2007/2008×1/2009+2007/2009÷2009/2008+1/2008=
Co ai lam on giup toi voi
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
2006/2007 + 2007/2008 + 2008/2009 + 2009/2006
= 1 - 1/2007 + 1 - 1/2008 + 1 - 1/2009 + 1 + 3/2006
= (1 + 1 + 1 + 1) - (1/2007 + 1/2008 + 1/2009) + 3/2006
= 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006
Vì 1/2007 < 1/2006
1/2008 < 1/2006
1/2009 < 1/2006
=> 1/2007 + 1/2008 + 1/2009 < 3/2006
=> -(1/2007 + 1/2008 + 1/2009) + 3/2006 > 0
=> 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006 > 4 - 0 = 4
=> 2006/2007 + 2007/2008 + 2008/2009 + 2009/2006 > 4
Ta có:
2006/2007 + 2007/2008 + 2008/2009 + 2009/2006
= 1 - 1/2007 + 1 - 1/2008 + 1 - 1/2009 + 1 + 3/2006
= (1 + 1 + 1 + 1) - (1/2007 + 1/2008 + 1/2009) + 3/2006
= 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006
Vì 1/2007 < 1/2006
1/2008 < 1/2006
1/2009 < 1/2006
=> 1/2007 + 1/2008 + 1/2009 < 3/2006
=> -(1/2007 + 1/2008 + 1/2009) + 3/2006 > 0
=> 4 - (1/2007 + 1/2008 + 1/2009) + 3/2006 > 4 - 0 = 4
=> 2006/2007 + 2007/2008 + 2008/2009 + 2009/2006 > 4
So sánh: x = 2006/2007 - 2007/2008 + 2008/2009 - 2009/2010.
y = - 1/(2006 × 2007) - 1/(2007 × 2008).
Ta có:
\(x=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(=\dfrac{2006.2008-2007^2}{2007.2008}+\dfrac{2008.2010-2009^2}{2009.2010}\)
\(=\dfrac{2006.2007+2006-2007^2}{2007.2008}+\dfrac{2008.2009+2008-2009^2}{2009.2010}\)
\(=\dfrac{2007\left(2006-2007\right)+2006}{2007.2008}+\dfrac{2009\left(2008-2009\right)+2008}{2009.2010}\)
\(=\dfrac{-1}{2007.2008}+\dfrac{-1}{2008.2010}< \dfrac{-1}{2006.2007}+\dfrac{1}{2007.2008}\)
\(\Rightarrow x< y\)
Vậy x < y
bạn sai rồi đề bài là y = \(\dfrac{-1}{2006.2007}-\dfrac{1}{2008.2009}\)
chứ ko phải là \(\dfrac{-1}{2006.2007}+\dfrac{1}{2008.2009}\)
suy ra bài làm của bạn là sai hoặc bạn kia chép sai đề bài
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)