Cho: a = \(\frac{2008}{2009}\) ; b = \(\frac{2009}{2008}\) ; c = \(\frac{1}{2009}\) ; d = \(\frac{2007}{2008}\)
Tìm : a - b+c+d
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a-b+c+d=\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}=\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)=1-\frac{2}{2008}=\frac{2006}{2008}=\frac{1003}{1004}\)
\(a-b+c+d=\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)
\(=\left(\frac{2008}{2009}+\frac{1}{2009}\right)+\left(\frac{2007}{2008}-\frac{2009}{2008}\right)=\frac{2009}{2009}+\frac{-2}{2008}\)
\(=1+\frac{-1}{1004}=\frac{1004}{1004}+\frac{-1}{1004}=\frac{1003}{1004}\)
\(A=28\left(\frac{7}{2008}+\frac{2001}{2008}\right)+\frac{2008}{2009}+\frac{1}{2009}=28+1+1=30\)
\(A=28\frac{7}{2008}+\frac{2008}{2009}+\frac{2001}{2008}+\frac{1}{2009}=\left(28+\frac{7}{2008}+\frac{2001}{2008}\right)+\left(\frac{2008}{2009}+\frac{1}{2009}\right)=29+1=30\)
Có :\(a-b=\frac{2008}{2009}-\frac{2009}{2008}\)\(=\frac{2008^2-2009^2}{2008\cdot2009}=\frac{\left(2008-2009\right)\left(2008+2009\right)}{2008\cdot2009}\)
\(=\frac{-2008-2009}{2008\cdot2009}=-\frac{1}{2009}-\frac{1}{2008}\)
=>a-b+c+d=\(-\frac{1}{2009}-\frac{1}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)
\(=-\frac{1}{2008}+\frac{2007}{2008}=\frac{2006}{2008}=\frac{1003}{1004}\)