[tex]pt\leftrightarrow (x+1).\dfrac{2x+1}{\sqrt{x^2+6}+\sqrt{x^2+2x+7}}+(2x+1)\sqrt{x^2+6}=13(2x+1)[/tex]
[tex]\leftrightarrow \left\{\begin{matrix} x=\dfrac{-1}{2} & \\ 5(\dfrac{x+1}{\sqrt{x^2+6}+\sqrt{x^2+2x+7}}+\sqrt{x^2+6})=13 (*)& \end{matrix}\right.[/tex]
[tex](*)\leftrightarrow 5(\dfrac{(x^2+2x+7)-(x^2+6)-x}{\sqrt{x^2+6}+\sqrt{x^2+2x+7}}+\sqrt{x^2+6})=13[/tex]
[tex]\leftrightarrow 5(\sqrt{x^2+2x+7}-\dfrac{x}{\sqrt{x^2+6}+\sqrt{x^2+2x+7}})=13[/tex]
[tex]\leftrightarrow 5(\dfrac{1}{\sqrt{x^2+6}+\sqrt{x^2+2x+7}}+\sqrt{x^2+6}+\sqrt{x^2+2x+7})=26[/tex] [tex]\rightarrow \sqrt{x^2+6}+\sqrt{x^2+2x+7}=5[/tex] hoặc [tex]=\dfrac{1}{5}[/tex]
Giải ra vẫn được $x=\dfrac{-1}{2}$