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\(\cfrac{1}{(x-1)(x-2)} +\cfrac{1}{(x-2)(x-3)} \) \(+\cfrac{1}{(x-3)(x-4)} =\cfrac{1}{6}\)
বা, \(\cfrac{(x-1)-(x-2)}{(x-1)(x-2)} +\cfrac{(x-2)-(x-3)}{(x-2)(x-3)}\) \(+\cfrac{(x-3)-(x-4)}{(x-3)(x-4)} =\cfrac{1}{6}\)
বা, \(\cfrac{(x-1)}{(x-1)(x-2)} -\cfrac{(x-2)}{(x-1)(x-2)} \)
\(+\cfrac{(x-2)}{(x-2)(x-3)} -\cfrac{(x-3)}{(x-2)(x-3)}\)
\(+\cfrac{(x-3)}{(x-3)(x-4)} -\cfrac{(x-4)}{(x-3)(x-4)} =\cfrac{1}{6}\)
বা, \(\cfrac{1}{(x-2)}-\cfrac{1}{(x-1)}+\cfrac{1}{(x-3)} -\cfrac{1}{(x-2)}\)
\(+\cfrac{1}{(x-4)}-\cfrac{1}{(x-3)}=\cfrac{1}{6}\)
বা, \(-\cfrac{1}{(x-1)}+\cfrac{1}{(x-4)}=\cfrac{1}{6}\)
বা, \(\cfrac{1}{(x-4)}-\cfrac{1}{(x-1)}=\cfrac{1}{6}\)
বা, \(\cfrac{(x-1)-(x-4)}{(x-4)(x-1)} =\cfrac{1}{6}\)
বা, \(\cfrac{x-1-x+4}{x^2-x-4x+4}=\cfrac{1}{6}\)
বা, \(\cfrac{3}{x^2-5x+4}=\cfrac{1}{6}\)
বা,\( x^2-5x+4=18\)
বা,\( x^2-5x+4-18=0\)
বা,\( x^2-5x-14=0\)
বা,\( x^2-(7-2)x-14=0\)
বা,\( x^2-7x+2x-14=0\)
বা, \(x(x-7)+2(x-7)=0\)
বা, \((x-7)(x+2)=0\)

অর্থাৎ,হয় \( (x-7)=0∴ x=7\)
নয়, \((x+2)=0 ∴x=-2\)

∴ নির্ণেয় সমাধান \(x=7\) ও \(x=-2\)