\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)

\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{3}.\frac{1}{3}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{6}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{2}.\frac{1}{1}.\frac{7}{5}...\frac{209}{210}\)

\(B=\frac{7}{10}...\frac{209}{210}\)

\(B=\frac{62}{210}\)

C=2/3.5/6.9/10...209/210

C=4/6.10/12.18/20...418/420 là do nhân với 2

C=1.4/2.3.2.5/3.4.3.6/4.5...19.22/20.21

C=1.2.3....19/2.3.4...20.4.5.6...22/3.4.5...21

C=1/20.22/3

C=11/30

Dễ ấy mà hiểu chưa 

sửa đề :

\(C=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{105}\right)\)

\(C=\left(1+1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\)

\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{210}\)

\(A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{210}\right)\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)

\(A=2.\frac{13}{30}\)

\(A=\frac{13}{15}\)

Thay A vào ta được :

B = \(\left(1+1+1+1+...+1\right)-\frac{13}{15}\)

B = \(14-\frac{13}{15}\)( có 14 số 1 )

B = \(\frac{197}{15}\)

\(=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot\cdot\cdot\cdot\frac{-35}{36}\)

\(=\frac{-4}{6}\cdot\frac{-10}{12}\cdot\frac{-18}{20}\cdot\cdot\cdot\cdot\frac{-70}{72}\)

\(=\frac{-1.4}{2.3}\cdot\frac{-2.5}{3.4}\cdot\frac{-3.6}{4.5}\cdot\cdot\cdot\cdot\frac{-7.10}{8.9}\)

\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-7\right)}{2.3.4....8}\cdot\frac{4.5.6....10}{3.4.5....9}\)

\(=\frac{\left(-1\right).2.3...7}{2.3.4....8}\cdot\frac{10}{3}\)

\(=\frac{-1}{8}\cdot\frac{10}{3}=\frac{-5}{12}\)

a) =\(\frac{1}{2}.\frac{2}{3}.....\frac{2017}{2018}=\frac{1.2.....2017}{2.3.4.....2018}=\frac{1}{2018}\)

a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2018}\right)\)

\(=\frac{1}{2}.\frac{2}{3}...\frac{2017}{2018}\)

\(=\frac{1.2...2017}{2.3...2018}\)

\(=\frac{1}{2018}\)

b) \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{190}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{189}{190}\)

\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{378}{380}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{7.4}{5.6}...\frac{18.21}{19.20}\)

\(=\frac{\left(1.2.3...18\right).\left(4.5.6...21\right)}{\left(2.3.4...19\right).\left(3.4.5...20\right)}\)

\(=\frac{1.21}{19.3}\)

\(=\frac{21}{57}\)

c) \(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)...\left(1+\frac{7}{2009}\right)\)

\(=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.\frac{56}{48}...\frac{2016}{2009}\)

mk ko bít làm câu c ! xin lỗi bn nha! bn tự nghĩ cách làm câu c giúp mk nhé!