(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)...
Đọc tiếp
(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (10)\(\sqrt{2020+2\sqrt{2019}}-\sqrt{2020-2\sqrt{2019}}\) (11)\(\sqrt{7+2\sqrt{12}}\) Các bạn giúp mình với ,Mình xin cảm ơn trước
NGuyễn Văn Tuấn mik ko bảo bn mik bảo tth cơ
tth làm sau có để thì để tên khác đi