C= \(\frac{3}{3\times5}+\frac{3}{5\times7}+\frac{3}{7\times9}+....+\frac{3}{47\times49}\)
K
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19 tháng 9 2019
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
VV
6 tháng 5 2017
m = 1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
m = 1/3-1/99=32/99
Sorry chị em ko làm đc câu b vì em mới học lớp 4
k em ha
6 tháng 5 2017
a) \(M=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{97\times99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow M=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
b) \(N=\frac{3}{5\times7}+\frac{3}{7\times9}+\frac{3}{9\times11}+...+\frac{3}{197\times199}\)
\(\Rightarrow N=3\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\)
\(\Rightarrow N=3\times\left[2\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\right]\)
\(\Rightarrow N=3\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{197\times199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\frac{194}{995}=\frac{582}{995}\)
----Chúc em học giỏi !----
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
NP
2
28 tháng 3 2016
\(C=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{97x99}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(C=\frac{1}{3}-\frac{1}{97}\)
\(C=\frac{94}{291}\)
M
19 tháng 7 2017
a=1/3x5+1/5x7+...+1/2003x2005
a=1x2/3x5x2+1x2/5x7x2+...+1x2/2003x2005x2
a=1/2(2/3x5+2/5x7+...+2/2003x2005)
a=1/2x(1/3-1/5+1/5-1/7+...+1/2003-1/2005)
a=1/2x(1/3-1/2005)
a=1/2x2002/6015
a=1001/6015
8 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{48}\)
=> 2x + 3 = 48
=> 2x = 48 - 3
=> 2x = 45
=> x = 45/2
4 tháng 4 2019
Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(\Leftrightarrow M=\frac{2}{3.\left(3+2\right)}+\frac{2}{5.\left(5+2\right)}+...+\frac{2}{97\left(97+2\right)}\)
\(\Leftrightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Leftrightarrow M=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
( Dòng thứ 2 mik làm để bạn hiểu mik đã áp dụng công thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) nên bạn ghi hay ko cx được)
4 tháng 4 2019
\(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)=\(\frac{32}{99}\)
PM
2
14 tháng 4 2017
gọi biểu thức đó là A
\(A=\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{2008}{6033}\right)\)
\(A=\frac{1004}{6033}\)
mink nghĩ vậy bạn ạ
9 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
\(\Rightarrow\)C . \(\frac{2}{3}\) =\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)....+\(\frac{2}{47.49}\)
\(\Rightarrow C.\frac{2}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow C=\left(\frac{1}{3}-\frac{1}{49}\right)\div\frac{2}{3}\)
\(\Rightarrow C=\frac{46}{147}\div\frac{2}{3}\)
\(\Rightarrow C=\frac{23}{49}\)
Vậy C = \(\frac{23}{49}\)
nếu đúng từ mk xin 1 chữ đúng và