Question

1/(x-1)(x-2) + 1/(x-2)(x-3) + 1/(x-3)(x-4) =1/6

Answer 1

$1/((x - 1)(x - 2)) + 1/((x - 2)(x - 3)) + 1/((x - 3)(x - 4)) = 1/6$ 
[(x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2)] /((x - 1)(x - 2)(x - 3)(x - 4)) = 1/6
(x^2 - 7x + 12 + x^2 - 5x + 4 + x^2 - 3x + 2)/((x - 1)(x - 2)(x - 3)(x - 4)) = 1/6
$(3x^2 - 15x + 18) /((x - 1)(x - 2)(x - 3)(x - 4)) = 1/6$ 
$[3(x - 3)(x - 2)] /((x - 1)(x - 2)(x - 3)(x - 4)) = 1/6$ 
$3 /((x - 1)(x - 4)) = 1/6$ 
$x^2 - 5x + 4 = 18$  
$x^2 - 5x - 14 = 0$  
$(x + 2)(x - 7) = 0$  
$x = -2 , 7$