Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\ =\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\ =-\frac{1.2.3.4....998.999}{2.3.4...1000}\\ =-\frac{1}{1000}\)

Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.

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$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$

$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$

$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$

$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$

$D=\frac{2^2.3^2.4^2....50^2}{1.3.2.4.3.5....49.50}$

$=\frac{(2.3.4...50)(2.3.4...50)}{(1.2.3...49)(3.4....50)}$
$=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4....50}{3.4....50}$
$=50.2=100$

help

Bài 2:

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{999\cdot1000}\)

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

=1-1/1000

=999/1000

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

\(=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4...100}\)

\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)

\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)