Giải thích các bước giải:
$\frac{1}{x^2-7x+12}+$ $\frac{-1}{x^2-6x+8}+$ $\frac{1}{x^2-5x+6}$
=$\frac{1}{x^2-3x-4x+12}-$ $\frac{1}{x^2-4x-2x+8}+$ $\frac{1}{x^2-2x-3x+6}$
=$\frac{1}{x(x-3)-4(x-3)}-$ $\frac{1}{x(x-4)-2(x-4}+$ $\frac{1}{x(x-2)-3(x-2)}$
=$\frac{1}{(x-4)(x-3)}-$ $\frac{1}{(x-2)(x-4)}+$ $\frac{1}{(x-3)(x-2)}$
=$\frac{x-2}{(x-2)(x-3)(x-4)}-$ $\frac{x-3}{(x-2)(x-3)(x-4)}$+$\frac{x-4}{(x-2)(x-3)(x-4)}$
=$\frac{x-2-x+3+x-4}{(x-2)(x-3)(x-4)}$
= $\frac{x-3}{(x-2)(x-3)(x-4)}$
= $\frac{1}{(x-2)(x-4)}$