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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1.3.5...79}{2.4.6...80}\)= \(\frac{1.3.5...79}{\left(1.2\right).\left(2.2\right).\left(3.2\right)...\left(40.2\right)}\).\(\frac{1.3.5...79}{\left(1.2.3.4...40\right).\left(2.2.2.2...2.2\right)}\)=\(\frac{1.3.5...79}{\left(1.3.5...39\right).\left(2.4.6...40\right).2^{40}}\)<1/9
![](https://rs.olm.vn/images/avt/0.png?1311)
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow A< 1-\frac{1}{9}=\frac{8}{9}\)(1)
Lại có: \(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(2)
Từ (1) và (2), suy ra: \(\frac{2}{5}< A< \frac{8}{9}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{79}{80}