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NH
0
TH
2
TM
24 tháng 6 2018
a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)
\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(a=\frac{1}{3}-\frac{1}{39}\)
\(a=\frac{12}{39}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)
\(=0\)
24 tháng 6 2018
a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)
\(=0\)
HN
1
DD
5 tháng 8 2016
=（2-1）*（2+1）+（4-1）*（4+1）+ ...+(2n-1)*(2n+1) =(2^2-1)+(4^2-1)+...+(4n^2-1) =(2^2+4^2+...+4n^2)-(1+1+...+1) =4(1^2+2^2+...n^2)-n n(n+1)(2n+1)/6： 1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n^2=n 1x3+3x5+5x7+7x9+...+17x19 =4(1^2+2^2+...n^2)-n =4*n(n+1)(2n+1)/6-n； n=10，1x3+3x5+5x7+7x9+...+17x19=1530
21 tháng 3 2016
M=1/3.7+1/7.11+1/11.15+...+1/43.47
M=1/3-1/7+1/7-1/11+1/11-1/15+...+1/43-1/47
M=1/3-1/47
CÒN LẠI TỰ TÍNH NHA BN
AI THẤY ĐÚNG THÌ ỦNG HỘ NHA
DD
21 tháng 3 2016
M=\(\frac{1}{3x7}+\frac{1}{7x11}+\frac{1}{11x15}+...+\frac{1}{43x47}\)
=>4M=\(\frac{4}{3x7}+\frac{4}{7x11}+...+\frac{4}{43x47}\)
=>4M=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\)
=>4M=\(\frac{1}{3}-\frac{1}{47}\)
=>4M=\(\frac{44}{141}\)
=>M=\(\frac{44}{141}:4\)
=>M=\(\frac{11}{141}\)
D
4
S
27 tháng 2 2017
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)
=> \(\frac{5}{11}y=\frac{2}{3}\)
=>y = \(\frac{2}{3}:\frac{5}{11}\)
=> y = \(\frac{22}{15}\)
3 tháng 4 2021
cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm
HH
11 tháng 6 2018
Giải:
\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)y=-\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{10}{11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow\dfrac{5}{11}=-\dfrac{2}{3y}\)
\(\Leftrightarrow15y=-22\)
\(\Leftrightarrow y=-\dfrac{22}{15}\)
Vậy ...
NL
1
21 tháng 8 2019
\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)
\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\frac{10}{11}.y=\frac{2}{3}\)
\(\frac{20}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{11}{30}\)
Study well
\(A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{91\cdot93}\)
\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{91}-\dfrac{1}{93}\)
\(=\dfrac{1}{5}-\dfrac{1}{93}=\dfrac{88}{93\cdot5}=\dfrac{88}{465}\)
\(B=\dfrac{1}{3\cdot7}+\dfrac{1}{7\cdot11}+...+\dfrac{1}{97\cdot101}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=\dfrac{1}{4}\cdot\dfrac{98}{303}=\dfrac{49}{606}\)