a. Điều kiện: $x\ge -1$.
Đặt: $u=\sqrt{x+1};v=\sqrt{x^2-x+1}, u,v\ge0$, phương trình trở thành:
$5uv=2(u^2+v^2)$
$\Leftrightarrow (2u-v)(u-2v)=0$
$\Leftrightarrow \left[\begin{array}{l}2u=v\\u=2v\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4(x+1)=x^2-x+1\\x+1=4(x^2-x+1)\end{array}\right.$
$\Leftrightarrow x=\dfrac{5\pm\sqrt{37}}{2}$