Ta có $y'=\frac{(x-1)'.\sqrt{3x^2+6x+7}-(\sqrt{3x^2+6x+7})'.(x-1)}{(\sqrt{3x^2+6x+7})^2}$
$=\frac{\sqrt{3x^2+6x+7}-\frac{6(x-1)(x+1)}{2\sqrt{3x^2+6x+7}}}{3x^2+6x+7}$
$=\frac{\sqrt{3x^2+6x+7}-\frac{3x^2-3}{\sqrt{3x^2+6x+7}}}{3x^2+6x+7}=\frac{\frac{3x^2+6x+7-3x^2+3}{\sqrt{3x^2+6x+7}}}{3x^2+6x+7}=\frac{\frac{6x+10}{\sqrt{3x^2+6x+7}}}{3x^2+6x+7}=\frac{6x+10}{(3x^2+6x+7)\sqrt{3x^2+6x+7}}=\frac{6x+10}{\sqrt{(3x^2+6x+7)^3}}$