Answer:
[tex]\large\boxed{2\sqrt{27}+\sqrt{12}-3\sqrt3-2\sqrt{12}=7\sqrt3}[/tex]
Step-by-step explanation:
[tex]2\sqrt{27}+\sqrt{12}-3\sqrt3-2\sqrt{12}\\\\=2\sqrt{(9)(3)}+\sqrt{(4)(3)}+3\sqrt3-2\sqrt{(4)(3)}\qquad\text{use}\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\\\\=2\sqrt9\cdot\sqrt3+\sqrt4\cdot\sqrt3+3\sqrt3-2\sqrt4\cdot\sqrt3\\\\=(2)(3)\sqrt3+2\sqrt3+3\sqrt3-(2)(2)\sqrt3\\\\=6\sqrt3+2\sqrt3+3\sqrt3-4\sqrt3\\\\=(6+2+3-4)\sqrt3\\\\=7\sqrt3[/tex]
