\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(=\frac{1}{5}-\frac{1}{108}\)
\(=\frac{108}{540}-\frac{5}{540}=\frac{103}{540}\)
Ta có:
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
= \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
= \(\frac{1}{5}-\frac{1}{108}\)
= \(\frac{103}{540}\)
#)Giải :
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(A=\frac{1}{5}-\frac{1}{108}\)
\(A=\frac{103}{540}\)
Đặt \(C=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(\Leftrightarrow C=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(\Leftrightarrow C=\frac{1}{5}-\frac{1}{108}=\frac{103}{540}\)
Vậy .........................
~ Hok tốt ~
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{105\cdot108}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(=\frac{1}{5}-\frac{1}{108}\)
