Tính \(\frac{-1}{3}\)+\(\frac{-1}{15}\)+\(\frac{-1}{35}\)+\(\frac{-1}{63}\)+........+\(\frac{-1}{9999}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MP
25 tháng 1 2016
A=1/3.5+1/5.7+1/7.9+...+1/99.101
2A= 2/3.5+2/5.7+2/7.9+...+2/99.101
2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101
2A=1/3-1/101=98/303
A=(98/303)/2=49/303
DD
5 tháng 1 2016
\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)
A.2=2/3.5+2/5.7+2/7.9+…+2/99.101
A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101
Vậy
A.2=1/3-1/101=98/303
A=98/303/2=49/303
Đúng không
4 tháng 1 2016
A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101
A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101
= 1/3 - 1/101 = 98/303
Vậy A = 98/303 : 2 = 49/303
PY
2
BN
22 tháng 5 2016
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{98}{303}\)
\(A=\frac{49}{303}\)
22 tháng 5 2016
A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
2A=\(\frac{1}{3}-\frac{1}{101}\)
2A=\(\frac{98}{303}\)
A=\(\frac{98}{303}.\frac{1}{2}\)
A=\(\frac{49}{303}\)
Chúc bạn học tốt!
LH
2
NT
10
IW
9 tháng 7 2016
\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)
\(\Rightarrow A=\frac{44}{303}\)
MH
9 tháng 7 2016
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
=> A = 98/203 : 2 = 49/303
27 tháng 6 2017
= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101
=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)
=1/2 x (1/3 - 1/99)
=1/2 x (1/3 - 1/101)
=1/2 x (98/303)
=1/15 + 1/35 + 1/63 +1/99+...+1/9999
=49/303
27 tháng 6 2017
\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)
\(=\frac{98}{303}\)
2 tháng 2 2016
1/
A= 1/15+1/35+1/63+1/99+ ... + 1/9999
A=1/3.5+1/5.7+1/7.9+ ... +1/99.101
2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101
2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101
2A=1/3-1/101
A=49/303
Sai thì thôi nhé
2 tháng 2 2016
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
25 tháng 12 2016
siêu tốc
\(2000+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=2000+\frac{50}{3.101}\)
LA
25 tháng 12 2016
Ta có: \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
=\(2000+\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
Đặt A=\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
=> 2xA =\(\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{99x101}\right)\)
2xA = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
2xA = \(\frac{1}{3}-\frac{1}{101}\)
2xA = \(\frac{98}{303}\)
A = \(\frac{98}{606}=\frac{49}{303}\)
=> \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)=2000+\frac{49}{303}=\frac{606049}{303}\)
Bạn xem rút gọn được thì rút nhé
WH
29 tháng 3 2018
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99\cdot101}\right)\)
\(2A=-\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{99}-\frac{2}{101}\right)\)
\(2A=-\left(2-\frac{2}{101}\right)\)
\(2A=-\frac{200}{101}\)
\(\Rightarrow A=-\frac{100}{101}\)
Đặt biểu thức trên là A, ta có:
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(\Rightarrow A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow A=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\frac{100}{101}\)
\(\Rightarrow A=-\frac{100}{101}\div2=-\frac{50}{101}\)