[tex]\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-\sqrt{2}}}[/tex]
Đề như thế này sẽ hợp lí hơn
Đặt $A=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}<0$
[tex]A^2=\left ( \frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}} \right )^{2}\\=\frac{(\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}})^{2}}{\sqrt{7}-2}\\=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}}{\sqrt{7}-2}\\=\frac{2\sqrt{7}-4}{\sqrt{7}-2}\\=\frac{2(\sqrt{7}-2)}{\sqrt{7}-2}\\=2\\\Rightarrow A=-\sqrt{2}[/tex]
Còn nếu như theo đề của bạn thì
Đặt $A=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-\sqrt{2}}}<0$
[tex]A^2=\left ( \frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-\sqrt{2}}} \right )^{2}\\=\frac{(\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}})^{2}}{\sqrt{7}-\sqrt{2}}\\=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}}{\sqrt{7}-\sqrt{2}}\\=\frac{2\sqrt{7}-4}{\sqrt{7}-\sqrt{2}}\\=\frac{2(\sqrt{7}-2)}{\sqrt{7}-\sqrt{2}}\\\Rightarrow A=-\sqrt{\frac{2(\sqrt{7}-2)}{\sqrt{7}-\sqrt{2}}}[/tex]
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