Ta có :
`(25 + 25/2 +25/3 + ... + 25/99 + 1/4)|x-2021| =1 + 102/2 + 103/3 + ... + 199/99 +200/100 `
`-> (25+25/2+25/3 + ... +25/99 +1/4)|x-2021| = 1 + 100/2 + 1 + 100/3 + ... + 100/100 +1`
`-> 25( 1 +1/2 +1/3 + ... +1/100 )|x-2021| = (1 + 1 + 1 + .... + 1) + (100/2 +100/3 + ...+100/100 )`
`-> 25(1+1/2+1/3+ ... +1/100)|x-2021| =100(1 + 1/2 + 1/3 + ... +1/100 )`
`-> |x-2021| = {100(1 + 1/2 + 1/3 + ... +1/100 )}/{25(1+1/2+1/3+ ... +1/100)}`
`-> |x-2021| = 100/25 =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 `x \in {2025 ; 2017}`
