Giải
ĐK: $x \neq 0; \dfrac{\sqrt{x^2 + 4356} + x}{x}, x\sqrt{4356} - x^2 \geq 0$
Nhận xét:
$\sqrt{x\left (\sqrt{x^2 + 4356} - x\right )} = \sqrt{\dfrac{4356x}{\sqrt{x^2 + 4356} + x}} = 66\sqrt{\dfrac{x}{\sqrt{x^2 + 4356} + x}}$
Vậy, đặt: $t = \sqrt{\dfrac{\sqrt{x^2 + 4356} + x}{x}} \geq 0$, ta được:
$t - \dfrac{66}{t} = 5 \Leftrightarrow t^2 - 5t - 66 = 0 \Leftrightarrow \left\{\begin{matrix}t = 11\\t = -6\end{matrix}\right.$
Vì $t \geq 0$ nên $t = 11$. Suy ra:
$\dfrac{\sqrt{x^2 + 4356} + x}{x} = 121 \Leftrightarrow \sqrt{x^2 + 4356} = 120x$
$\Leftrightarrow \left\{\begin{matrix}x \geq 0\\14399x^2 = 4356\end{matrix}\right. \Leftrightarrow x = \sqrt{\dfrac{4356}{14399}}$