B=\(\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+...+\frac{9}{1999\cdot2009}\)
Tính B = \(\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+....+\frac{9}{1999\cdot2009}\)
Tính giá trị của biểu thức sau:
A = \(\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+......+\frac{9}{1999\cdot2009}\)
Ta có: \(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
\(\Rightarrow A=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow A=\frac{200}{2009}\)
Vậy \(A=\frac{200}{2009}.\)
Giá trị của biểu thức :
A= \(\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+...+\frac{9}{1999\cdot2009}\) là...
( Nhập kết quả dưới dạng phân số tối giản )
CÂU 1:
\(a)A=\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{36\cdot\frac{1}{5}\cdot\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(b)B=\frac{1}{19}+\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+\frac{9}{39\cdot49}+....+\frac{9}{2009.2019}\)
HELP ME, AI ĐÚNG MÌNH TICK CHO
\(B=\frac{1}{19}+\frac{9}{19\times29}+\frac{9}{29\times39}+.......+\frac{9}{1999\times2009}\)
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)
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 \(B=\frac{200}{2009}\)
tính A=\(\frac{1}{19}+\frac{9}{19\times29}+\frac{9}{29\times39}+...+\frac{9}{1999\times2009}\)
\(A=\frac{1}{19}+\frac{9}{19.29}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
đến đay bn tự tính nha
\(A=\frac{1}{19}+\frac{9}{19\times29}+\frac{9}{29\times39}+...+\frac{9}{1999+2009}la\)
viet dưới dang số thập phân tối giản
\(\frac{5\cdot4^{15}-9^9-4\cdot3^{20}\cdot89}{5\cdot29\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
Tính: \(\frac{1}{19}\)+\(\frac{9}{19\times29}\)+\(\frac{9}{29\times39}\)+.....+\(\frac{9}{2009\times2019}\)
\(\text{Ta có:}\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{2009.2019}\)
\(=\frac{9}{10}.\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+.....+\frac{10}{2009.2019}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2019}\right)\)
\(=\frac{9}{10}.\frac{670}{6057}\)