Câu hỏi:

Ta có \(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=\frac{2021}{1}+\frac{2020}{2}+\frac{2019}{3}+...+\frac{1}{2021}\).

\(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=2021+\frac{2020}{2}+\frac{2019}{3}+...+\frac{1}{2021}\)

\(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=\left( 1+\frac{2020}{2} \right)+\left( 1+\frac{2019}{3} \right)+\left( 1+\frac{2018}{4} \right)+...+\left( 1+\frac{1}{2021} \right)+1\)

\(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=\left( 1+\frac{2020}{2} \right)+\left( 1+\frac{2019}{3} \right)+\left( 1+\frac{2018}{4} \right)+...+\left( 1+\frac{1}{2021} \right)+\frac{2022}{2022}\)

\(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=\frac{2022}{2}+\frac{2022}{3}+\frac{2022}{4}+...+\frac{2022}{2021}+\frac{2022}{2022}\)

\(\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)x=2022\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2022} \right)\)

\(x=2022\)

\(\frac{2}{7}+\left( -\frac{3}{5} \right)+\frac{3}{5}=\frac{2}{7}+\left( \frac{3}{5}-\frac{3}{5} \right)=\frac{2}{7}.\)

Tốc độ của vận động viên này là: \(100:12,5=8\) (m/s).

Tia Ok là tia phân giác của \(\widehat{xOm}\).

Tia Ou là tia phân giác của \(\widehat{nOy}\).

\({{(\frac{2}{5})}^{n}}=\frac{8}{125}\)

\({{(\frac{2}{5})}^{n}}={{(\frac{2}{5})}^{3}}\)

Vậy \(n=3\).