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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
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(3A=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
Đặt A=\(\frac{9}{10}+\frac{39}{40}+...+\frac{1119}{1120}\)
=>A=\(\frac{10-1}{10}+\frac{40-1}{40}+...+\frac{1120-1}{1120}\)
=>A=\(1-\frac{1}{10}+1-\frac{1}{40}+...+1-\frac{1}{1120}\)
=>A=\(11-\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\right)\)
Đặt B=\(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\)
=>3B=\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{1120}\)
=>3B=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)
=>3B=\(\frac{33}{70}\)
=>B=\(\frac{11}{70}\)
=>A=11-\(\frac{11}{70}\)
=>A=\(\frac{759}{70}\)
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\(\frac{1}{10}\)+\(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)
=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)
=\(\frac{1}{3}\)(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))
=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+\(\frac{1}{17}\)-\(\frac{1}{20}\))
=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{20}\))
=\(\frac{1}{3}\).\(\frac{9}{20}\)
=\(\frac{3}{20}\)
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
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.16}+\frac{1}{16.20}\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.16}+\frac{3}{16.20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(A=\frac{1}{3}.\frac{9}{20}\)
\(A=\frac{3}{20}\)
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=\(\frac{3}{20}\)
bạn ơi tách ra thừa số chung rồi làm như bình thường nha
1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)