`\text{a)}`
`(x-1)^2023 = (x-1)^2021`
`-> (x-1)^2023 - (x-1)^2021 = 0`
`-> (x-1)^2021[ (x-1)^2 - 1 ] = 0`
`->` \(\left[ \begin{array}{l}(x-1)^{2021} =0\\(x-1)^2-1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=1\\x=0\\x=2\end{array} \right.\)
Vậy `x \in {1 ; 0 ; 2}`
$\\$
`\text{b)}`
`x - 5 - 1/3 = 3 . x/4 + 1/5`
`-> x - 5 - 1/3 = {3x}/4 + 1/5`
`-> x - {3x}/4 = 1/5 +1/3 +5`
`-> {4x}/{4} - {3x}/4 = 83/15`
`-> x/4 = 83/15`
`-> x = 332/15`
Vậy `x =332/15`
