Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow A=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000}{100000}+\frac{1000}{100000}+\frac{100}{100000}+\frac{10}{100000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000+1000+100+10+1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{11111}{100000}\right)\)
\(\Rightarrow A=\frac{-11111}{100000}\)
\(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(=-\frac{10000}{10000}-\frac{1000}{10000}-\frac{100}{10000}-\frac{10}{10000}-\frac{1}{10000}\)
\(=\frac{-10000-1000-100-10-1}{10000}\)
\(=-\frac{11111}{10000}=-1,1111\)
\(=-\left(1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}\right)\)
\(=-\left(\frac{10000}{10000}+\frac{1000}{10000}+\frac{100}{10000}+\frac{1}{10000}\right)\)
\(=-\left(\frac{10000+1000+100+10+1}{10000}\right)\)
\(=-\left(\frac{11111}{10000}\right)\)
Vậy.....
\(-1-\left(\frac{1}{10}\right)-\frac{1}{100}-\frac{1}{1000}+\frac{1}{10000}\)
\(=1,1111\)
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
\(B=\frac{\left(1,09-0,29\right).1\frac{1}{4}}{\left(18,9-16\frac{13}{20}\right).\frac{8}{9}}=\frac{0,8.\frac{5}{4}}{\left[\left(18+0,9\right)-\left(16+\frac{13}{20}\right)\right].\frac{8}{9}}\)
\(B=\frac{\frac{4}{5}.\frac{5}{4}}{\left(18+\frac{9}{10}-16-\frac{13}{20}\right).\frac{8}{9}}\)
\(B=\frac{1}{\left[\left(18-16\right)+\left(\frac{9}{10}-\frac{13}{20}\right)\right].\frac{8}{9}}\)
\(B=\frac{1}{\left(2+\frac{1}{4}\right).\frac{8}{9}}\)
\(B=\frac{1}{\frac{9}{4}.\frac{8}{9}}=\frac{1}{2}\)
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 = \(-\frac{11111}{100000}\)