Mình tính đc kết quả 22 nhưng hơi khó hiểu mong mọi người giải dùm

đặt A=(1-1/3)........

Ta có A=\(\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}\cdot...\cdot\frac{209}{210}=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}\cdot...\cdot\frac{418}{420}=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{19\cdot22}{20\cdot21}\)

=\(\frac{1\cdot4\cdot2\cdot5\cdot3\cdot6\cdot...\cdot19\cdot22}{2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot...\cdot20\cdot21}=\frac{\left(1\cdot2\cdot3\cdot...\cdot19\right)\cdot\left(4\cdot5\cdot6\cdot...\cdot22\right)}{\left(2\cdot3\cdot4\cdot...\cdot20\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot21\right)}\)

=\(\frac{1\cdot22}{20\cdot3}=\frac{11}{30}\)

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

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

=>\(A=\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}.\frac{14.2}{15.2}...\frac{209.2}{210.2}\)

=>\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{418}{420}\)

=>\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{19.22}{20.21}\)

=>\(A=\frac{\left(1.4\right).\left(2.5\right).\left(3.6\right).\left(4.7\right)...\left(19.22\right)}{\left(2.3\right).\left(3.4\right).\left(4.5\right).\left(5.6\right)...\left(20.21\right)}\)

=>\(A=\frac{\left(1.2.3.4...19\right).\left(4.5.6.7...22\right)}{\left(2.3.4.5...20\right).\left(3.4.5.6...21\right)}\)

=>\(A=\frac{1.22}{20.3}\)

=>\(A=\frac{22}{60}=\frac{11}{30}\)

Vậy \(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)=\frac{11}{30}\)

11/30 là đúng bạn ạ k hộ cái nha

a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)

b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)

c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)

\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)

Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)

  \(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)

  \(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

   \(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)

   \(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)

 \(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)

\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)

Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{100}{101}\cdot\frac{1}{2}=\frac{50}{101}\)

Ta có:

a)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

b)

 \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{210}\)

\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)

a) Gọi biểu thức trên là A

 \(A=\frac{1}{1\times3}+\frac{1}{5\times7}+...+\frac{1}{99\times101}\)

\(2A=\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{99\times101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{100}{101}\div2\)

\(A=\frac{50}{101}\)

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