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Cho \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{47.49}=\frac{1}{x}\). Tìm |x|
\(1\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)
\(1\frac{1}{3}+\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}.\frac{46}{147}=\frac{1}{x}\)
\(\frac{4}{3}+\frac{23}{147}=\frac{1}{x}\)
\(\frac{73}{49}=\frac{1}{x}\)
=>\(x=\frac{49.1}{73}=\frac{49}{73}\Rightarrow\)I x I= \(\frac{49}{73}\)
\(1\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)
\(1\frac{1}{3}+\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{4}{3}+\frac{1}{2}.\frac{46}{147}=\frac{1}{x}\)
\(\frac{4}{3}+\frac{23}{147}=\frac{1}{x}\)
\(\frac{73}{49}=\frac{1}{x}\)
=>\(x=\frac{49.1}{73}=\frac{49}{73}\Rightarrow\)I x I= \(\frac{49}{73}\)