Ta có: \(\frac{2}{3.5}=\frac{5-3}{3.5}=\frac{5}{3.5}-\frac{3}{3.5}=\frac{1}{3}-\frac{1}{5}\)
\(\frac{2}{5.7}=\frac{7-5}{5.7}=\frac{7}{5.7}-\frac{5}{5.7}=\frac{1}{5}-\frac{1}{7}\)
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\(\frac{2}{97.99}=\frac{99-97}{97.99}=\frac{99}{97.99}-\frac{97}{97.99}=\frac{1}{97}-\frac{1}{99}\)
Vậy \(M=\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{95.97}+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-...-\left(\frac{1}{95}-\frac{1}{95}\right)-\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)
=\(\frac{33}{99}-\frac{1}{99}\)
=\(\frac{23}{99}\)
M = 2/3.5+2/5.7+2/7.9+...+2/97.99
= 1/2(2/3-2/5+2/5-2/7+2/7-2/9+...+2/97-2/99)
= 1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
= 1/3 - 1/99
= 33/99-1/99
= 32/99
đúng nha
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