Question

Иррациональные числа. Оба варианта . 50 балллв

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\frac{6}{\sqrt{3}} \cdot\frac{\sqrt{3} }{\sqrt{3} }=\frac{6\sqrt{3} }{3}=2\sqrt{3}\\\\\\\frac{3}{\sqrt{5}+\sqrt{2} } \cdot\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{3(\sqrt{5}-\sqrt{2})}{5-2} =\sqrt{5}-\sqrt{2} \\\\\\\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3} }\cdot\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3} }= \frac{(\sqrt{7}-\sqrt{3})^2 }{7-3}=\frac{7-2\sqrt{21}+3 }{4} =\frac{5-\sqrt{21} }{2}$

$\displaystyle\\\frac{8}{\sqrt{2} } \cdot\frac{\sqrt{2} }{\sqrt{2} }=\frac{8\sqrt{2} }{2} =4\sqrt{2} \\\\\\\frac{2}{\sqrt{3}-1 }\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1} =\frac{2({\sqrt{3}+1})}{3-1}=\sqrt{3} +1 \\\\\\\frac{\sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} \cdot \frac{\sqrt{3}+\sqrt{2} }{\sqrt{3}+\sqrt{2}} =\frac{(\sqrt{3}+\sqrt{2})^2}{3-2} =3+2\sqrt{6} +2=5+2\sqrt{6}$