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Đặt \(A=\dfrac{2003.2004-1}{2003.2004}\) và \(B=\dfrac{2004.2005-1}{2004.2005}\)
Ta có : \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}\)
\(=1-\dfrac{1}{2003.2004}\)
\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}\)
\(=1-\dfrac{1}{2004.2005}\)
Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)
Nên \(A< B\)
Vậy \(\dfrac{2003.2004-1}{2003.2004}< \dfrac{2004.2005-1}{2004.2005}\)
~ Học tốt ~
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)\(n+1=4008\Rightarrow n=4007\)
\(2004A=\frac{2004^{2004}+2004}{2004^{2004}+1}=1+\frac{2003}{2004^{2004}+1}\)
\(2004B=\frac{2004^{2005}+2004}{2004^{2005}+1}=1+\frac{2003}{2004^{2005}+1}\)
\(\frac{2003}{2004^{2004}+1}>\frac{2003}{2004^{2005}+1}\)
\(\Rightarrow2004A>2004B\)
\(\Rightarrow A>B\)
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 \(A>B\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}.\frac{n+1}{n+2}\)
\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)
\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)
a) Ta có: \(\frac{n}{n-3}\)có tử số lớn hơn mẫu số. \(\Rightarrow\frac{n}{n-3}>1\)
Ta lại có: \(\frac{\left(n+1\right)}{n+2}< 1\)( vì \(\frac{\left(n+1\right)}{n+2}\) có tử bé hơn mẫu)
\(\Rightarrow\frac{n}{n-3}>\frac{\left(n+1\right)}{n+2}\)
b)
Mà: \(\frac{2003.2004-1}{2003.2004}=1\)( Loại hai số giống nhau ở cả tử và mẫu: 2003 , 2004)
Còn: \(\frac{2004.2005-1}{2004.2005}=1\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}=\frac{2004.2005-1}{2004.2005}\)
P/s: Mình không chắc câu b) Nhé
Ta thấy : n > n - 3
=> \(\frac{n}{n-1}>1\)
Có : n + 1 < n + 2
=> \(\frac{n+1}{n+2}< 1\)
=> \(\frac{n}{n-3}>\frac{n+1}{n+2}\)