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

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

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

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

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

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

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

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

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

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

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

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

      \(=\frac{1}{100}\)

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

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

\(M=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{\frac{18.19.20...36}{2.3...19}}{\frac{20.21.22...36}{2.3...17}}=\frac{\frac{18.19}{18.19}}{1}=\frac{1}{1}=1\)

 \(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)...\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right)}\)

\(=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{18.19.20...36}{1.2.3...19}:\frac{20.21.22...36}{1.2.3...17}\)

\(=\frac{18.19.20...36}{1.2.3...19}.\frac{1.2.3...17}{20.21.22....36}=\frac{1.2.3...17.18...36}{1.2.3...19.20...36}=1\)

A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

A = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)...\left(-\frac{2013.2015}{2014.2014}\right)\)

A = \(-\left[\frac{\left(1.2....2013\right)\left(3.4....2015\right)}{\left(2.3....2014\right)\left(2.3...2014\right)}\right]\)

A = \(-\left(\frac{2015}{2014.2}\right)\)

A = \(-\frac{2015}{4028}\)

còn câu b thì sao z mấy bn?