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\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\frac{9}{20}\)
\(A=\frac{3}{20}\)
\(A=\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\)
\(A\times3=\frac{3}{2\times5}+\frac{3}{5\times8}+...+\frac{3}{17\times20}\)
\(A\times3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A\times3=\frac{1}{2}-\frac{1}{20}\)
\(A\times3=\frac{9}{20}\)
\(A=\frac{3}{20}\)
Bạn tách mẫu số ra kiểu 2 x 5
5 x 8
........
Cứ như thế
Sau đó rút gọn
Thực hiện một phép tính nữa
Vậy là ra kết quả
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(3A=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
\(\frac{1}{10}\)+\(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)
=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)
=\(\frac{1}{3}\)(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))
=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+\(\frac{1}{17}\)-\(\frac{1}{20}\))
=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{20}\))
=\(\frac{1}{3}\).\(\frac{9}{20}\)
=\(\frac{3}{20}\)
Ta có: S = 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340
=> S = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 +1/14.17 +1/17.20
Nhân 2 vế với 3 và áp dụng công thức tách 1 phân số thành hiệu 2 phân số: x/n.(n + x) = 1/n - 1/(n + x)
=> 3.S = 3.(1/2.5 + 1/5.8 + 1/8.11 +1/11.14 +1/14.17 +1/17.20)
=> 3.S = 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 +3/14.17 +3/17.20
=> 3.S = 1/2 - 1/ 5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20
=> 3.S = 1/2 - 1/20
=> 3.S = 9/20
=> S = 3/20
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Ta có:
\(Coi\) \(A=\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\right).x=1\frac{1}{5}\)
\(=\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right).x=\frac{6}{5}\)
\(\Rightarrow3A=\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right).x=\frac{6}{5}.3=\frac{18}{5}\) \(=\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right).x\)
\(=\left(\frac{1}{2}-\frac{1}{20}\right).x\)
\(=\frac{9}{20}.x=\frac{18}{5}\)
\(\Rightarrow x=\frac{18}{5}:\frac{9}{20}=8\)
Vậy \(x=8\).
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.16}+\frac{1}{16.20}\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.16}+\frac{3}{16.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\frac{9}{20}\)
\(A=\frac{3}{20}\)
a, \(A=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{340}\)
\(\Leftrightarrow A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{17.20}\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{17.20}\right)\)
\(\Leftrightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{6}-\frac{1}{60}=\frac{3}{20}\)
b, \(2004^{10}+2004^9=2004^9\left(2014+1\right)=2014^9+2005\)
\(2015^{10}=2015^9.2015\)
-Vậy: \(2004^{10}+2004^9< 2005^{10}\)
Sửa đề chút : \(\frac{1}{138}\) thành \(\frac{1}{238}\)
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\frac{9}{20}\)
\(=\frac{3}{20}\)
Ukm Nuzi Sửa đề như này mới làm được : \(\frac{1}{138}\) thành \(\frac{1}{238}\)
\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\cdot\frac{9}{20}\)
\(=\frac{3}{20}\)
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= \(\frac{1}{3}\).(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))
S= \(\frac{1}{3}\).(\(\frac{1}{2}\)-\(\frac{1}{20}\))
S= \(\frac{1}{3}\).\(\frac{9}{20}\)
S=\(\frac{3}{20}\)
chắc chắn nhé