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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
Ta có :
\(N=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(N=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3N=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3N=\frac{1}{2}-\frac{1}{20}\)
\(3N=\frac{9}{20}\)
\(N=\frac{9}{20}:3\)
\(N=\frac{3}{20}\)
Vậy \(N=\frac{3}{20}\)
Chúc bạn học tốt ~
\(N=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{238}+\frac{1}{340}\)
\(N=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{14.17}+\frac{1}{17.20}\)
\(N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(N=\frac{1}{2}-\frac{1}{20}\)
\(N=\frac{10}{20}-\frac{1}{20}\)
\(N=\frac{9}{20}\)
Ta có:\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{340}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)
= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{17}-\frac{1}{20}\right)=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)
vậy S>0,1
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
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
8,112