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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
Vì khi tang tu so va mau so len 1 don vi voi so thu nhat ta duoc:
2008+1/2009+1=2008/2010 + 1/2010
tuong tu cac phan so khac neu ban muon so sanh cac phan so voi nhau.
a) =\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+....+\(\frac{1}{10\cdot11}\)=\(\frac{1}{2}\)-\(\frac{1}{11}\)=\(\frac{9}{22}\)
b)\(\frac{2008}{2009}\)=1 - \(\frac{1}{2009}\); \(\frac{2007}{2008}\)=1 - \(\frac{1}{2008}\)Do \(\frac{1}{2009}\)<\(\frac{1}{2008}\)nen 1 - \(\frac{1}{2009}\)>1 - \(\frac{1}{2008}\)
=> 2008/2009>2007/2008
8 x 427 x 3 + 6 x 573 x 4
= 8 x 3 x 427 + 6 x 4 x 573
= 24 x 427 + 24 x 573
= 24 x (427 + 573)
= 24 x 1000
= 24000
2008 x 867 + 2009 x 133
= 2008 x 867 + 2008 x 133 + 133
= 2008 x (867 + 133) + 133
= 2008 x 1000 + 133
= 2008133
a. kết quả = 401/402
b. Ta có: 1-2004/2009=5/2009 , 1--2005/2010=5/2010 . Vì 5/2009 > 5/2010 nên 2004/2009 < 2005/2010.
Đấy phần b. mk ko quy đồng nha!
Nhớ Tích cho mk đấy
\(\frac{2008\times2009+1000}{2009\times2009-1009}=\frac{2008\times2009+1000}{2008\times2009+2009-1009}=\frac{2008\times2009+1000}{2008\times2009+1000}=1\)