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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 \(A=\frac{200}{2009}.\)
\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{9.19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(=\frac{200}{2009}\)
Gọi \(B=\frac{9}{19}+A\)
\(A=\frac{9}{19\cdot29}+\frac{9}{29\cdot39}+...+\frac{9}{1999\cdot2009}\)
\(\frac{A}{9}=\frac{1}{19\cdot29}+\frac{1}{29\cdot39}+...+\frac{1}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{10}{19+29}+\frac{10}{29\cdot39}+...+\frac{10}{1999\cdot2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\)
\(\frac{A\cdot10}{9}=\frac{1}{19}-\frac{1}{2009}\)
\(A=\frac{1791}{38171}\)
\(\Rightarrow B=\frac{1}{19}+\frac{1791}{38171}\)
\(\Rightarrow B=\frac{200}{2009}\)
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
\(\frac{a^2+2b^2-m^2}{a^2+3b^2-6m^2}=\frac{\left(4m\right)^2+2\left(5m\right)^2-m^2}{\left(4m\right)^2+3\left(5m\right)^2-6m^2}\)
\(=\frac{4^2.m^2+2.5^2.m^2-m^2}{4^2.m^2+3.5^2.m^2-6.m^2}=\frac{16.m^2+50.m^2-m^2}{16.m^2+75.m^2-6.m^2}\)
\(=\frac{m^2.\left(16+50-1\right)}{m^2.\left(16+75-6\right)}=\frac{65}{85}=\frac{13}{17}\)
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(\Leftrightarrow x\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow\frac{3}{7}x=\frac{1}{21}\)
\(\Leftrightarrow x=\frac{1}{9}\)