Giải phương trình sau: 

- Hướng dẫn giải

Điều kiện: $x\in ℝ$

$\begin{array}{l}\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\\ \iff \frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{2x\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\sqrt{5}+1\\ \iff \frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\\ \iff \frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{2x\left(\sqrt{3}-1\right)}{2}=\sqrt{5}+1\\ \iff x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\\ \iff x\sqrt[]{5}+x\sqrt{3}-x\sqrt{3}+x=\sqrt{5}+1\\ \iff x\sqrt[]{5}+x=\sqrt{5}+1\\ \iff x\left(\sqrt{5}+1\right)=\sqrt{5}+1\\ \iff x=1\end{array}$

Vậy phương trình có nghiệm x=1