Giải thích các bước giải:

Ta có:

$\dfrac{b-c}{(a-b)(a-c)}+\dfrac{c-a}{(b-a)(b-c)}+\dfrac{a-b}{(c-a)(c-b)}=2024$

$\to \dfrac{c-b}{(a-b)(c-a)}+\dfrac{a-c}{(a-b)(b-c)}+\dfrac{b-a}{(c-a)(b-c)}=2024$

$\to \dfrac{(c-a)+(a-b)}{(a-b)(c-a)}+\dfrac{(a-b)+(b-c)}{(a-b)(b-c)}+\dfrac{(c-a)+(b-c)}{(c-a)(b-c)}=2024$

$\to \dfrac1{a-b}+\dfrac1{c-a}+\dfrac1{b-c}+\dfrac1{a-b}+\dfrac1{b-c}+\dfrac1{c-a}=2024$

$\to 2(\dfrac1{a-b}+\dfrac1{b-c}+\dfrac1{c-a})=2024$

$\to 2Q=2024$

$\to Q=1012$