Đáp án: $x\in\{-5,11\}$
Giải thích các bước giải:
ĐKXĐ: $x\ne 1,2,3,4,5$
Ta có:
$\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}=\dfrac{1}{15}$
$\to\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}+\dfrac{1}{(x-3)(x-4)}+\dfrac{1}{(x-4)(x-5)}=\dfrac{1}{15}$
$\to\dfrac{(x-1)-(x-2)}{(x-1)(x-2)}+\dfrac{(x-2)-(x-3)}{(x-2)(x-3)}+\dfrac{(x-3)-(x-4)}{(x-3)(x-4)}+\dfrac{(x-4)-(x-5)}{(x-4)(x-5)}=\dfrac{1}{15}$
$\to -\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}$
$\to -\dfrac{1}{x-1}+\dfrac{1}{x-5}=\dfrac{1}{15}$
$\to \dfrac{-(x-5)+(x-1)}{(x-1)(x-5)}=\dfrac{1}{15}$
$\to \dfrac{4}{x^2-6x+5}=\dfrac{1}{15}$
$\to x^2-6x+5=60$
$\to x^2-6x-55=0$
$\to (x-11)(x+5)=0$
$\to x\in\{-5,11\}$
