Phương trình đã cho tương đương với:
$7x^2+5\sqrt2x+7=5\sqrt{x^4+2x^2+1-2x^2}$
$\Leftrightarrow 7x^2+5\sqrt2x+7=5\sqrt{(x^2+1)^2-2x^2}$
$\Leftrightarrow 6(x^2+\sqrt2x+1)+(x^2-\sqrt2x+1)=5\sqrt{(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)}$
$\Leftrightarrow [2\sqrt{x^2+\sqrt2x+1}-\sqrt{x^2-\sqrt2x+1}][3\sqrt{x^2+\sqrt2x+1}-\sqrt{x^2-\sqrt2x+1}]=0$
$\Leftrightarrow \left[\begin{array}{l}2\sqrt{x^2+\sqrt2x+1}=\sqrt{x^2-\sqrt2x+1}\\3\sqrt{x^2+\sqrt2x+1}=\sqrt{x^2-\sqrt2x+1}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4(x^2+\sqrt2x+1)=x^2-\sqrt2x+1\\9(x^2+\sqrt2x+1)=x^2-\sqrt2x+1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=-\dfrac{5+\sqrt7}{3\sqrt2}\\x=\dfrac{\sqrt7-5}{3\sqrt2}\end{array}\right.$