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

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

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

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

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

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

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=\(\frac{-125}{627}\)

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)

\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)

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

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

1 , ủng hộ mk nha

Vân Anh Ta ơi cậu phải giải thích ra tại sao lại bằng một thì mk mới ủng hộ đc chứ

=> x- (\(\frac{20}{11.13}\) + \(\frac{20}{13.15}\) +...+ \(\frac{20}{53.55}\)) = \(\frac{3}{11}\)

=> x - 10.(\(\frac{2}{11.13}\) + \(\frac{2}{13.15}\) +...+ \(\frac{2}{53.55}\)) = \(\frac{3}{11}\)

=> x - 10.( \(\frac{1}{11}\) - \(\frac{1}{13}\) + \(\frac{1}{13}\) -  \(\frac{1}{15}\) +...+ \(\frac{1}{53}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)

=> x - 10. (\(\frac{1}{11}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)

=> x - 10.\(\frac{4}{55}\) = \(\frac{3}{11}\) 

=> x - \(\frac{8}{11}\) = \(\frac{3}{11}\)=> x=1 Vậy x=1

giải được xin cảm ơn

thank you

bạn vào đây nhé:olm.vn/hoi-dap/question/53723.html

Nguyên Thị Nami có thể vào để hỏi bài !

\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-......-\frac{20}{53\cdot55}=\frac{3}{11}\)

\(\Leftrightarrow x-10\left(\frac{2}{11\cdot13}-\frac{2}{13\cdot15}-\frac{2}{15\cdot17}-.....-\frac{2}{53\cdot55}\right)=\frac{3}{11}\)

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

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

\(\Rightarrow x=1\)