Rất dễ
Cho $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=A$
Ta được:
$A.\dfrac{1}{b-c}=\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)(b-c)}+\dfrac{c}{(b-c)(a-b)}=
\dfrac{a}{(b-c)^2}+\dfrac{ab-b^2}{(a-b)(b-c)(c-a)}+\dfrac{c^2-ca}{(a-b)(b-c)(c-a)}=0$
$A.\dfrac{1}{c-a}=\dfrac{a}{(b-c)(c-a)}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(c-a)(a-b)}=
\dfrac{b}{(c-a)^2}+\dfrac{a^2-ab}{(a-b)(b-c)(c-a)}+\dfrac{bc-c^2}{(a-b)(b-c)
(c-a)}=0$
$A.\dfrac{1}{a-b}=\dfrac{a}{(b-c)(a-b)}+\dfrac{b}{(c-a)(a-b)}+\dfrac{c}{(a-b)^2}=
\dfrac{c}{(a-b)^2}+\dfrac{ac-a^2}{(a-b)(b-c)(c-a)}+\dfrac{b^2-bc}{(a-b)(b-c)
(c-a)}=0$
Cộng 3 vế:
$(\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2})+\dfrac{ab-b^2+c^2-
ca+a^2-ab+bc-c^2+ac-a^2+b^2-bc}{(a-b)(b-c)(c-a)}=0$
$\leftrightarrow \dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}+0=0$
$\leftrightarrow \dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}
{(a-b)^2}=0(dpcm)$
P/S:Đánh toát cả mồ hôi