`(0,2)/(|2x-1|) - 20/11.13 - 20/13.15 - ...- 20/53.55= 3/11`
`(0,2)/(|2x-1|) - (20/11.13 + 20/13.15 +...+ 20/53.55)= 3/11`
Đặt `A= 20/11.13 + 20/13.15 +...+ 20/53.55`
`A=10 ( 2/11.13 + 2/13.15 + ....+ 2/53.55)`
`A= 10 ( 1/11 - 1/13 + 1/13 - 1/15 +...+ 1/53 - 1/55)`
`A= 10 ( 1/11 - 1/55)`
`A= 10. 4/55`
`A= 8/11`
`=> (0,2)/(|2x-1|) - 8/11 = 3/11`
`=> (0,2)/(|2x-1|) = 3/11 + 8/11`
`=> (0,2)/(|2x-1|) = 1`
`=> |2x-1| = 0,2`
`=>` \(\left[ \begin{array}{l}2x-1=0,2\\2x-1=-0,2\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x= 1,2\\2x=0,8\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0,6\\x=0,4\end{array} \right.\)
Vậy `x= 0,6` hoặc `x= 0,4`
