Tính nhanh:
a) S= \(\frac{3}{1.3}\)+\(\frac{3}{3.5}\)+\(\frac{3}{5.7}\)+...+\(\frac{3}{2013.2015}\)
b)B=\(\frac{1}{120}\)-\(\frac{2}{30.33}\)-\(\frac{2}{33.36}\)-...-\(\frac{2}{117.120}\)
K
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MC
4 tháng 5 2016
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
10 tháng 7 2019
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
10 tháng 7 2019
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
3 tháng 3 2016
= 1/2. ( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +........+ 1/2013 - 1/2015)
= 1/2 . ( 1- 1/2015)
= 1007/2015
NT
0
12 tháng 7 2015
\(A=\frac{12}{3.5}+\frac{12}{5.7}+...+\frac{12}{2013.2015}\)
\(2A=\frac{24}{3.5}+\frac{24}{5.7}+...+\frac{24}{2013.2015}\)
\(2A=\frac{24}{3}-\frac{24}{5}+\frac{24}{5}-\frac{24}{7}+...+\frac{24}{2013}-\frac{24}{2015}\)
\(2A=8-\frac{24}{2015}\)
\(2A=\frac{8}{1}-\frac{24}{2015}\)
\(2A=\frac{16120}{2015}-\frac{24}{2015}\)
\(2A=\frac{16096}{2015}\)
\(=>A=\frac{16096}{2015}:2\)
\(=>A=\frac{16096}{4030}\)
31 tháng 3 2019
1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(C=1-\frac{1}{101}\)
\(C=\frac{100}{101}\)
d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)
\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)
\(D=\frac{1}{2}-\frac{1}{95}\)
\(D=\frac{95-2}{190}=\frac{93}{190}\)
Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)
12 tháng 7 2015
a/
S = 1-2+3-4+5-6+...+2001-2002+2003
= [-1] +[-1] +...+[-1] +2003
------------------------
1001 số -1
= -1001 +2003 = 1002
b/
A = \(6.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)=6.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)=6.\left(\frac{1}{3}-\frac{1}{2015}\right)=\frac{6.2012}{6045}=\frac{4024}{2015}\)
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
TG
14 tháng 5 2019
Mk giải ko chép lại đề nhá!
Bài 3:
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}\)\(-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{50}{50}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Vậy: M < 1
TG
14 tháng 5 2019
Bài 2:
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(=\frac{1}{1}-\frac{1}{2015}\)
\(=\frac{2015}{2015}-\frac{1}{2015}\)
\(=\frac{2014}{2015}\)
Ta có S = \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{2013.2015}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2015}\right)=\frac{3}{2}.\frac{2014}{2015}=\frac{3021}{2015}\)
b) B = \(\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{3}{30.33}+\frac{3}{33.36}+..+\frac{3}{117.120}\right)\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{1}{30}-\frac{1}{120}\right)=\frac{1}{120}-\frac{2}{3}.\frac{1}{40}=\frac{1}{120}-\frac{2}{120}=-\frac{1}{120}\)
Trả lời:
a, \(S=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{2013.2015}\)
\(\Rightarrow S=\frac{3.2}{1.3.2}+\frac{3.2}{3.5.2}+\frac{3.2}{5.7.2}+...+\frac{3.2}{2013.2015.2}\)
\(\Rightarrow S=\frac{3}{2}\cdot\frac{2}{1.3}+\frac{3}{2}\cdot\frac{2}{3.5}+\frac{3}{2}\cdot\frac{2}{5.7}+...+\frac{3}{2}\cdot\frac{2}{2013.2015}\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\frac{2014}{2015}=\frac{3021}{2015}\)
b, \(B=\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)
\(\Rightarrow B=\frac{1}{120}-\left(\frac{2}{30.33}+\frac{2}{33.36}+...+\frac{2}{117.120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\left(\frac{2.3}{30.33.3}+\frac{2.3}{33.36.3}+...+\frac{2.3}{117.120.3}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{3}{30.33}+\frac{3}{33.36}+...+\frac{3}{117.120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\frac{1}{40}=\frac{1}{120}-\frac{1}{60}=-\frac{1}{120}\)