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\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
\(\frac{\frac{4}{3}+\frac{4}{7}-\frac{2}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{4.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{-4}{3}\)
\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
\(A=\frac{2^9-1}{2^9}\)
Vậy \(A=\frac{2^9-1}{2^9}\)
Chúc bạn học tốt ~
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{4}{3}.\frac{99}{202}\)
\(=\frac{66}{101}\)
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\)
\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\)
\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\)
\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)
\(\frac{4}{2\cdot5}\)+\(\frac{4}{5\cdot8}\)+\(\frac{4}{8\cdot11}\)+.......+\(\frac{4}{8\cdot83}\)=\(\frac{4}{3}\) (\(\frac{3}{2\cdot5}\)+\(\frac{3}{5\cdot8}\) +......+\(\frac{3}{80\cdot83}\) )
=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{5}\) +\(\frac{1}{5}\) -\(\frac{1}{8}\) +..........+\(\frac{1}{80}\) -\(\frac{1}{83}\) )
=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{83}\) )
=\(\frac{4}{3}\)*\(\frac{81}{166}\)
=\(\frac{54}{83}\)
Sai đề => Sửa: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{20}\)
\(\Rightarrow\frac{9}{20}\)
Gọi biểu thức đó là A
Ta có: \(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{17.20}\)
\(A:4.3=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(A:4.3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(A:4.3=\frac{1}{2}-\frac{1}{20}\)
\(A:4.3=\frac{9}{20}\)
\(A=\frac{3}{5}\)
k mình nha