a.

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$

$=1-\frac{1}{1000}=\frac{999}{1000}$

b.

$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$

$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$

$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$

$=1-\frac{1}{500}=\frac{499}{500}$

$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$

c.

$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$

$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$

$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$

Đặt Q = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{997.998}+\frac{1}{999.1000}\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{997.999}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{997}-\frac{1}{999}\)

\(2A=1-\frac{1}{999}\)

\(2A=\frac{998}{999}\)

\(\Leftrightarrow A=\frac{499}{999}\)

Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{998.1000}\)

\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{998}-\frac{1}{1000}\)

\(2B=\frac{1}{2}-\frac{1}{1000}\)

\(B=\frac{499}{1000}\)

Vậy Q = A + B = \(\frac{499}{999}+\frac{499}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{999.1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}=\frac{999}{1000}\)

Đặt 

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{999.1000}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(\Leftrightarrow A=1-\frac{1}{1000}\)

\(\Leftrightarrow A=\frac{999}{1000}\)

Đặt:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow A=1-\frac{1}{100}\)

\(\Leftrightarrow A=\frac{99}{100}\)

bạn viết sai đề rồi

Đặt biểu thức là A.

A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

A=\(\frac{1}{1}-\frac{1}{1000}\)

A=\(\frac{999}{1000}\)

Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)

\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)

Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)\

\(A=1-\frac{1}{1000}=\frac{999}{1000}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\)

\(A=1-\frac{1}{1000}\)

\(A=\frac{999}{1000}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(A=1-\frac{1}{1000}=\frac{999}{1000}\)

1: \(A=4\dfrac{7}{1000}\cdot\dfrac{1}{999}-1\dfrac{1}{500}\cdot\dfrac{4}{999}+\dfrac{1001}{999\cdot1000}\)

Đặt 1/1000=a; 1/999=b

\(A=\left(4+7a\right)\cdot b-\left(1+2a\right)\cdot4b+b\cdot\dfrac{1001}{1000}\)

\(=4b+7ab-4b-8ab+b\cdot\left(1+a\right)\)

=-ab+b+b+ba=2b=2/999

2: Đặt 1/4587=a;1/3897=b

\(B=a\cdot\left(7+b\right)-\left(3+1-a\right)\cdot2b-7a-3ab\)

=7a+ab-8a+2ab-7a-3ab

=-8a=-8/4587