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LT
0
HT
1
TA
21 tháng 12 2018
Đặt \(B=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)
\(\Leftrightarrow B=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(\Leftrightarrow2B=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{13}\right)=1-\frac{3}{13}=\frac{10}{13}\)
\(\Leftrightarrow A=1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}=1+\frac{10}{13}=\frac{23}{13}\)
L
4
26 tháng 7 2018
Đặt \(A=\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+...+\frac{3}{143}\)
\(A=\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{11\cdot13}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...-\frac{1}{13}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{13}\right)\)
\(A=\frac{3}{2}\cdot\frac{12}{13}=\frac{18}{13}\)
Vậy ...........
26 tháng 7 2018
\(\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+...+\frac{3}{143}.\)
\(=\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+\frac{3}{7\times9}+...+\frac{3}{11\times13}\)
\(=\frac{2}{3}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\right)\)
\(=\frac{2}{3}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{13}\right)\)
\(=\frac{2}{3}\times\frac{12}{13}\)
\(=\frac{8}{13}\)
14 tháng 6 2021
Giải:
\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{1}-\dfrac{1}{13}\)
\(=\dfrac{12}{13}\)
Chúc em học tốt!
14 tháng 6 2021
2/3+2/15+2/35+2/63+2/99+2/143
=2(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)
=2(1-1/3+1/3-1/5+1/5-....+1/13)
=2(1-1/13)
=2.12/13=24/13
VT
25 tháng 6 2023
a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
DA
2
24 tháng 4 2021
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
\(\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+...+\frac{3}{143}\)
\(=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{11.13}\)
\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\frac{12}{13}\)
\(=\frac{18}{13}\)
_Chúc bạn học tốt_
\(\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+...+\frac{3}{143}\)
\(=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{11.13}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{11.13}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\frac{12}{13}=\frac{18}{13}\)