Ta có:

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=2001\)

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}}\)

\(=\frac{\left[\frac{2001}{1}+1\right]+\left[\frac{2001}{2}+1\right]+...+\left[\frac{2001}{2000}+1\right]+2001}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}\)

\(=\frac{2001\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}\right]}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}=2001\)

$\ge $ 

là sao?

cho mình hỏi và đề đâu

\(D=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1998.1999}+\dfrac{1}{1997.1998}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{1997.1998}+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\right)\)

\(D=\dfrac{1}{1999.2000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)\(D=\dfrac{1}{1999.2000}-\dfrac{1999}{2000}\)

1, S1 = (-2) +  (-2) +..+ (-2).

Có SS (-2) là :

(1997 - 1) : 4 +1 = 500 (số ).

Tổng số (-2) là: 500 x (-2) = (-1000)

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Đặt A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+....+\frac{1}{1999}}\)

Xét mẫu số:

\(\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+\frac{1996}{4}+....+\frac{1}{1999}\)

=\(\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+\left(\frac{1996}{4}+1\right)+....+\left(\frac{1}{1999}+1\right)+1\)

=\(\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{1999}+\frac{2000}{2000}\)

= 2000\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{1999}+\frac{1}{2000}\right)\)

=> A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}\right)}\)

=> A = \(\frac{1}{2000}\)

a) \(A=1+\left(-3\right)+5+\left(-7\right)+...+\left(-1999\right)+2001\)

Số số hạng của tổng trên là: \(\frac{2001-1}{2}+1=1001\).

\(A=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[1997+\left(-1999\right)\right]+2001\)

\(A=-2.500+2001\)

\(A=1001\)

b) \(1+\left(-2\right)+\left(-3\right)+4+5+\left(-6\right)+\left(-7\right)+8+...+1997+\left(-1998\right)+\left(-1999\right)+2000\)

\(=\left\{\left[1+\left(-2\right)\right]+\left[\left(-3\right)+4\right]\right\}+...+\left\{\left[1997+\left(-1998\right)\right]+\left[\left(-1999\right)+2000\right]\right\}\)

\(=\left(-1+1\right)+\left(-1+1\right)+...+\left(-1+1\right)\)

\(=0+0+...+0=0\)