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TV
1
3 tháng 5 2018
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
\(\frac{1}{2}\)\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)\(=\frac{15}{93}\)
\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\rightarrow2x=90\rightarrow x=45\)
S
4 tháng 3 2017
\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
= \(\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)
24 tháng 5 2017
Bài giải:
\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(=\frac{8}{51}\)
NV
1
HT
25 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+.....+\frac{1}{129.15}=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+....+\frac{1}{1935}\)
=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{43.45}\)
=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{43}-\frac{1}{45}\right)\)
=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{45}\right)\)
=\(\frac{1}{2}.\frac{43}{90}\)
=\(\frac{43}{180}\)
NH
1
4 tháng 5 2016
\(S=1:3+1:15+1:35+...+1:9999\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2S=1-\frac{1}{101}\)
\(2S=\frac{100}{101}\)
\(S=\frac{100}{101}:2\)
\(S=\frac{50}{101}\)
7 tháng 8 2016
A= 1/1.3+1/3.5+1/5.7+...+1/13.15
2A=2/1.3+2/3.5+2/5.7+...+2/13.15
2A=1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15
2A=1-1/15
2A=14/15
A=7/15
S
30 tháng 7 2016
Ta có: 1/3 ; 1/15 ; 1/35;...
<=> 1/1.3 ; 1/3.5 ; 1/5.7
=> chữ số thứ 100 là: 1/199.201
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
XO
10 tháng 6 2020
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
PQ
2
4 tháng 8 2019
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
4 tháng 8 2019
A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\)
A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)
A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)
A=\(\frac{1}{1}-\frac{1}{11}\)
=>A=\(\frac{10}{11}\)
B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)
2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
2B=\(\frac{1}{3}-\frac{1}{21}\)
2B=\(\frac{2}{7}\)
B=\(\frac{2}{7}:2\)
=>B=\(\frac{1}{7}\)