A. \(\left[\begin{array}{l} x_{1}=\frac{\sqrt{6}+\sqrt{2}+2 \sqrt{5-\sqrt{3}}}{2} \\ x_{2}=\frac{\sqrt{6}-\sqrt{2}-2 \sqrt{5-\sqrt{3}}}{2} \end{array}\right.\)
B. \(\left[\begin{array}{l} x_{1}=\frac{-\sqrt{6}-\sqrt{2}+2 \sqrt{5-\sqrt{3}}}{2} \\ x_{2}=\frac{-\sqrt{6}-\sqrt{2}-2 \sqrt{5-\sqrt{3}}}{2} \end{array}\right.\)
C. \(\left[\begin{array}{l} x_{1}=\frac{\sqrt{6}-\sqrt{2}+2 \sqrt{5-\sqrt{3}}}{2} \\ x_{2}=\frac{\sqrt{6}-\sqrt{2}-2 \sqrt{5-\sqrt{3}}}{2} \end{array}\right.\)
D. Vô nghiệm.
C
Ta có
\(\begin{array}{l} \sqrt{2} x^{2}-2(\sqrt{3}-1) x-3 \sqrt{2}=0 \\ \Delta=[-2(\sqrt{3}-1)]^{2}-4 \cdot \sqrt{2} \cdot(-3 \sqrt{2})=40-8 \sqrt{3} \end{array}\)
Khi đó
\(\left[\begin{array}{l} x_{1}=\frac{2(\sqrt{3}-1)+\sqrt{40-8 \sqrt{3}}}{2 \cdot \sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}+2 \sqrt{5-\sqrt{3}}}{2} \\ x_{2}=\frac{2(\sqrt{3}-1)-\sqrt{40-8 \sqrt{3}}}{2 \cdot \sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}-2 \sqrt{5-\sqrt{3}}}{2} \end{array}\right.\)
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