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PT
3
T
6 tháng 6 2019
#)Giải :
\(P=1+\frac{9}{45}+\frac{9}{105}+\frac{9}{189}+...+\frac{9}{29997}\)
\(P=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(P=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101} \right)\)
\(P=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(P=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(P=\frac{3}{2}\times\frac{100}{101}\)
\(P=\frac{150}{101}\)
NA
0
LT
3
KN
4 tháng 5 2019
S = 1/9 + 1/45 + 1/105 + 1/189 + 1/297
=> S = 1/2 ( 6/27 + 6/135 + 6/315 + 6/567 + 6/891 )
=> S = 1/2 ( 6/3.9 + 6/9.15 + 6/15.21 + 6/21.27 + 6/27.33 )
=> S = 1/2 ( 1/3 - 1/9 + 1/9 - 1/15 + ... + 1/27 - 1/33 )
=> S = 1/2 ( 1/3 - 1/33 )
=> S = 1/2 . 10/33
=> S = 5/33
4 tháng 5 2019
\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)
\(S=\frac{1}{1.9}+\frac{1}{9.5}+\frac{1}{5.21}+\frac{1}{21.9}+\frac{1}{9.33}\)
\(5S=\frac{5}{1.9}+\frac{5}{9.5}+\frac{5}{5.21}+\frac{5}{21.9}+\frac{5}{9.33}\)
\(5S=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{5}+\frac{1}{5}+\frac{1}{21}+\frac{1}{21}-\frac{1}{9}+\frac{1}{9}-\frac{1}{33}\)
\(5S=1-\frac{1}{33}\)
\(5S=\frac{32}{33}\)
\(S=\frac{32}{33}:5\)
\(S=\frac{32}{165}\)
LT
5
XO
2 tháng 7 2020
\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}=\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=\frac{1}{6}\left(1-\frac{1}{11}\right)=\frac{1}{6}.\frac{10}{11}\)
\(=\frac{5}{33}\)
3 tháng 7 2020
Bài làm:
\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)
\(S=\frac{1}{6}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)
\(S=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(S=\frac{1}{6}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(S=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{6}\left(1-\frac{1}{11}\right)\)
\(S=\frac{1}{6}.\frac{10}{11}=\frac{5}{33}\)
Vậy \(S=\frac{5}{33}\)
Xin lỗi bạn Xyz nhé, mk ko có chép bài bạn đâu! với lại mk cx ko k sai bài bn đâu nhé!
BT
1
C
0
TN
1
19 tháng 7 2017
\(A=\frac{7}{9}+\frac{7}{45}+\frac{7}{105}+...+\frac{7}{27645}\)
\(=7\left(\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+...+\frac{1}{27645}\right)\)
\(=7.\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9215}\right)\)
\(=\frac{7}{3}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{95.97}\right)\)
Đặt \(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{95.97}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{95.97}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\)
\(=1-\frac{1}{97}=\frac{96}{97}\)
\(\Rightarrow S=\frac{96}{97}:2=\frac{96}{97.2}=\frac{48}{97}\). Thay vào \(A\) ta có:
\(A=\frac{7}{3}.\frac{48}{97}=\frac{112}{97}\)
Vậy \(A=\frac{112}{97}\).
EH
14