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K
Khách
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NT
4
G
30 tháng 8 2020
\(2^3+3.\left(\frac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2\div\frac{1}{2}.8\right]\)
\(=8+3.1-\frac{1}{2^2}.4+\left[4\div\frac{1}{2}.8\right]\)
\(=8+3-1+64\)
\(=74\)
NP
0
CT
11 tháng 5 2016
Đề của bạn sai rồi: Phải là B = \(\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\) chứ ?!
12 tháng 2 2018
Ta có :
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(B=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)}=\frac{1}{2009}\)
Vậy \(\frac{A}{B}=\frac{1}{2009}\)
12 tháng 2 2018
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{1007}+\frac{1}{2008}\)
\(B=\frac{2008}{1}+1+\frac{2007}{2}+1+\frac{2006}{3}+1+....+\frac{2}{2007}+1+\frac{1}{2008}+1-2008\)
\(B=\frac{2009}{1}+\frac{2009}{2}+\frac{2009}{3}+.....+\frac{2009}{2007}+\frac{2009}{2008}-\frac{2009.2008}{2009}\)
\(B=2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{2008}-\frac{2008}{2009}\right)\)
Từ đó suy ra \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}{2009\cdot\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2007}+\frac{1}{1008}+\frac{2008}{2009}\right)}\)
\(=\frac{\frac{1}{2009}}{2009\cdot\left(1+\frac{2008}{2009}\right)}\)
Bí òi
BB
31 tháng 8 2020
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...............+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+........+\left(1+\frac{1}{2008}\right)+1\)
\(B=\frac{2009}{2}+\frac{2009}{3}+..............+\frac{2009}{2008}+\frac{2009}{2009}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{2009}\right)\)
Khi đó: \(\text{}\text{}\text{}\frac{A}{B}=\frac{1}{2009}\)
Chuc bạn học tốt!!
30 tháng 8 2020
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)\)
Ta có: \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}}{2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)}\)
hay \(\frac{A}{B}=\frac{1}{2009}\)