Question

Solve for
$x:\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}=\frac{1}{6}$

Answer 1

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$

$\frac{(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)}{(x-1)(x-2)(x-3)(x-4)}=\frac{1}{6}$

$\frac{x^2-7x+12+x^2-5x+4+x^2-3x+2}{(x-1)(x-2)(x-3)(x-4)}\frac{1}{6}$

$\frac{3x^2-15x+18}{(x-1)(x-2)(x-3)(x-4)}=\frac{1}{6}$

$\frac{3(x^2-5x+6)}{(x-1)(x-2)(x-3)(x-4)}=\frac{1}{6}$

$\frac{3(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)}=\frac{1}{6}$

$18=(x-1)(x-4)$

$x^2-5x+4=18$

$x^2-5x-14=0$

$x^2-7x+2x-14=0$

$x(x-7)+2(a-7)$

$x=7$ or $-2$