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a) chắc bạn cũng biết
b) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005.1-1}{2004+2005.2006}=\frac{2005.2006+2004}{2004+2005.2006}=1\)
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 \(\frac{1975}{2005}>\frac{1975}{2006}>\frac{1974}{2006}\)
=>\(\frac{1975}{2005}>\frac{1974}{2006}\)
a)=(3/8+10/16)+(7/12+10/24)
=1+1=2
c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)
=3+2+2=7
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
a) Ta có:
\(1-\frac{2005}{2006}=\frac{1}{2006}\)
\(1-\frac{2006}{2007}=\frac{1}{2007}\)
Vì \(\frac{1}{2006}>\frac{1}{2007}\)nên \(\frac{2005}{2006}>\frac{2006}{2007}\)
b) Ta có:
\(\frac{2008}{2007}-1=\frac{1}{2007}\)
\(\frac{2007}{2006}-1=\frac{1}{2006}\)
Vì \(\frac{1}{2006}>\frac{1}{2007}\)nên \(\frac{2008}{2007}< \frac{2007}{2006}\)
a, \(\frac{2005}{2006}v\text{à}\frac{2006}{2007}\)= \(\frac{2005\cdot2007}{2006\cdot2007}\)và \(\frac{2006\cdot2006}{2007\cdot2006}\)
= \(\frac{4024035}{4026042}\)< \(\frac{4024036}{4026042}\)
b, \(\frac{2008}{2007}v\text{à}\frac{2007}{2006}\)= \(\frac{2008\cdot2006}{2007\cdot2006}v\text{à}\frac{2007\cdot2007}{2006\cdot2007}\)
=\(\frac{4028048}{4026042}\)< \(\frac{4028049}{4026042}\)
Ta có: \(A=\dfrac{2005}{2006}+\dfrac{2006}{2007}+\dfrac{2007}{2005}=\dfrac{2006-1}{2006}+\dfrac{2007-1}{2007}+\dfrac{2005}{2005}+\dfrac{1}{2005}+\dfrac{1}{2005}\)\(=1+1+1+\left(\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
\(=3+\left(\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
Ta thấy: \(\dfrac{1}{2005}>\dfrac{1}{2006};\dfrac{1}{2005}>\dfrac{1}{2007}\) \(\Rightarrow\dfrac{1}{2005}-\dfrac{1}{2006}+\dfrac{1}{2005}-\dfrac{1}{2007}>0\)
\(\Rightarrow A>3\)