Đáp án:
`a)\sqrt{12+2\sqrt{35}}=\sqrt{7+5+2\sqrt5.\sqrt7}=\sqrt{(sqrt7+sqrt5)^2}=|\sqrt7+\sqrt5|=sqrt7+sqrt5`
`b)\sqrt{16+6sqrt{7}}=sqrt{9+2.3.sqrt7+7}=sqrt{(3+sqrt7)^2}=|3+sqrt7|=3+sqrt7`
`c)\sqrt{31-12\sqrt3}=\sqrt{27-2.3sqrt3. 2+4}=sqrt{(3sqrt3-2)^2}=|3sqrt3-2|=3sqrt3-2(do \ 3sqrt3>2)`.