`(x - 2)/13 + (x - 3)/12 + (x - 4)/11 + (x - 5)/10 + (x - 30)/5 = 1`
`=> (x - 2)/13 + (x - 3)/12 + (x - 4)/11 + (x - 5)/10 + (x - 30)/5 - 1 = 1 - 1`
`=> (x - 2)/13 - 1 + (x - 3)/12 - 1 + (x - 4)/11 - 1 + (x - 5)/10 - 1 + (x - 30)/5 + 3 = 0`
`=> (x - 15)/13 + (x - 15)/12 + (x - 15)/11 + (x - 15)/10 + (x - 30)/5 + 3 = 0`
`=> (x - 15)/13 + (x - 15)/12 + (x - 15)/11 + (x - 15)/10 + (x - 15)/15 = 0`
`=> (x - 15) . (1/13 + 1/12 + 1/11 + 1/10 + 1/15) = 0`
`=>` \(\left[ \begin{array}{l}x - 15 = 0\\\dfrac{1}{13} + \dfrac{1}{12} + \dfrac{1}{11} + \dfrac{1}{10} +\dfrac{1}{15} = 0 \end{array} \right.\)
Mà `1/13 + 1/12 + 1/11 + 1/10 + 1/15 \ne 0`
`=> x - 15 = 0`
`=> x = 0 + 15`
`=> x = 15`
Vậy `x = 15`
