Đáp án:
$\dfrac{6}{x(x-6)}$
Giải thích các bước giải:
$\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}$
$=\dfrac{1}{x(x-1)}+\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}+\dfrac{1}{(x-3)(x-4)}+\dfrac{1}{(x-4)(x-5)}+\dfrac{1}{(x-5)(x-6)}$
$=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}$
$=-\dfrac{1}{x}+\dfrac{1}{x-6}$
$=\dfrac{6-x+x}{x(x-6)}=\dfrac{6}{x(x-6)}$
