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2 tháng 3 2019

\(x-\dfrac{20}{11.13}-\dfrac{20}{23.15}-....-\dfrac{20}{53.55}=\dfrac{3}{11}\)

\(x-\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+....+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)

\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(x-\dfrac{8}{11}=\dfrac{3}{11}\)

=> \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)

20 tháng 5 2019

Ta có:

x-\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)=\(\frac{3}{11}\)

\(\Rightarrow\)\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)= x-\(\frac{3}{11}\)

\(\frac{1}{10}\).\((\frac{20}{11.13}.\frac{20}{13.15}.\frac{20}{15.17}...\frac{20}{53.57})\)= x-\(\frac{3}{11}\)

\(\frac{2}{11.13}\).\(\frac{2}{13.15}\).\(\frac{2}{15.17}\)...\(\frac{2}{53.57}\)= x-\(\frac{3}{11}\)

\(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}\)-\(\frac{1}{17}\)+...+\(\frac{1}{53}\)-\(\frac{1}{57}\)=x-\(\frac{3}{11}\)

\(\frac{1}{11}-\frac{1}{57}\)=x-\(\frac{3}{11}\)

\(\frac{46}{627}\)=x-\(\frac{3}{11}\)

x=\(\frac{46}{627}\)-\(\frac{3}{11}\)

Vậy x=\(\frac{-125}{627}\)