Đáp án:
Giải thích các bước giải:
`(25+25/2+25/3+...+25/99+1/4)|x-2021|=1+102/2+103/3+...+199/99+200/100`
`<=>(25/1+25/2+25/3+...+25/99+1/4)|x-2021|=101/1+102/2+103/3+...+199/99+200/100 -100`
`<=>[25(1+1/2+1/3+...+1/99)+1/4]|x-2021|=(101/1-1)+(102/2-1)+(103/3-1)+...+(199/99-1)+(200/100-1)`
`<=>[25(1+1/2+1/3+...+1/99)+1/4]|x-2021|=100/1+100/2+100/3+...+100/99+1`
`<=>[25(1+1/2+1/3+...+1/99)+1/4]|x-2021|=100(1+1/2+1/3+...+1/99)+1`
Đặt `1+1/2+1/3+...+1/99=A`
`<=>(25A+1/4)|x-2021|=100A+1`
`<=>|x-2021|=(100A+1)/(25A+1/4)`
`<=>|x-2021|=4`
`<=>` \(\left[ \begin{array}{l}x-2021=4\\x-2021=-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2025\\x=2017\end{array} \right.\)
Vậy `S={2025,2017}`
