Question

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

Answer 1

$\begin{array}{c}\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}=\frac{1}{6}\\ \Rightarrow \frac{\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)+\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=\frac{1}{6}\\ \Rightarrow 6\left(x^2-7x+12+x^2-5x+4+x^2-3x+2\right)-\left(x^2-3x+2\right)\left(x^2-7x+12\right)\\ \Rightarrow 6\left(3x^2-15x+18\right)=\left(x^4-10x^3-35x^2-50x+24\right)\\ Divisor=x^2-5x+6\\ Quotient=x^2-5x+4\\ Remainde=0\\ \Rightarrow 18=x^2-5x+4\\ \Rightarrow x^2-5x-14=0\\ \Rightarrow x^2+2x-7x-14=0\\ \Rightarrow x\left(x+2\right)-7\left(x+2\right)=0\\ \Rightarrow \left(x-7\right)\left(x+2\right)=0\\ x=7or-2\end{array}$