So sánh 2 biểu thức:
A = 2006/2007 + 2007/2008 + 2008/2009
B = 2006 + 2007 + 2008/2007 + 2008 + 2009
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A>b
Cách làm: Bạn tách |B ra rồi so sánh với từng ps ở A, sau đó Kết luận
2006/2007<1
2007/2008<1
2008<2009<1
2009/2006>1
A=2006/2007+2007/2008+2008/2009+2009/2006\(\approx\)3+1=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
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 : \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
\(=\frac{2007-1}{2007}+\frac{2008-1}{2008}+\frac{2009-1}{2009}+\frac{2006+3}{2006}\)
\(=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
\(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
\(< 4-\left(\frac{1}{2009}+\frac{1}{2009}+\frac{1}{2009}-\frac{3}{2009}\right)\)
\(=4\)
=> A < 4
Vậy A < 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
Ta có A= (1 -1/2007) +(1-1/2008)+(1-1/2009)+(1+3/2006)= 4-(1/2007+1/2008+1/2009-3/2006) <4
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}\)
\(=3-\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>1\).
\(B=\frac{2006+2007+2008}{2007+2008+2009}< \frac{2007+2008+2009}{2007+2008+2009}=1\).
Suy ra \(A>B\).