A=\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)
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Bài này có 2 cách, nhưng cách nào thì cách cx phải dùng tới máy tính, ai cs cách hay show hộ kham khảo !
Cách 1 : cầm máy tính lên bấm
Cách 2 : \(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{1}{340}\)
\(C=3\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}\right)+\frac{1}{340}\)
\(C=3.\frac{3}{22}+\frac{1}{340}=\frac{9}{22}+\frac{1}{340}=\frac{1541}{3740}\)
S = 1/2-1/5+1/5-1/8+1/8-...-1/20
S = 1/2-1/20
S = 9/20 nha
\(S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
\(A=\frac{50}{101}\)
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)
\(A=\frac{3^2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=3.\frac{9}{20}\)
\(A=\frac{27}{20}\)
k nhá bn!
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{5}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{50}{101}\)
\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{17.20}\)
\(\Rightarrow A=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(A=3\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=3.\frac{9}{20}\)
\(A=\frac{27}{20}\)
a) \(\frac{1}{10}-\frac{1}{40}-\frac{1}{88}-\frac{1}{154}-\frac{1}{238}-\frac{1}{340}\)
\(=\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)
\(\Rightarrow\left[\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\right]\cdot\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\cdot\frac{9}{20}\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\frac{3}{20}\cdot20=\frac{2x}{10}\)
\(\Rightarrow3\cdot20=\frac{2x}{10}\Leftrightarrow60=\frac{2x}{10}\)
=> 2x = 60*10
=> 2x = 600
=> x = 300
\(\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\right).20=\frac{2x}{10}\)
\(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right).20=\frac{2x}{10}\)
\(\left[3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)
\(\left[3.\left(\frac{1}{2}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)
\(\left(3.\frac{9}{20}\right).20=\frac{2x}{10}\)
\(\frac{27}{20}.20=2x\div10\)
\(27=2x\div10\)
\(x=27\times10\div2\)
\(\Rightarrow x=135\)
\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\frac{9}{20}\)
\(A=\frac{3}{20}\)
\(A=\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\)
\(A\times3=\frac{3}{2\times5}+\frac{3}{5\times8}+...+\frac{3}{17\times20}\)
\(A\times3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A\times3=\frac{1}{2}-\frac{1}{20}\)
\(A\times3=\frac{9}{20}\)
\(A=\frac{3}{20}\)
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
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(3A=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
\(A=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)
= \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{20}\right)\)
= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
= 1. \(\frac{9}{20}\)
= \(\frac{9}{20}\)