Sprendimas:
23+3=\frac{2}{3+\sqrt {3}} = 3+32= 2⋅(3−3)(3+3)⋅(3−3)=\frac{2\cdot (3-\sqrt {3})}{(3+\sqrt {3})\cdot (3-\sqrt {3})} = (3+3)⋅(3−3)2⋅(3−3)= 2⋅(3−3)32−32=\frac{2\cdot (3-\sqrt {3})}{3^{2}-\sqrt {3}^{2}} = 32−322⋅(3−3)= 2⋅(3−3)9−3=\frac{2\cdot (3-\sqrt {3})}{9-3} = 9−32⋅(3−3)= 2⋅(3−3)6=\frac{2\cdot (3-\sqrt {3})}{6} = 62⋅(3−3)= 3−33\frac{3-\sqrt {3}}{3}33−3
Atsakymas: B 3−33\frac{3-\sqrt {3}}{3}33−3
