Gía trị của biểu thức:
\(A=\frac{1}{19}+\frac{1}{19\times29}+\frac{1}{29\times39}+...+\frac{1}{1999\times2009}\) là...
(Nhập kết quả dưới dạng phân số tối giản)
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\(A=\frac{1}{19}+\frac{9}{19.29}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
đến đay bn tự tính nha
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)
Ta có: \(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(B=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)\)
\(B=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)
\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(B=\frac{9}{10}.\frac{2000}{18081}\)
\(B=\frac{200}{2009}\)
Vậy \(B=\frac{200}{2009}\)
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
\(\text{Ta có:}\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{2009.2019}\)
\(=\frac{9}{10}.\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+.....+\frac{10}{2009.2019}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2019}\right)\)
\(=\frac{9}{10}.\frac{670}{6057}\)
\(\frac{x+1}{x-1}=\frac{7}{3}\)
=> \(3.\left(x+1\right)=7.\left(x-1\right)\)
=> \(3x+3=7x-7\)
=> \(3x+10=7x\)
=> \(4x=10\)
=> \(x=\frac{10}{4}=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
Phân số thứ 20 à , hơi khó đó
Nhưng kết quả là:\(\frac{1}{1599}\)
\(\frac{-1}{2009}\)
ta có 1/19 x 29 + 1/29x39+.........+1/1999x2009
=1/19 - 1/29 . 1/29 - 1/39 ........ 1/1999-1/2009
=1/2009-1/19
=-1990/38171
=>1/19+-1990/38171
=1/2009
K MK MK K LAI