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Cho A=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\); B=\(\dfrac{2003+2004}{2004+2005}\)
Ta có: B=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Vì: \(\dfrac{2003}{2004+2005}< \dfrac{2003}{2004}\)
\(\dfrac{2004}{2004+2005}< \dfrac{2004}{2005}\)
=>\(\dfrac{2003}{2004+2005}+\dfrac{2004}{2004+2004}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>\(\dfrac{2003+2004}{2004+2005}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>B<A
Vậy B<A
\(2004A=\frac{2004^{2004}+2004}{2004^{2004}+1}=1+\frac{2003}{2004^{2004}+1}\)
\(2004B=\frac{2004^{2005}+2004}{2004^{2005}+1}=1+\frac{2003}{2004^{2005}+1}\)
\(\frac{2003}{2004^{2004}+1}>\frac{2003}{2004^{2005}+1}\)
\(\Rightarrow2004A>2004B\)
\(\Rightarrow A>B\)
2004A=\(\frac{2004^{2004}+2004}{2004^{2004}+1}\)
\(\frac{2004^{2004}+2004}{2004^{2004}+1}-1=\frac{2003}{2004^{2004}+1}\)
2004B=\(\frac{2004^{2005}+2004}{2004^{2005}+1}\)
\(\frac{2004^{2005}+2004}{2004^{2005}+1}-1=\frac{2003}{2004^{2005}+1}\)
Ta thấy :\(\frac{2003}{2004^{2004}+1}>\frac{2003}{2004^{2005}+1}\)
=> \(2004A>2004B\)
Vậy \(A>B\)
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
Bạn tham khảo nhé
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\) \(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{2004^{2004}+1}{2004^{2005}+1}< \frac{2004^{2004}+1+2003}{2004^{2005}+1+2003}=\frac{2004^{2004}+2004}{2004^{2005}+2004}=\frac{2004\left(2004^{2003}+1\right)}{2004\left(2004^{2004}+1\right)}=\frac{2004^{2003}+1}{2004^{2004}+1}\)
Lại có :
\(A=\frac{2004^{2003}+1}{2004^{2004}+1}\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
A > B nhé
A = 20042005 / 20042005 - 2004 + 1 / 20042005 - 2004
B = 20042005 / 20042005 +2004
Ta có B < 20042005 / 20042005 - 2004 ( tử bằng nhau, mẫu B lớn hơn) >> A > B ( ng` ta thêm 1 vào hack não hs thôi )
Tuy mk chỉ học lớp 5 nhưng mk cũng sẽ thử đoán nha !
Chắc là A = B
nếu đúng thì tk cho mk nha !