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\(C=-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(10C=-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(10C-C=\left(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\right)-\left(\frac{-1}{10}-\frac{1}{100}-...-\frac{1}{100000}\right)\)
\(9C=-1+\frac{1}{100000}\)
\(C=\frac{\frac{1}{100000}-1}{9}\)
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
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 = \(-\frac{11111}{100000}\)
\(A=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow A=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000}{100000}+\frac{1000}{100000}+\frac{100}{100000}+\frac{10}{100000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000+1000+100+10+1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{11111}{100000}\right)\)
\(\Rightarrow A=\frac{-11111}{100000}\)
a) Ta có: \(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=\frac{-1}{2}.\frac{-2}{3}...\frac{-9}{10}=\frac{-\left(1.2.3...9\right)}{2.3.4...10}=-\frac{1}{10}\)
b) Ta có : \(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}....\frac{-99}{100}=-\frac{3.8....99}{\left(2.3...10\right)\left(2.3...10\right)}\)
\(=-\frac{1.3.2.4...9.11}{\left(2.3....10\right)\left(2.3...10\right)}=\frac{\left(1.2.3...10\right).\left(3.4..10.11\right)}{\left(2.3...10\right).\left(2.3.4...10\right)}=\frac{11}{2}=5,5\)
c) Ta có : \(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)=\frac{1}{2}.\frac{2}{3}...\frac{n}{n+1}=\frac{1.2...n}{2.3...\left(n+1\right)}=\frac{1}{n+1}\)
=-(0,1+0,01+0,001+0,0001+0,00001)
=-0,11111
Đây là cách đơn giản nhất
= \(-\frac{1111}{10000}\)
=-1111/10000