\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)

\(A=\frac{1.2.3...99}{2.3.4...100}\)

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

\(B=1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{72}\)

\(B=1+1+...+1+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}\right)\)

\(B=5.1+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)

\(B=5+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(B=5+\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(B=5+\frac{2}{9}=\frac{47}{9}\)

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

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

     \(=\frac{1.2.3.4....99}{2.3.4.5...100}\)

      \(=\frac{1}{100}\)

Câu A thì dễ rồi

bạn trừ đi là dc

câu b thì hơi khó

Ta có

\(\frac{1}{12}+\frac{1}{20}+....+\frac{1}{72}=\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{8}{9}\)

Ta thấy khi cộng tổng mới cho và  thì kết quả sẽ là  1+1+1+1+1+1=6 ( VÌ cứ ghếp từng đôi sẽ thấy)

=>B=\(6-\frac{8}{9}=5\frac{1}{9}\)

\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)

\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)

\(x=\frac{1}{2011}:\frac{1}{2011}=1\)

Vậy x=1

\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)

\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)

\(\frac{1}{2011}.x=\frac{2}{4022}\)

\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)

Ai thấy đún thì ủng hộ mink nha !!!

Thanks you very much !!

Chúc các bạn luôn học giỏi !!!

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

A = ( 1 - 1/2) . ( 1 - 1/3 ) . (1-1/4) ....(1-1/2015) . (1-1/2016)

A= 1/2 . 2/3 . 3/4...2014/2015 . 2015/2016

A = 1 . 2 . 3 . 4 ... 2014 . 2015/ 2 . 3 . 4 ... 2015 . 2016

A = 1/ 2016

\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)

\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)

\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)

<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)

<=> x=-40

Vậy x=-40

=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

k mk nha

1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010

=> -1/x+3 = 1/2010

=> 1/x+3 = 1/-2010

=> x+3 = -2010

=> x = -2010-3 = -2013

\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)

=> \(\frac{-1}{x+3}=\frac{1}{2010}\)

=> \(-x-3=2010\)

=> \(x=2007\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)

\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)

\(\Leftrightarrow x=\frac{2010}{10}=201\)

Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)

=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)

\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)

\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)

\(\Rightarrow x=\frac{2010}{10}=201\)

\(=\frac{1}{10}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{x}\right)=\frac{1}{2010}.\)

\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{x-1}{x}=\frac{1}{2010}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2010}\)

\(\Leftrightarrow x=2010\)

Vậy x=2010.