tính tổng \(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
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22 tháng 2 2022
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
22 tháng 3 2023
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
PT
13 tháng 8 2017
a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)
\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)
\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)
b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}\)
c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)
\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)
\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)
\(A=\dfrac{7}{2}.\dfrac{100}{101}\)
\(A=\dfrac{256}{101}\)
TM
19 tháng 1 2022
a. \(\dfrac{-3}{12}+\dfrac{1}{-4}=\dfrac{-3}{12}+\dfrac{-3}{12}=\dfrac{-3-3}{12}=\dfrac{-6}{12}=\dfrac{-1}{2}\)
b. \(\dfrac{5}{12}+\dfrac{-3}{28}=\dfrac{35}{84}+\dfrac{-9}{84}=\dfrac{35+\left(-9\right)}{84}=\dfrac{26}{84}=\dfrac{13}{42}\)
c. \(\dfrac{-7}{15}+\dfrac{3}{35}=\dfrac{-49}{105}+\dfrac{9}{105}=\dfrac{-49+9}{105}=\dfrac{-40}{105}=\dfrac{-8}{21}\)
d. \(\dfrac{-5}{7}+\dfrac{-3}{4}=\dfrac{-20}{28}+\dfrac{-21}{28}=\dfrac{-20+\left(-21\right)}{28}=\dfrac{-41}{28}\)
18 tháng 6 2018
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
PJ
2
Q
5 tháng 4 2017
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(\Rightarrow-A=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)
\(-A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(-2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(-2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(-2A=1-\dfrac{1}{101}\)
\(-2A=\dfrac{100}{101}\)
\(-A=\dfrac{100}{101}:2\)
\(-A=\dfrac{50}{101}\)
\(\Rightarrow A=\dfrac{-50}{101}\)
Chúc bạn học tốt!
BN
5 tháng 4 2017
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+...+\dfrac{-1}{9999}\)
\(A=-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\)
Đặt \(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{9999}\)
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2B=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(B=\dfrac{100}{101}:2=\dfrac{50}{101}\)
\(\Rightarrow A=-B=-\dfrac{50}{101}\)
B
1
NG
22 tháng 8 2017
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=50-\left(1-\dfrac{1}{101}\right)=50-\dfrac{100}{101}\)
\(=\dfrac{4950}{101}\)
13 tháng 3 2023
Sửa đề bài xíu nhé !
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}.\dfrac{98}{99}=\dfrac{49}{99}\)
\(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{9603}+\dfrac{2}{9999}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{96}{505}=\dfrac{150}{101}\)