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ta có 1 - 1/3 + 1/3 - 1/5 +1/5 - 1/7 +... +1/n-2 - 1/n=60/61
rút gọn ta còn 1-1/n=60/61
n-1/n=60/61
vậy n= 61
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{223}+\frac{4}{437}+\frac{2}{575}\)
\(=\frac{2}{5.7}+\frac{2}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}\)
\(=\frac{4}{25}\)
P/S: Dấu "." là dấu nhân bạn nhé
Theo đầu bài ta có:
\(\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{25}\)
\(\Leftrightarrow\frac{4}{25}\)
Chuyển vế tất cả số hạng tự do sang phải, ta được \(x=1931\)bạn nhé!
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!
\(C=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)
\(C=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{13}\)
\(C=1-\frac{1}{13}\)
\(C=\frac{12}{13}\)
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{1}{1}-\frac{1}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{1}{1}-\frac{1}{13}\)
\(C=\frac{12}{13}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+2}{x+2}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+1}{x+2}=\frac{332}{323}\)
=>332.(x+2)=323.(x+1)
=>332x+664=323x+323
=>332x-323x=323-664
=>x.(332-323)=-323
=>9x=-323
=>x=-323/9
vậy n=-323/9 .(-323/9+2)=98515/81