A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}-\dfrac{1}{20}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{7}-\dfrac{1}{20}=1-\dfrac{27}{140}=\dfrac{113}{140}\)
Giải:
A=1/2+1/6+1/12+1/30+1/42
A=1/1.2+1/2.3+1/3.4+1/5.6+1/6.7
A=1/1-1/2+1/2-1/3+1/3-1/4+1/5-1/6+1/6-1/7
A=1/1-1/4+1/5-1/7
A=113/140
Chúc bạn học tốt!
Sửa đề: \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
Ta có: \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}++\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}=\dfrac{6}{7}\)
A = \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{42}\)
A = \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{42}\right)+\dfrac{1}{12}\)
A = \(\left(\dfrac{21}{42}+\dfrac{7}{42}+\dfrac{1}{42}\right)+\dfrac{1}{12}\)
A = \(\dfrac{29}{42}+\dfrac{1}{12}\)
A = \(\dfrac{58}{84}\) \(+\dfrac{7}{84}\)
A= \(\dfrac{65}{84}\)
Vậy A = \(\dfrac{65}{84}\)