\(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{x\times\left(x+2\right)}=\frac{32}{99}\)\(\frac{32}{99}\)
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27 tháng 7 2015
\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)
\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)
30 tháng 4 2019
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
NH
10 tháng 8 2017
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\right)\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{101}{303}-\frac{3}{303}=\frac{98}{303}\)
T
10 tháng 8 2017
Đặt A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Leftrightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.100}\)
\(=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{7}+\frac{1}{9}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
9 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
3 tháng 6 2015
(4/1*3+4/3*5+4/5*7+4/7*9)*10-x=0
=4*2/1*3+4*2/3*5+4*2/5*7+4*2/7*9
=1/1+1/3+1/5+1/7+1/9
=1/1-1/9
=8/9
8/9*10-x=0
89-x=0
x=89-0
x=89
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
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 !----
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97