E = 1/7 + 1/91 + 1/247 + 1/475 + 1/ 755 + 1/ 1147
tính tổng nha
K
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PB
5
17 tháng 5 2019
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)
\(=\frac{1}{7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\frac{36}{37}\)
\(=\frac{6}{37}\)
17 tháng 5 2019
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{755}+\frac{1}{1147}\)
\(=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{35.37}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}.\frac{36}{37}=\frac{6}{37}\)
~ Hok tốt ~
DT
26 tháng 2 2017
1/7 + 1 / 91 + 1/247 + 1/475 + 1/755 + 1/1147
= 1/1.7 + 1/7.13 + 1/13.19 + 1/19.25 + 1/25.31 + 1/31.37
= 1/6. 6/1.7 + 1/6 . 6/7.13 + 1/6 . 6/13.19 + 1/6 . 6/19.25 + 1/6. 25.31 + 1/6. 31.37
= 1/6 ( 6/1.7 + 6/7.13 + 6/13.19 + 6/19.25 + 6/25.31 + 6/31.37 )
= 1/6. ( 1/1 - 1/7 + 1/7 - 1/13 +1/13 - 1/19 + 1/19 - 1/25 + 1/25 - 1/31 + 1/31 - 1/37
= 1/6 . ( 1/1 - 1/37 )
= 1/6. ( 37/37 - 1/37)
= 1/6 . 36/37
= 6 / 37
Chúc bn học tốt!
16 tháng 10 2023
1/7 + 1 / 91 + 1/247 + 1/475 + 1/755 + 1/1147
= 1/1.7 + 1/7.13 + 1/13.19 + 1/19.25 + 1/25.31 + 1/31.37
= 1/6. 6/1.7 + 1/6 . 6/7.13 + 1/6 . 6/13.19 + 1/6 . 6/19.25 + 1/6. 25.31 + 1/6. 31.37
= 1/6 ( 6/1.7 + 6/7.13 + 6/13.19 + 6/19.25 + 6/25.31 + 6/31.37 )
= 1/6. ( 1/1 - 1/7 + 1/7 - 1/13 +1/13 - 1/19 + 1/19 - 1/25 + 1/25 - 1/31 + 1/31 - 1/37
= 1/6 . ( 1/1 - 1/37 )
= 1/6. ( 37/37 - 1/37)
= 1/6 . 36/37
= 6 / 37
NN
5
14 tháng 3 2017
= 6/37 nha
đúng 10000000000000000000000000% đó nếu có sai thì trách cái máy tính của mk nha vì k pải mk tính mờ là nó tính....
tk mk nha đg âm 439 kìa....
T_T
28 tháng 1
a: \(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+\dfrac{6}{19\cdot25}+\dfrac{6}{25\cdot31}+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)
b: Sửa đề:\(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{80}{31\cdot101}+\dfrac{99}{101\cdot200}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)
\(=\dfrac{1}{5}-\dfrac{1}{200}=\dfrac{39}{200}\)
VN
Võ Ngọc Phương
VIP
6 tháng 9 2023
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}.\dfrac{36}{37}\)
\(=\dfrac{6}{37}\)
\(#Wendy.Dang\)
CU
12 tháng 2 2019
\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+...+\frac{1}{31\cdot37}\)
\(=\frac{1}{6}\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\frac{6}{13\cdot19}+...+\frac{6}{31\cdot37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...-\frac{1}{31}-\frac{1}{37}\right)\)
\(=\frac{1}{6}\left(1-\frac{1}{37}\right)\)
\(=\frac{1}{6}\cdot\frac{36}{37}=\frac{6}{37}\)
12 tháng 2 2019
Tổng cần tính bằng:\(\frac{1}{1.7}\)+\(\frac{1}{7.13}\)+\(\frac{1}{13.19}\)+\(\frac{1}{19.25}\)+\(\frac{1}{25.31}\)+\(\frac{1}{31.37}\)=(\(\frac{1}{1}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{13}\)+...+\(\frac{1}{31}\)\(\frac{1}{37}\)):3 =(\(1\)-\(\frac{1}{37}\)):3=\(\frac{12}{37}\)