\(\frac{1}{2000}\)+2001+\(\frac{1}{2001}\)+ 2002+\(\frac{1}{2002}\)+2003+...+\(\frac{1}{2009}\)+2010

2001,0005+2002,0005+2003,0005+...+2010,0005

Số số hạng là:

(2010,0005-2001,0005)+1=10( số)

Số cặp số hạng là:

10:2= 5 ( cặp)

Tổng từng cặp là: 2001,0005+2010,0005=2002,0005+2009,0005=...=4011,001

Tổng của các số hạng trên là :

4011,001x5=20055,005

\(\frac{1}{2000+2001}+\frac{1}{2001+2002}+\frac{1}{2002+2003}+...+\frac{1}{2009+2010}\)

\(=\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2002}-...+\frac{1}{2009}-\frac{1}{2010}\)

\(=\frac{1}{2000}-\frac{1}{2010}\)

\(=\frac{1}{402000}\)

\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)

\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)

\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)

B = \(\frac{2001}{2002}+\frac{2002}{2003}\)

có: \(\frac{2000}{2001}>\frac{2000}{2001}+2002\)

\(\frac{2001}{2002}>\frac{2001}{2001}+2002\)

Vậy A>B

minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.