a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\cdot\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot98\cdot99}{2\cdot3\cdot4\cdot...\cdot99\cdot100}=\frac{1}{100}\)

b) \(\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot...\cdot\left(1+\frac{1}{99}\right)\cdot\left(1+\frac{1}{100}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\cdot\frac{101}{100}\)

\(=\frac{3\cdot4\cdot5\cdot...\cdot100\cdot101}{2\cdot3\cdot4\cdot...\cdot99\cdot100}=\frac{101}{2}\)

\(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\)

\(=\left(1+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{99}\right)+....+\left(\frac{1}{50}+\frac{1}{51}\right)\)

\(=\frac{101}{1.100}+\frac{101}{2.99}+....+\frac{101}{50.51}\)

\(=101.\left(\frac{1}{1.100}+\frac{1}{2.99}+...+\frac{1}{50.51}\right)\)

Vế mẫu :

 \(\frac{1}{1.100}+\frac{1}{2.99}+......+\frac{1}{1.100}\)

\(=2\left(\frac{1}{1.100}+\frac{1}{2.99}+....+\frac{1}{50.51}\right)\)

Vậy kết quả là :

 \(\frac{101}{2}\)

Tử số = 1 + 1/2 + 1/3 + 1/4 + ... + 1/100

= (1 + 1/100) + (1/2 + 1/99) + ... + (1/50 + 1/51)

= 101/1.100 + 101/2.99 + ... + 101/50.51

= 101.(1/1.100 + 1/2.99 + ... + 1/50.51)

Mẫu số = 1/1.100 + 1/2.99 + 1/3.98 + ... + 1/99.2 + 1/100.1

= 2.(1/1.100 + 1/2.99 + ... + 1/50.51)

=> phân số đề bài cho = 101/2

có thể biến đổi mẫu số cho mình hông 

Xét VT:

\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)

\(VT=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=VP\)

=>đpcm

Ta xét vế trái:

\(vt=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(VT=VP\)

xét đi

=A*(1/100-1/10^2)

=0

1 - 1/2 + 1/3 - 1/4 +...+ 1/99 - 1/100

= (1 + 1/3 +...+ 1/99) - (1/2 + 1/4 +...+ 1/100)

= (1+1/2+1/3+...+1/100) - 2(1/2+1/4+...+1/100)

= (1+1/2+1/3+...+1/100) - (1+1/2+...+1/50)

= 1/51+1/52+...+1/100 (đpcm)

Bạn đã được chuyển khoản số tiền 1.000.000.000 VND 

\(=\left(\dfrac{1}{100}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{10^2}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)\)

\(=0\cdot\left(\dfrac{1}{100}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{400}\right)=0\)