\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}=\frac{50}{51}\)
2/1.3 + 2/3.5 + 2/5.7 + ... + 2/49 . 51
= 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/49 . 51
= 1 + 51 = 52
2/1.3+2/3.5+...+2/49.51
=2(1/1.3+1/3.5+...+1/49.51)
=2(1/1-1/3+1/3-1/5+...+1/49-1/51)
=2(1/1-1/51)
=2.50/51=100/51
\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+...+\(\frac{2}{49.51}\)
= \(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+...+\(\frac{1}{49}\)-\(\frac{1}{51}\)
=\(\frac{1}{1}\)-\(\frac{1}{51}\)=\(\frac{51}{51}\)-\(\frac{1}{51}\)=\(\frac{49}{51}\)Vậy