tính tổng các phân số sau:
a)\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+❓+\(\dfrac{1}{2003.2004}\)
b)\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+❓\(\dfrac{1}{2003.2005}\)
Những câu hỏi liên quan
\(\dfrac{1}{1.3}\) +\(\dfrac{1}{3.5}\) +\(\dfrac{1}{5.7}\) +.....+\(\dfrac{1}{2003.2005}\)
Đặt biểu thức là A
\(2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2005-2003}{2003.2005}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}:2=\dfrac{1002}{2005}\)
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Gọi tổng trên là A. Ta có
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{2005}=\dfrac{2005}{2005}-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
⇒ A= \(\dfrac{2004}{2005}:2=\dfrac{2004}{2005}.\dfrac{1}{2}=\dfrac{1002}{2005}\)
Vậy tổng trên bằng \(\dfrac{1002}{2005}\)
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1.Tính
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
B=\(\dfrac{3}{5.6}+\dfrac{3}{6.7}+\dfrac{3}{7.8}+.....+\dfrac{3}{101.102}\)
C=\(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}\)
D=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
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1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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G= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{2003.2004}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=\dfrac{2003}{2004}\)
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tính tổng: A dfrac{2}{1.3}+dfrac{2}{3.5}+dfrac{2}{5.7}+...+dfrac{2}{99.101} B dfrac{5}{1.3}+dfrac{5}{3.5}+dfrac{5}{3.7}+...+dfrac{5}{99.101}
C dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{99.100} D dfrac{5}{1.4}+dfrac{5}{4.7}+...+dfrac{5}{100.103} E dfrac{1}{15}+dfrac{1}{35}+...+dfrac{1}{2499}
Đọc tiếp
tính tổng: A= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\) B= \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{3.7}+...+\dfrac{5}{99.101}\)
C= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) D= \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\) E= \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
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
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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
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C=1/2+1/3+1/3+1/4+....+1/99+1/100
=1/2+1/100
=50/100+1/100
=51/100
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Tính
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
Đặt \(U=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
\(\Rightarrow U=\dfrac{1.2}{1.2.3}+\dfrac{1.2}{3.2.5}+\dfrac{1.2}{5.2.7}+...+\dfrac{1.2}{2003.2.2005}\)
\(\Rightarrow U=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(\Rightarrow U=\dfrac{1}{2}.\left(1-\dfrac{1}{2005}\right)\Rightarrow U=\dfrac{1}{2}.\dfrac{2004}{2005}\Rightarrow U=\dfrac{1002}{2005}\)
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1.Tính :
a ) Sdfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+....+dfrac{1}{2018.2019}
b ) Sdfrac{1}{1.3}+dfrac{1}{3.5}+dfrac{1}{5.7}+....+dfrac{1}{2017.2019}
c) Sdfrac{1}{2.4}+dfrac{1}{4.6}+dfrac{1}{6.8}+....+dfrac{1}{2018.2020}
d) Sdfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+....+dfrac{1}{2017.2018.2019}
2. Tính tổng:
a) S1.2+2.3+3.4+...+2018.2019
b) S3.5+5.7+7.9+...+2017.2019
c) S2.4+4.6+6.8+...+2018.2020
d) S1.2.3+2.3.4+3.4.5+...+2017.2018.2019
3.Tính
a ) Sdfrac{1}{1.4}+dfrac{1}{4....
Đọc tiếp
1.Tính :
a ) \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)
b ) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2017.2019}\)
c) \(S=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+....+\dfrac{1}{2018.2020}\)
d) \(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2017.2018.2019}\)
2. Tính tổng:
a) \(S=1.2+2.3+3.4+...+2018.2019\)
b) \(S=3.5+5.7+7.9+...+2017.2019\)
c) \(S=2.4+4.6+6.8+...+2018.2020\)
d) \(S=1.2.3+2.3.4+3.4.5+...+2017.2018.2019\)
3.Tính
a ) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2017.2020}\)
b ) \(S=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{2017.2019.2021}\)
Ai có công thức không cho mình xin với ????
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
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tìm x:
a,\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right).\left(x-1\right)+\dfrac{1}{10}.x=x-\dfrac{9}{10}\)
b,\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right).\left(x-2\right)+x=\dfrac{149}{99}.x-\dfrac{98}{99}\)
tính tổng có quy luật :
H=\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+......+\(\dfrac{1}{47.49}\)+\(\dfrac{1}{49.51}\)
\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)
\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)
\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)
\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2H=1-\dfrac{1}{51}\)
\(2H=\dfrac{50}{51}\)
\(H=\dfrac{25}{51}\)
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23 tháng 4 2021 lúc 13:31
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)