\(\sqrt{x^2+2x+2}+\sqrt{x^2-2x+5}=\sqrt{13}\)
\(pt\Leftrightarrow\sqrt{x^2+2x+2}-\dfrac{\sqrt{13}}{3}+\sqrt{x^2-2x+5}-\dfrac{2\sqrt{13}}{3}=0\)
\(\Leftrightarrow\dfrac{x^2+2x+2-\dfrac{13}{9}}{\sqrt{x^2+2x+2}+\dfrac{\sqrt{13}}{3}}+\dfrac{x^2-2x+5-\dfrac{52}{9}}{\sqrt{x^2-2x+5}+\dfrac{2\sqrt{13}}{3}}=0\)
\(\Leftrightarrow\dfrac{\dfrac{\left(3x+1\right)\left(3x+5\right)}{9}}{\sqrt{x^2+2x+2}+\dfrac{\sqrt{13}}{3}}+\dfrac{\dfrac{\left(3x+1\right)\left(3x-7\right)}{9}}{\sqrt{x^2-2x+5}+\dfrac{2\sqrt{13}}{3}}=0\)
\(\Leftrightarrow\dfrac{\left(3x+1\right)}{9}\left(\dfrac{3x+5}{\sqrt{x^2+2x+2}+\dfrac{\sqrt{13}}{3}}+\dfrac{3x-7}{\sqrt{x^2-2x+5}+\dfrac{2\sqrt{13}}{3}}\right)=0\)
Dễ thấy: \(\dfrac{3x+5}{\sqrt{x^2+2x+2}+\dfrac{\sqrt{13}}{3}}+\dfrac{3x-7}{\sqrt{x^2-2x+5}+\dfrac{2\sqrt{13}}{3}}>0\)
\(\Rightarrow3x+1=0\Rightarrow x=-\dfrac{1}{3}\)
