Question

нуждаюсь в помощи
ответ: (2-√3)^-1​

Answer 1

$\frac{2}{\sqrt3-1}-\frac{1}{\sqrt3+2}+\frac{6}{\sqrt3+3}=\\\\\\=\frac{2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}-\frac{\sqrt3-2}{(\sqrt3+2)(\sqrt3-2)}+\frac{6(\sqrt3-3)}{(\sqrt3+3)(\sqrt3-3)}=\\\\\\=\frac{2(\sqrt3+1)}{3-1}-\frac{\sqrt3-2}{3-4}+\frac{6(\sqrt3-3)}{3-9}=(\sqrt3+1)+(\sqrt3-2)-(\sqrt3-3)=\\\\\\=\sqrt3+2=\frac{(2+\sqrt3)(2-\sqrt3)}{2-\sqrt3}=\frac{4-3}{2-\sqrt3}=\frac{1}{2-\sqrt3}=(2-\sqrt3)^{-1}$