\(ĐK:x\ge0\\ \Leftrightarrow\sqrt{\left(x\sqrt{5}-1\right)^2}-\sqrt{\left(2x+\sqrt{5}\right)^2}=0\\ \Leftrightarrow\left(x\sqrt{5}-1\right)-\left(2x+\sqrt{5}\right)=0\\ \Leftrightarrow x\sqrt{5}-2x=1+\sqrt{5}\\ \Leftrightarrow x\left(\sqrt{5}-2\right)=1+\sqrt{5}\\ \Leftrightarrow x=\dfrac{1+\sqrt{5}}{\sqrt{5}-2}=\dfrac{\left(1+\sqrt{5}\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{7+3\sqrt{5}}{3}\left(tm\right)\)
Lời giải:
PT $\Leftrightarrow \sqrt{(x\sqrt{5}-1)^2}-\sqrt{(2x+\sqrt{5})^2}=0$
$\Leftrightarrow |x\sqrt{5}-1|=|2x+\sqrt{5}|$
\(\Rightarrow \left[\begin{matrix} x\sqrt{5}-1=2x+\sqrt{5}\\ x\sqrt{5}-1=-2x-\sqrt{5}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=7+3\sqrt{5}\\ x=-7+3\sqrt{5}\end{matrix}\right.\)