Ta dùng phương pháp triệt tiêu sẽ được kết quả cuối cùng là :
1 - \(\frac{1}{15}\) = \(\frac{14}{15}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=1-\frac{1}{15}\)
\(A=\frac{14}{15}\)
A=\(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+ \(\frac{2}{5x7}\)+ .... + \(\frac{2}{13x15}\)
A = \(\frac{3-1}{1x3}\)+ \(\frac{5-3}{3x5}\)+ \(\frac{7-5}{5x7}\)+ \(\frac{15-13}{13x15}\)
A = \(\frac{3}{1x3}\)- \(\frac{1}{1x3}\)+ \(\frac{5}{3x5}\)- \(\frac{3}{3x5}\)+ \(\frac{7}{5x7}\)- \(\frac{5}{5x7}\)+ ... + \(\frac{15}{13x15}\)- \(\frac{13}{13x15}\)
A = \(1\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ... + \(\frac{1}{13}\)- \(\frac{1}{15}\)
A = \(1\)- \(\frac{1}{15}\)= \(\frac{14}{15}\)
Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
=14/15
A=1-1/15=14/15
AI K MIK, MIK K LẠI
k mik nhađặng thị hà vi
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{15.13}\)
\(A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}\)
\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{1}-\frac{1}{15}\)
\(A=\frac{14}{15}\)