nhưng cách giải hơi khác bạn chút xíu
HK
0
18 tháng 10 2017
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-...-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)\(=\frac{1}{99\cdot97}-\left(\frac{1}{97\cdot95}+\frac{1}{95\cdot93}+...+\frac{1}{3\cdot1}\right)\)
\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-\frac{1}{95}+\frac{1}{95}-\frac{1}{93}+...+\frac{1}{3}-1\right)\)\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-1\right)=\frac{1}{9603}-2\cdot\left(-\frac{96}{97}\right)\)\(\frac{1}{9603}-\frac{-192}{97}\)phần còn lại tự làm
Đặt: \(A=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-....-\dfrac{1}{3\cdot1}\)
\(2A=\dfrac{2}{99}-\dfrac{2}{99\cdot97}-\dfrac{2}{97\cdot95}-...-\dfrac{2}{3\cdot1}\)
\(2A=\dfrac{2}{99}-\left(\dfrac{2}{99\cdot97}+\dfrac{2}{97\cdot95}+...+\dfrac{2}{3\cdot1}\right)\)
\(2A=\dfrac{2}{99}-\left(\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{95}-\dfrac{1}{97}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
\(2A=\dfrac{2}{99}-\left(-\dfrac{1}{99}+1\right)\)
\(2A=\dfrac{2}{99}-\dfrac{98}{99}\)
\(2A=-\dfrac{439}{99}\)
\(A=-\dfrac{439}{99}:2\)
\(A=-\dfrac{439}{198}\)
1/99 - 1/99.97 - 1/97.95 - ... - 1/3.1
= 1/99 - 1/2.(1/97 - 1/99 + 1/95 - 1/97 + ... + 1 - 1/3)
= 1/99 - 1/2.(1 - 1/99)
= 1/99 - 1/2 . 98/99
= 1/99 - 49/99
= -48/99