\(\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+...+\frac{6}{97\cdot100}\)
\(=2\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(=2\cdot\frac{24}{100}=\frac{24}{50}=\frac{12}{25}\)
Đặt \(A=\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+...+\frac{6}{97\cdot100}\)
\(A=\)\(2\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=2\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=2\left(\frac{25}{100}-\frac{1}{100}\right)\)
\(A=2\cdot\frac{24}{100}=2\cdot\frac{6}{25}=\frac{12}{25}\)
Đặt biểu thức là A.Từ biểu thức, ta suy ra:
\(A=6\cdot\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}\right)\)
\(3A=6\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\right)\)
\(3A=6\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(3A=6\cdot\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(3A=6\cdot\frac{6}{25}\)
\(A=\frac{6}{25}:3=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
