Đáp án:
Giải thích các bước giải:
`a)\sqrt{12}-3\sqrt{3}+\sqrt{48}=\sqrt{2^2 .3}-3\sqrt{3}+\sqrt{4^2 . 3}=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}`
`b)4\sqrt{(2-\sqrt{3})^2}+2\sqrt{(2+\sqrt{3})^2}+2\sqrt{3}=2.|2-\sqrt{3}|+ 2|2+\sqrt{3}|+2\sqrt{3}=4-2\sqrt{3}+4+2\sqrt{3}+2\sqrt{3}=8+2\sqrt{3}`
`c)1/{2-\sqrt{3}}+1/{2+\sqrt{3}}={2+\sqrt{3}}/{4-3}+{2-\sqrt{3}}/{4-3}=4`
`d)\sqrt{5-2\sqrt{6}}+\sqrt{7-2\sqrt{6}}+\sqrt{2}-\sqrt{6}+1=\sqrt{(\sqrt{2})^2+(\sqrt{3)^2)-2.\sqrt{2}.\sqrt{3}}+ \sqrt{(\sqrt{6})^2+1-2.\sqrt{6}}+\sqrt{2}-\sqrt{6}+1=\sqrt{(\sqrt{3}-\sqrt{2})^2}+\sqrt{(\sqrt{6}-1)^2)+\sqrt{2}-\sqrt{6}+1=\sqrt{3}-\sqrt{2}+\sqrt{6}-1+\sqrt{2}-\sqrt{6}+1=\sqrt{3}`
