$\Leftrightarrow \frac{x-a}{b+c}-1+\frac{x-b}{c+a}-1+\frac{x-c}{b+a}-1=\frac{3x}{a+b+c}-3$
$\Leftrightarrow \frac{x-a-b-c}{b+c}+\frac{x-a-b-c}{a+c}+\frac{x-a-b-c}{a+b}=3(\frac{x}{a+b+c}-1)$
$\Leftrightarrow (x-a-b-c)(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=3(\frac{x-a-b-c}{a+b+c})$
Đến đây thì dễ rồi nhé!