Question

Tính A/B biết

A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)

B=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)

Answer 1

Ta có:  1/1.2 = 1- 1/2 

           1/3.4 = 1/3 - 1/4 

          ...............

             1/19.20 = 1/19 - 1/20 

Cộng vế với vế ta đc: 

    A = 1- 1/20 = 19/20   

Answer 2

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)

\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)

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

Answer 3

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{19.20}\)

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

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)

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

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}=B\)

\(\Rightarrow\frac{A}{B}=1\)

Answer 4

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}=B\)

\(\Rightarrow\frac{A}{B}=1\)

Answer 5

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{17.18}+\frac{1}{19.20}\)

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

\(=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)