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B = \(-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-...-\frac{1}{1000000}\)
B = \(-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^6}\right)\)
Đặt A = \(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^6}\)
10A = \(1+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^5}\)
9A = 10A - A = \(1-\frac{1}{10^6}\)
=> A = \(\frac{1-\frac{1}{10^6}}{9}\)
=> B = \(-\left(\frac{1-\frac{1}{10^6}}{9}\right)\)
C=(0,1+0,01+0,001+...+0,000001)=-0,111111
mình ko chép đề bài
A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)...\(\dfrac{9999}{10000}\)
A = \(\dfrac{1.3.2.4..3.5......99.101}{2.2.3.3.4.4....100.100}\)
A = \(\dfrac{1.2.3..4.5.....99}{2.3.4.5.....99.100}\).\(\dfrac{3.4.5....100.101}{2.3.4.5...100}\)
A = \(\dfrac{1}{100}\).\(\dfrac{101}{2}\)
A = \(\dfrac{101}{200}\)
2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))
Xem lại đề bài.
Ta có : \(B=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow B=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
Đặt \(A=\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\)
\(\Rightarrow10A=1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}\)
\(\Rightarrow10A-A=1-\frac{1}{100000}\)
\(\Rightarrow9A=\frac{99999}{100000}\)
\(\Rightarrow A=\frac{99999}{100000}.\frac{1}{9}=\frac{11111}{100000}\)
=> B = \(-\frac{11111}{100000}\)
\(A=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow A=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000}{100000}+\frac{1000}{100000}+\frac{100}{100000}+\frac{10}{100000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000+1000+100+10+1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{11111}{100000}\right)\)
\(\Rightarrow A=\frac{-11111}{100000}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
c/
C = 1/100-1/100-1/99-1/99-1/98-1/98-1/97-..........-1/3-1/2-1/2-1/1
C = 1/100-1/100-1/1
C = 0-1/1
C = -1
\(C=-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(10C=-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(10C-C=\left(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\right)-\left(\frac{-1}{10}-\frac{1}{100}-...-\frac{1}{100000}\right)\)
\(9C=-1+\frac{1}{100000}\)
\(C=\frac{\frac{1}{100000}-1}{9}\)
cảm ơn bạn nhiều lắm Bonking mai mình cần nếu bạn giúp được mình có câu hổi mới đăng bạn giúp mình được ko