Răspuns :
[tex]\it 4\sqrt3=\sqrt{4^2\cdot3}=\sqrt{16\cdot3}=\sqrt{48}<\sqrt{49}=7 \Rightarrow 4\sqrt3<7 \Rightarrow \\ \\ \Rightarrow4\sqrt3-7<0 \Rightarrow |4\sqrt3-7|=-4\sqrt3+7\\ \\ Prin\ urmare:\\ \\ |4\sqrt3-7|+4\sqrt3=-4\sqrt3+7+4\sqrt3=7[/tex]