quá dễ

\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x........x\frac{99}{100}\)

= \(\frac{1x2x3x4x........x99}{2x3x4x5x.......x100}\)

=> \(\frac{1}{100}\)

Ta có: 

\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1}{2}.\frac{4}{6}.\frac{9}{12}....\frac{9801}{9900}.\frac{10000}{10100}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}=\frac{1.2.3...99.100}{2.3.4...100.101}=\frac{1}{101}\)(Tối giản)

biết làm bài 1 thôi

\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)

= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)

lượt bỏ đi còn :

\(\frac{1000}{2}=500\)

Bài 2:

B = (1/2 - 1)(1/3 -1).(1/4 -1)...(1/1000 - 1)

B = (1/2 - 2/2).(1/3 - 3/3)...(1/1000 - 1000/1000)

B = (-1/2).(-2/3)...(-999/1000)

Xét dãy số: 1; 2 ;3;...999

Dãy số trên có 999 số hạng vậy B là tích của 999 số âm

B = - 1/1000

A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)

=3/2x4/3x...............x100/99

=2-1/99

=197/99

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

A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)

A=\(\frac{100}{2}=50\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%

\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}...\cdot\frac{98}{99}\cdot\frac{99}{100}\)

\(=\frac{1}{100}\)

#

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{98}{99}.\frac{99}{100}\)

\(=\frac{1.2.3.4.....98.99}{2.3.4.5.....99.100}\)

\(=\frac{1}{100}\)

em hỏi thầy cô đây là toán chuyên