\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
=> 3A = \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
=> 3A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
=> 3A = \(\frac{1}{2}-\frac{1}{98}\)
=> 3A = \(\frac{24}{49}\)
=> A = \(\frac{8}{49}\)
\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)
\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)
Sửa 95.98 thành 1/(95.98) nhá
Ta có
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)
\(3A=\frac{1}{2}-\frac{1}{98}\)
\(3A=\frac{49}{98}-\frac{1}{98}\)
\(3A=\frac{48}{98}=\frac{24}{49}\)
\(A=\frac{24}{49}\div3\)
\(A=\frac{8}{49}\)
Hok Tốt !!!!!!!!!!!!!!!!!!!!!!
Bài làm :
Ta có :
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(\Leftrightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(\Leftrightarrow3A=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)
\(\Rightarrow A=\frac{24}{49}\div3=\frac{8}{49}\)
Vậy A=8/49
