Tính frac2-sqrt3sqrt2+sqrt2+sqrt3

Ta có:

$\frac{2-\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}=\frac{(2-\sqrt{3}) \cdot \sqrt{2}}{(\sqrt{2}+\sqrt{2+\sqrt{3}}) \cdot \sqrt{2}}$

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

$=\frac{2 \sqrt{2}-\sqrt{6}}{2+1+\sqrt{3}}=\frac{2 \sqrt{2}-\sqrt{6}}{3+\sqrt{3}}=\frac{(2 \sqrt{2}-\sqrt{6})(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}$

$=\frac{6 \sqrt{2}-2 \sqrt{6}-3 \sqrt{6}+3 \sqrt{2}}{(3+\sqrt{3})(3-\sqrt{3})}$

$=\frac{(6 \sqrt{2}+3 \sqrt{2})-(2 \sqrt{6}+3 \sqrt{6})}{9-3}=\frac{9 \sqrt{2}-5 \sqrt{6}}{6}$