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4 tháng 11 2018
\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)
\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(S=\frac{5}{11}\)
\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)
\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(Q=\frac{1007}{2015}\)
~ Đấng Ed :) ~
5 tháng 1 2016
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99\cdot101}\right)\)
\(2A=-\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{99}-\frac{2}{101}\right)\)
\(2A=-\left(2-\frac{2}{101}\right)\)
\(2A=-\frac{200}{101}\)
\(\Rightarrow A=-\frac{100}{101}\)
28 tháng 2 2018
Đặt biểu thức trên là A, ta có:
\(A=\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(\Rightarrow A=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(\Rightarrow A=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(\Rightarrow2A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\left(1-\frac{1}{101}\right)\)
\(\Rightarrow2A=-\frac{100}{101}\)
\(\Rightarrow A=-\frac{100}{101}\div2=-\frac{50}{101}\)
6 tháng 7 2019
a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)
=\(\frac{1}{50}\)
BK
6 tháng 7 2019
\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)
\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)
\(\)
17 tháng 9 2016
\(B=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
\(B=\frac{1}{5.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\frac{670}{671}=\frac{134}{2013}\)
17 tháng 9 2016
Nguyễn Huy Thắngsoyeon_Tiểubàng giảiSilver bulletLê Nguyên HạoPhương AnVõ Đông Anh Tuấnsoyeon_Tiểubàng giảiLê Thị Linh ChiNguyễn Huy Tú
17 tháng 8 2018
C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9
C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9
C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9
C = 20/57 + -28/36 + 10/15 + -2/9
C = 20/57 + -7/9 + 2/3 + -2/9
C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )
C = 58/57 + -1
C = 1/57
D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41
D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41
D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41
D = 1 + -1 + 1/41
D = 1/41
E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7
E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7
E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7
E = -1 + 5/7 + 1/257 + 2/7
E = -1 + ( 5/7 + 2/7 ) + 1/127
E = -1 + 1 + 1/127
E = 1/127
PN
2
Z
31 tháng 7 2019
Lời giải:
Ta có: \(G=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}\)
\(\Rightarrow2.G=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\Rightarrow G=\frac{50}{101}\) . Vậy: \(\\G=\frac{50}{101}\)
Chúc bạn học tốt!
Tick cho mình nhé!
31 tháng 7 2019
\(G=\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{1}{99.101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\frac{100}{101}\)
\(\Leftrightarrow G=\frac{50}{101}\)
Vậy : \(G=\frac{50}{101}\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{99\times101}\)
\(=\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\right)\div2\)
\(=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\div2\)
\(=\left(1-\frac{1}{101}\right)\div2\)
\(=\frac{100}{101}\div2\)
\(=\frac{50}{101}\)
Vậy \(S=\frac{50}{101}\)
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