$A=[ab(a-b)+bc(b-c)+ca(c-a)][\dfrac{1}{ab(a-b)}+\dfrac{1}{bc(b-c)}+\dfrac{1}{ca(c-a)}]$
Ta có $ab(a-b)+bc(b-c)+ca(c-a)$
$=ab(a-b)+bc(b-a+a-c)+ca(c-a)$
$=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)$
$=b(a-b)(a-c)-c(a-b)(a-c)$
$=(a-b)(a-c)(b-c)$
$\Rightarrow A=(a-b)(b-c)(a-c)[\dfrac{1}{ab(a-b)}+\dfrac{1}{bc(b-c)}+\dfrac{1}{ca(c-a)}]\\ \Rightarrow A=\dfrac{(b-c)(a-c)}{ab}+\dfrac{(a-b)(a-c)}{bc}-\dfrac{(a-b)(b-c)}{ca}$
Ta có:
$\dfrac{(b-c)(a-c)}{ab}=\dfrac{ab-bc-ca-c^2}{ab}=\dfrac{ab-c(a+b)-c^2}{ab}=\dfrac{ab-c(-c)-c^2}{ab}=\dfrac{ab}{ab}=1$
Chứng minh tương tự: $\dfrac{{a-b}{a-c}}{bc}=1;\dfrac{(a-b)(b-c)}{ca}=1$
Cộng vế theo vế
$\Rightarrow A=1+1-1=1$
