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$109+12\sqrt3=109+2\cdot 6\sqrt3=109+2\sqrt{6^2\cdot 3}=\\\\=109+2\cdot \sqrt{108}=1+108+2\cdot \sqrt{1}\cdot \sqrt{108}=\\\\=(1+\sqrt{108})^2=(1+6\sqrt3)^2\\\\\\\sqrt{109+12\sqrt3} - \sqrt{109-12\sqrt3} =\sqrt{(1+6\sqrt3)^2}-\sqrt{(1-6\sqrt3)^2}=\\\\=|1+6\sqrt3|-|1-6\sqrt3|=1+6\sqrt3-(6\sqrt3-1)=2\\\\\\P,S.\; \; \sqrt{A^2}=|A|= \left [ {{+A,\; esli\; A \geq 0} \atop {-A,\; esli\; A\ \textless \ 0}} \right. \; \; \Rightarrow \\\\|1-6\sqrt3|=-(1-6\sqrt3)=6\sqrt3-1$