Có:
- [tex]\sqrt{\frac{-34+13\sqrt{7}}{2\sqrt{7}-5}}=\sqrt{\frac{13(2\sqrt{7}-5)-3}{2(2\sqrt{7}-5)}}=\sqrt{\frac{13}{2}-\frac{3}{2(2\sqrt{7}-5)}}=\sqrt{\frac{13}{2}-\frac{3(2\sqrt{7}+5)}{2.3}}=\sqrt{\frac{13}{2}-\frac{2\sqrt{7}+5}{2}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}[/tex]
- [tex]\sqrt{\frac{15+6\sqrt{7}}{\sqrt{7}+2}}=\sqrt{\frac{6(\sqrt{7}+2)+3}{\sqrt{7}+2}}=\sqrt{6+\frac{3}{\sqrt{7}-2}}=\sqrt{6+\frac{3(\sqrt{7}-2)}{3}}=\sqrt{6+\sqrt{7}-2}=\sqrt{4+\sqrt{7}}=\sqrt{\frac{8+2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}+1)^2}{2}}=\frac{\sqrt{7}+1}{\sqrt{2}}[/tex]
Vậy [tex]\sqrt{\frac{-34+13\sqrt{7}}{2\sqrt{7}-5}}-\sqrt{\frac{15+6\sqrt{7}}{\sqrt{7}+2}}=-\sqrt{2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
giúp em với ạ
