Question

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Answer 1

Ответ:

Объяснение:

а)$\frac{2}{\sqrt{3} } =\frac{2\sqrt{3}}{\sqrt{3}\sqrt{3}} =\frac{2\sqrt{3}}3$

б)    $\frac{16}{\sqrt{2} } =\frac{16\sqrt{2}}{\sqrt{2}\sqrt{2}} =\frac{16\sqrt{2}}2=8\sqrt{2}$

в)   $\frac{7}{3\sqrt{7} } =\frac{7\sqrt{7} }{3\sqrt{7} \sqrt{7} } =\frac{7\sqrt{7} }{3*7} =\frac{\sqrt{7} }{3}$

г)    $\frac{2}{3\sqrt{2} } =\frac{2\sqrt{2}}{3\sqrt{2}\sqrt{2}} =\frac{2\sqrt{2} }{3*2} =\frac{\sqrt{2} }{3}$

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

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