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\(M=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(M=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{99-97}{97\cdot99}\)
\(M=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+\frac{9}{7\cdot9}-\frac{7}{7\cdot9}+...+\frac{99}{97\cdot99}-\frac{97}{97\cdot99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{33}{99}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
Vậy \(M=\frac{32}{99}\)
Có 2/ 3.5 + 2/ 5.7 + 2/ 7.9 +...+ 2/ 97.99
= 1/3 -1/5 +1/5 -1/7 +1/7 -1/9 +...+ 1/ 97- 1/99
= 1/3 - 1/99
= 32/ 99
\(B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=2.\frac{98}{303}=\frac{196}{303}\)
Mk bik câu B nè!
2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99
2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99
2B = 1/3 - 1/99
2B = 32/99
=> B = 16/99
ta có 2/n(n+2)=1/n-1/(n+2)
nên 2/3.5=1/3-1/5
2^2/3.5+2^2/5.7+2^2/7.9+...+2^2/49.51
=2.{2/3.5+2/5.7+..+2/49.51}
=2{1/3-1/5+1/5-1/7+...+1/49-1/51}
=2{1/3-1/51}=32/51
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
\(M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)\(\)
\(M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\)
\(M=\frac{2}{3}-\frac{2}{99}\)
\(M=\frac{64}{99}\)
\(2M=\left(\frac{2}{3}-\frac{2}{5}\right)+\left(\frac{2}{5}-\frac{2}{7}\right)+\left(\frac{2}{7}-\frac{2}{9}\right)+...+\left(\frac{2}{97}-\frac{2}{99}\right)\)
\(2M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{97}-\frac{2}{99}\)
\(2M=\frac{2}{3}-\left(\frac{2}{5}-\frac{2}{5}\right)-\left(\frac{2}{7}-\frac{2}{7}\right)-...-\left(\frac{2}{97}-\frac{2}{97}\right)-\frac{2}{99}\)
\(2M=\frac{2}{3}-\frac{2}{99}\)
\(2M=\frac{64}{99}\)
\(M=\frac{32}{99}\)
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