a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)

\(=1-\dfrac{1}{256}\)

\(=\dfrac{255}{256}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{13}{14}\)

c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)

\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)

\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)

\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=\dfrac{1}{15}-\dfrac{1}{90}\)

\(=\dfrac{6}{90}-\dfrac{1}{90}\)

\(=\dfrac{5}{90}=\dfrac{1}{18}\)

tớ đang cần gấp

a)255/256
b)13/14
c) 1/18

Bài 3: 

= 1- 1/2 + 1/2 -1/3 +...+ 1/98 -1/99

= 1- 1/99

= 98/99

Bài 4:

= 1/2*3 + 1/3*4 + 1/4*5 +...+  1/10*11

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/10 - 1/11

= 1/2 - 1/11= 9/22

1/2 × 1/2 + 1/2 × 1/3 + 1/3 × 1/4 + 1/4 × 1/5 + 1/5 × 1/6

= 1/4 + 1/2×3 + 1/3×4 + 1/4×5 + 1/5×6

= 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

= 1/4 + 1/2 - 1/6

= 3/12 + 6/12 - 2/12

= 9/12 - 2/12

= 7/12

1/2×1/2+1/3×1/3+1/4×1/4+1/5×1/5+1/6×1/6

phải k ạ

\(\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)

\(=\frac{1}{2\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)

M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)

M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

M = \(1-\dfrac{1}{100}\)

M = \(\dfrac{99}{100}\)

\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(M=1-\dfrac{1}{100}\)

\(M=\dfrac{99}{100}\)

??????????

Ko bit làm 

1000,5

\(\left(\dfrac{2+1}{2}\right)\left(\dfrac{3+1}{3}\right)\left(\dfrac{4+1}{4}\right)...\left(\dfrac{2000+1}{2000}\right)\)

\(=\dfrac{3.4.5...2001}{2.3.4...2000}=\dfrac{2001}{2}\)

Ta có:

Vậy giá trị của A là:

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2001}{2000}=\dfrac{2001}{2}\)