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Sprendimas:

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

Atsakymas: B $\frac{3-\sqrt {3}}{3}$

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