A= ( 1/10-1) + ( 1/11 - 1 ) +...+ ( 1/100-1)

  = 9/10 + 10/11 +...+ 99/100

  = 9/100

^_^ ( have a good day)

\(=\frac{-9}{10}\cdot\frac{-10}{11}\cdot\frac{-11}{12}\cdot\cdot\cdot\cdot\frac{-99}{100}\)

\(=\frac{9}{-10}\cdot\frac{-10}{11}\cdot\frac{11}{-12}\cdot\cdot\cdot\cdot\frac{99}{-100}\)

\(=\frac{-9}{100}\)

cảm ơn bạn nhé ST kết bạn với mình nhé để mình có thể hỏi bài từ bạn, đi mình xin bạn đấy

\(\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)

\(=-\frac{9}{10}.\frac{-10}{11}.\frac{-11}{12}...\frac{-99}{100}\)

\(=-\frac{9}{100}\)

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\)    \(\left(1-\frac{1}{11}\right)\)\(\left(1-\frac{1}{13}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\)
 \(=1.\left(\frac{1}{3}-\frac{1}{2}\right).\left(\frac{1}{5}-\frac{1}{4}\right).\left(\frac{1}{7}-\frac{1}{6}\right)\)\(.\left(\frac{1}{11}-\frac{1}{10}\right).\frac{1}{13}\)
  \(=1.1.1.1.1.\frac{1}{13}\)
   \(=\frac{1}{13}\)
Cái này là chỉ do mình nghĩ thôi, ko biết đúng ko nữa
Có sai cko mình sorry nha!
Tk và kb hộ mình  với! thanks 

b)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-99}{100}=\frac{-1.\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}=-\frac{1}{100}\)

\(=\frac{-9}{10}.....\frac{-2011}{2012}\)

=\(\frac{-9.....\left(-2011\right)}{10.11.....2012}\)=\(\frac{9}{2012}\)

mình nhầm phải là \(\frac{-9}{2012}\)

đỡ hơn chưa??? mong các bn giúp mình vs

Vê trái: 

\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)

\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)

Vế phải:

\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)

\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải

=> đpcm