\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)

\(\frac{1.51}{50.2}=\frac{51}{100}\)

a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)

\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)

Vậy \(A=\frac{1}{999}\)

a) x.(x+1) = (2+2500)[(2500-2):2+1]:2 =2502.625=1563750

  x(x+1) = 1250(1250+1) 

=> x = 1250

b) 1+2+3+4+.......+x = 820

 (1+x)x:2 =820

x(x+1) = 1640 =40(40+1)

=> x = 40

ko ai trả lời thì để mình

C/M : n/n+1 < n+1/n+2

1 - n/n+1 = 1/n+1

1 - n/n + 2 = 1/n+2

Vì 1/n+1 > 1/n+2 nên n/n+1 < n+1/n+2

1/2 . 3/4 . 5/6 ... 2499/2500 < 1/2 . 2/3 . 3/4 ... 2501/2502

=1/2501 < 1/2500 (1/50) ²

1/50 < 1/49 => A <1/49

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

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

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

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