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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
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
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
M = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ..... + 1/97 - 1/99
M = 1/3 - 1/99
M = 32/99
M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)
M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99
M=1/3-1/99
M=32/99
ta co M =2/3.5+2/5.7+...+2/97.99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99=32/99
vay M =32/99
\(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
có bài toán nào nũa cứ hỏi mình mình sẽ giải đáp
2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99
m=/3.5+2/5.7+2/7.9+.....+2/97.99
=m=1/3-1/5+1/5-1/7+.......+1/97-1/99
m=1/3-1/99
=32/99
\(M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)\(\)
\(M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\)
\(M=\frac{2}{3}-\frac{2}{99}\)
\(M=\frac{64}{99}\)
\(2M=\left(\frac{2}{3}-\frac{2}{5}\right)+\left(\frac{2}{5}-\frac{2}{7}\right)+\left(\frac{2}{7}-\frac{2}{9}\right)+...+\left(\frac{2}{97}-\frac{2}{99}\right)\)
\(2M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{97}-\frac{2}{99}\)
\(2M=\frac{2}{3}-\left(\frac{2}{5}-\frac{2}{5}\right)-\left(\frac{2}{7}-\frac{2}{7}\right)-...-\left(\frac{2}{97}-\frac{2}{97}\right)-\frac{2}{99}\)
\(2M=\frac{2}{3}-\frac{2}{99}\)
\(2M=\frac{64}{99}\)
\(M=\frac{32}{99}\)
k mình nha mình đang cần