\(\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\) \(\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\) \(\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
K
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Những câu hỏi liên quan
KN
27 tháng 4 2019
1.
a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
KN
27 tháng 4 2019
b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}\)
\(=\frac{200}{101}\)
NH
5
23 tháng 3 2020
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)
=> \(2P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{29.31}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{31-29}{29.31}\)
\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{29}-\frac{1}{31}\right)\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
=> \(P=\frac{30}{31}:2=\frac{15}{31}\)
A
23 tháng 3 2020
Nếu đề là tính thì bạn làm như sau nhé :
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)
\(\Rightarrow2P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{29.31}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
\(\Rightarrow P=\frac{30}{31}\div2=\frac{15}{31}\)
QD
2 tháng 6 2016
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh
2 tháng 6 2016
A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
=1-\(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=1-\(\frac{1}{101}\)
=\(\frac{100}{101}\)
B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)
=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))
=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)
Chúc bạn học tốt
MC
4 tháng 5 2016
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
19 tháng 2 2017
lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé
a) \(\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}\)
\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}\times\frac{100}{101}=\frac{250}{101}\)
b) \(\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}\)
\(=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\times\frac{3}{25}=\frac{9}{25}\)
19 tháng 2 2017
Ta có \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(\Rightarrow3.\left(1-\frac{1}{200}\right)\)
\(\Rightarrow3.\frac{199}{200}=\frac{597}{200}\)
7 tháng 5 2016
\(a,=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
7 tháng 5 2016
a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)
M
10 tháng 7 2015
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
ML
2 tháng 4 2017
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}\)\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
TA CÓ:
Đặt C = \(\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\)
\(\Rightarrow\frac{2}{5}C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)
\(\Rightarrow\frac{2}{5}C=\frac{1}{3}-\frac{1}{31}=\frac{28}{93}\)
\(\Rightarrow C=\frac{28}{93}:\frac{2}{5}=\frac{70}{93}\)
Tương tự, đặt \(D=\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\)
Mà \(D=C\)
\(\Rightarrow D=\frac{70}{93}\)
Đặt \(B=\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{15.16}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{15.16}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
\(\Rightarrow B=\frac{7}{16}:\frac{1}{3}=\frac{21}{16}\)
Nên \(\frac{5}{3.5}+...+\frac{5}{29.31}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}+\frac{3}{2.3}+..+\frac{3}{15.16}\)
\(\Rightarrow C+D+B=C.2+B=\frac{140}{93}+\frac{21}{16}=2,81\)