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Đặt \(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
=> \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\) (dấu . có nghĩa là nhân)
=> \(3A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Đây là kiến thức lớp 6 nhá =)) bạn mà có chỗ nào ko hiểu thì hỏi ng thầy cô giạy bạn ý
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\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)
\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)
\(=\frac{1}{2}-\frac{1}{23}\)
\(=\frac{21}{46}\)
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\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)
vậy S>0,1
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S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)
S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)
S = \(\frac{3}{20}\)
S = 0,15 > 0,1
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Ta có:\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{340}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)
= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{17}-\frac{1}{20}\right)=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
\(\frac{2}{10}+\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(=\frac{2}{3}\cdot\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)
\(=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\right)\)
\(=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{17}\right)\)
\(=\frac{2}{3}\cdot\frac{15}{34}=\frac{2.15}{3.34}=\frac{1.5}{1.17}=\frac{5}{17}\)
Chú ý nè: Dấu "." là dấu "x" đó!
\(\frac{2}{10}+\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}=\frac{1}{4}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}=\frac{3}{11}+\frac{2}{154}+\frac{2}{238}=\frac{2}{7}+\frac{2}{238}=\frac{5}{17}\)