Question

solve: $\frac{1}{(x-1)(x+2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3}$

Answer 1

$\begin{array}{c}\frac{1}{(x-1)(x+2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3}\\ \frac{(x-2)(x-3)+(x+2)(x-1)}{(x-1)(x-2)(x+2)(x-3)}=\frac{2}{3}\\ \frac{x^2-3x-2x+6+x^2-2-x-2}{(x^2-x-4)(x^2-5x+6)}=\frac{2}{3}\\ 6x^2-18x+6=2(x^2-x-4)(x^2-5x+6)\\ 6x^2-18x+6=2(x^4-5x^3+6x^2-x^3+5x^2-6x-4x^2+20x-24)\\ 6x^2-18x+6=2(x^4-6x^3+7x^2+14x-24)\\ 6x^2-18x+6=2x^4-12x^3+14x^2+28x-48\\ 2x^4-12x^3+14x^2+28x-48-6x^2+18x-6=0\\ 2x^4-12x^3+8x^2+38x-54=0\end{array}$