Bài 1: Tính tổng các p/s sau :
a, 1/18 + 1/54 + 1/108 +....+ 1/990
Giải đúng và chuẩn nhoa các bn! Mk like cho.
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TB
3
5 tháng 5 2015
\(\frac{1}{18}\)+\(\frac{1}{54}\)+\(\frac{1}{108}\)+...+\(\frac{1}{990}\)
=\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+\(\frac{1}{9.12}\)+...+\(\frac{1}{30.33}\)
=\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\)\(\frac{1}{30}-\frac{1}{33}\)
=\(\frac{1}{3}-\frac{1}{33}\)
=\(\frac{10}{33}\)
GS
17 tháng 4 2016
=1/3x6+1/6x9+1/9x12+...+1/30x33
=1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33
=1/3-1/33
=10/33
17 tháng 4 2016
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)=\frac{1}{3}.\left(\frac{6-3}{3.6}+\frac{9-6}{6.9}+\frac{12-9}{9.12}+...+\frac{33-30}{30.33}\right)=\frac{1}{3}.\left(\frac{6}{3.6}-\frac{3}{3.6}+\frac{9}{6.9}-\frac{6}{6.9}+\frac{12}{9.12}-\frac{9}{9.12}+...+\frac{33}{30.33}-\frac{30}{30.33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}.\frac{10}{33}=\frac{10}{99}\)
30 tháng 3 2017
đặt A với biểu thức trên
A=\(\frac{1}{3x6}\)+\(\frac{1}{6x9}\)+......+\(\frac{1}{30x33}\)
Nhân cả 2 vế với 3 ta có
A x 3 = \(\frac{3}{3x6}\)+....+\(\frac{3}{30x33}\)
A x 3 = \(\frac{1}{3}\)-\(\frac{1}{6}\)+....+\(\frac{1}{30}\)-\(\frac{1}{33}\)
A x 3 = \(\frac{1}{3}\)-\(\frac{1}{33}\)
A x 3 = \(\frac{10}{33}\)
A = \(\frac{10}{33}\):3
A= \(\frac{10}{99}\)
23 tháng 7 2019
B)A*2=(1/2+1/4+....+1/256)*2
=1+1/2+1/4+....+1/128)
A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)
=1-1/256
=255/256
XO
23 tháng 7 2019
a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)
\(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)
Lấy \(A-\frac{1}{3}\times A\)theo vế ta có :
\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)
\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)
\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)
\(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)
\(\Rightarrow A=\frac{605}{162}\)
Vậy \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)
b) Đặt B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)
Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có :
\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)
\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)
\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)
\(\Rightarrow B=\frac{255}{256}\)
Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)
2 tháng 7 2022
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)
c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
KT
26 tháng 4 2018
\(S=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}=\frac{10}{99}\)
NH
3
11 tháng 5 2016
S=1/3.7+1/7.11+...+1/19.23 (1)
Nhân cả 2 vế của đẳng thức (1) với 4 ta được:
4S=4/3.7+4/7.11+...+4/19.23
4S=1/3.7+1/7.11+...+1/19.23
4S=1/3-1/7+1/7-1/11+..+1/19-1/23
4S=1/3-1/23
4S=20/69
S =20/69:4
S =5/69
Mọi người ủng hộ mik nha
11 tháng 5 2016
\(S=\frac{1.4}{3.7.4}+\frac{1.4}{7.11.4}+......+\frac{1.4}{19.23.4}\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+......+\frac{4}{19.23}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{20}\right)\)
\(=\frac{1}{4}.\frac{17}{60}=\frac{17}{240}\)
Đặt A = \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(3A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(3A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\)
\(3A=\frac{1}{3}-\frac{1}{33}\)
\(3A=\frac{10}{33}\)
\(A=\frac{10}{33}:3\)
\(A=\frac{10}{99}\)