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Ta có:
2006x2005-1/2004x2006+2005 = 2006x(2004+1)-1 / 2004x2006+2005
= 2006x2004 + 2006 -1 / 2004x2006+2005
= 2006x2004+2005 / 2004x2006+2005
= 1
2006*2005-1/2004*2006+2005
=4022030 - 168/167 + 2005
=4022028,994 + 2005
=4024033 , 994
2006*2005-1/2004*2006+2005
=2006*2005-1/(2005-1)*2006+2005
=2006*2005-1/2005*2006-2006+2005
=2006*2005-1/2005*2006-1
=1
\(=\frac{2006x\left(2004+1\right)-1}{2004x2006+2005}=\frac{2006x2004+2005}{2004x2006+2005}=1\)
\(=\frac{2006\cdot2004+2006.1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006+2005}\)
\(\Rightarrow1\)
\(\frac{2006\times2005-1}{2004\times2006+2005}\)
\(=\frac{2004\times2006+2006-1}{2004\times2006+2005}\)
\(=\frac{2004\times2006+2005}{2004\times2006+2005}\)
\(=1\)
\(=\frac{2006\cdot2005-1}{2005\cdot2006-2006\cdot1+2005}\)
\(=\frac{2006\cdot2005-1}{2005\cdot2006-2006-2005}\)
\(=\frac{2006\cdot2005-1}{2005\cdot2006-1}\)
\(\Rightarrow1\)
\(\frac{2005.2006-1}{2006.2004+2005}\)
\(=\frac{2004.2006+2006-1}{2006.2004+2005}\)
\(=\frac{2004.2006+2005}{2006.2004+2005}\)
\(=1\)
\(\frac{2005.2006-1}{2006.2004+2005}=\frac{\left(2004+1\right).2006-1}{2006.2004+2005}\)
\(=\frac{2004.2006+\left(2006-1\right)}{2006.2004+2005}\)
\(=\frac{2004.2006+2005}{2006.2004+2005}=1\)
mình ko ghi lại đề nha
a) \(1-\frac{2004}{2005}=\frac{1}{2005}\);\(1-\frac{2005}{2006}=\frac{1}{2006}\)
ta thấy \(\frac{1}{2005}>\frac{1}{2006}\)
=> \(\frac{2004}{2005}< \frac{2005}{2006}\)
b)\(\frac{2007}{2006}-1=\frac{1}{2006};\frac{2006}{2005}-1=\frac{1}{2005}\)
ta thấy \(\frac{1}{2006}< \frac{1}{2005}\)
=>\(\frac{2007}{2006}< \frac{2006}{2005}\)
c) ta có \(\frac{1975}{2005}>\frac{1975}{2006}>\frac{1974}{2006}\)
=>\(\frac{1975}{2005}>\frac{1974}{2006}\)
2005/2006 + 2006/2007 + 2007/2008 + 2008/2005
= 4,000001491
k minh di xin day
minh ko bietcach giai tra loi giup minh di ban minh can gap
4010
6015
k cho mk nha
mk dang bi am 285 diem
2006+2005-1=2010
2004+2006+2005=6015