Ta có: `(1)/(4).(2)/(6).(3)/(8).. . ..(2022)/(4046).(2023)/(4048)=(4^x)/(253)`

$\rightarrow$ `(1)/(2.2).(2)/(2.3).(3)/(2.4).. . ..(2022)/(2.2023).(2023)/(2.2024)=(4^x)/(253)`

$\rightarrow$ `(1.2.3.. . ..2022.2023)/(2.2.2.3.2.4.. . ..2.2023.2.2024)=(4^x)/(253)`

$\rightarrow$ `(1.2.3.. . ..2022.2023)/(2^{2023}.2.3.4.. . ..2023.2024)=(4^x)/(253)`

$\rightarrow$ `(1)/(2^{2023}.2024)=(4^x)/(253)`

$\rightarrow$ `(253)/(2^{2023}.2024}=4^x`

$\rightarrow$ `(1)/(2^{2023}.8}=4^x`

$\rightarrow$ `4^x=(1)/(2^{2026})`

$\rightarrow$ `4^x=(1)/(4^{1013})`

$\rightarrow$ $x=-1013$