Question

Tính: \(A=\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{19.20}}{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}\)

Answer 1

Tính tử số:

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{20}-\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)

\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(\Rightarrow A=\frac{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}=1\)