$\sqrt{(x-3)^2+(x+3)^2}-2\sqrt2=\sqrt{(x-5)^2+(x+5)^2}-4\sqrt2$
$\Leftrightarrow \sqrt{2(x^2+9)}-2\sqrt2=\sqrt{2(x^2+25)}-4\sqrt2$
$\Leftrightarrow \sqrt{x^2+9}-2=\sqrt{x^2+25}-4$
$\Leftrightarrow \sqrt{x^2+25}-\sqrt{x^2+9}=2 (1)$
Chú ý rằng $\sqrt{x^2+25}-\sqrt{x^2+9}=\frac{x^2+25-x^2-9}{\sqrt{x^2+25}+\sqrt{x^2+9}}=\frac{16}{\sqrt{x^2+25}+\sqrt{x^2+9}}$
Suy ra $\sqrt{x^2+25}+\sqrt{x^2+9}=8 (2)$
$(1) , (2)\Rightarrow \sqrt{x^2+25}=5 , \sqrt{x^2+9}=3$
Vậy $x=0$