Question

2 ×(√12-√3)+(√27-√48) × 3

Answer 1

avem:
2rad3-3rad3=-rad3

Answer 2

 [tex]\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt3\;;\;\sqrt{27}=\sqrt{3^2\cdot3}=3\sqrt3\;si\;\sqrt{48}=\sqrt{4^2\cdot3}=4\sqrt3\;\;;\;\\ \Rightarrow\;\;\;2\cdot(2\sqrt3-\sqrt3)+(3\sqrt3-4\sqrt3)\cdot3=2\cdot\sqrt3-3\sqrt3=\,-\sqrt3[/tex]