Giải thích các bước giải:
Ta có:
$A+B+C=\dfrac{x-y}{1+xy}+\dfrac{y-z}{1+yz}+\dfrac{z-x}{1+zx}$
$\to A+B+C=\dfrac{x-y}{1+xy}+\dfrac{y-z}{1+yz}+\dfrac{z-y+y-x}{1+zx}$
$\to A+B+C=\dfrac{x-y}{1+xy}+\dfrac{y-z}{1+yz}+\dfrac{z-y}{1+zx}+\dfrac{y-x}{1+zx}$
$\to A+B+C=\dfrac{x-y}{1+xy}+\dfrac{y-z}{1+yz}-\dfrac{y-z}{1+zx}-\dfrac{x-y}{1+zx}$
$\to A+B+C=(\dfrac{x-y}{1+xy}-\dfrac{x-y}{1+zx})+(\dfrac{y-z}{1+yz}-\dfrac{y-z}{1+zx})$
$\to A+B+C=(x-y)(\dfrac{1}{1+xy}-\dfrac{1}{1+zx})+(y-z)(\dfrac{1}{1+yz}-\dfrac{1}{1+zx})$
$\to A+B+C=(x-y).\dfrac{1+zx-(1+xy)}{(1+xy)(1+zx)}+(y-z).\dfrac{1+zx-(1+yz)}{(1+yz)(1+zx)}$
$\to A+B+C=(x-y).\dfrac{zx-xy}{(1+xy)(1+zx)}+(y-z).\dfrac{zx-yz}{(1+yz)(1+zx)}$
$\to A+B+C=(x-y).\dfrac{x(z-y)}{(1+xy)(1+zx)}+(y-z).\dfrac{z(x-y)}{(1+yz)(1+zx)}$
$\to A+B+C=\dfrac{(x-y)(y-z)}{(1+xy)(1+yz)(1+zx)}(-x(1+yz)+z(1+xy))$
$\to A+B+C=\dfrac{(x-y)(y-z)}{(1+xy)(1+yz)(1+zx)}(-x-xyz+z+xyz)$
$\to A+B+C=\dfrac{(x-y)(y-z)}{(1+xy)(1+yz)(1+zx)}(z-x)$
$\to A+B+C=ABC$
