Solve the following quadratic equations by factorization:

We have been given,

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

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

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

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

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

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

$x(x-7)+2(x-7)=0$

$(x+2)(x-7)=0$

Therefore,

$x+2=0$

$x=-2$

or,

$x-7=0$

$x=7$

Hence, $x=-2$ or $x=7$.