$\\$
`G = (2a-b-c)/( (a-b)(a-c) ) + (2b-a-c)/((b-c)(b-a)) + (2c-a-b)/((c-a)(c-b)`

`->G = (2a-b-c)/( (a-b)(a-c)) - (2b-a-c)/( (a-b)(b-c)) + (2c-a-b)/((a-c)(b-c))`

`->G = ((b-c)(2a-b-c))/((a-b)(b-c)(a-c)) - ((a-c)(2b-a-c))/((a-b)(b-c)(a-c)) + ((a-b)(2c-a-b))/((a-b)(a-c)(b-c))`

`->G = (2ab - b^2 - bc - 2ac + bc +c^2)/((a-b)(b-c)(a-c)) - (2ab - a^2 - ac - 2bc +ac +c^2)/((a-b)(b-c)(a-c)) + (2ac -a^2 - ab - 2bc + ab + b^2)/((a-b)(a-c)(b-c))`

`->G = (2ab-b^2-bc-2ac +bc+c^2 - 2ab + a^2 +ac+2bc -ac-c^2 +2ac-a^2-ab-2bc+ab+b^2)/((a-b)(a-c)(b-c))`

`->G=0/((a-b)(a-c)(b-c))`

`->G=0`

Vậy `G=0`

$G = \dfrac{2a-b-c}{(a-b)(a-c)} +  \dfrac{2b-a-c}{(b-c)(b-a)} + \dfrac{2c-a-b}{(c-a)(c-b)}$

$= \dfrac{2a-b-c}{(a-b)(a-c)} + \dfrac{a+c-2b}{(b-c)(a-b)} + \dfrac{2c-a-b}{(a-c)(b-c)}$

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

Đặt $a-b=x;\ a-c=y;\ b-c=z$

Ta có$G = \dfrac{2a-b-c}{(a-b)(a-c)} + 

\dfrac{2b-a-c}{(b-c)(b-a)} + \dfrac{2c-a-b}{(c-a)(c-b)}$

$= \dfrac{2a-b-c}{(a-b)(a-c)} + \dfrac{a+c-2b}{(b-c)(a-b)} + \dfrac{2c-a-b}{(a-c)(b-c)}$

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

Đặt $a-b=x;\ a-c=y;\ b-c=z$

Ta có
$ \dfrac{(a-b)+(a-c)}{(a-b)(a-c)} + \dfrac{(a-b)+(c-b)}{(b-c)(a-b)}+ \dfrac{(c-a)+(c-b)}{(a-c)(b-c)} = \dfrac{x+y}{xy}+ \dfrac{x-z}{xz} + \dfrac{-y-z}{yz}$
$= \dfrac{x+y}{xy}+ \dfrac{x-z}{xz} - \dfrac{y+z}{yz}$
$= \dfrac{xz+yz+xy-yz-xy-xz}{xyz} = \dfrac{0}{xyz}=0$
Vậy $G=0$