Question

Ex 2 va dau coroana sau 3​

Answer 1

Explicație pas cu pas:

A) amplifica frăția cu radical din 6=2radical din 6/6

Answer 2

Răspuns:

Explicație pas cu pas:

$Ex~2)\\a)~\dfrac{2}{\sqrt{6}}= \dfrac{2\sqrt{6} }{(\sqrt{6})^2}= \dfrac{2\sqrt{6} }{6}=\dfrac{\sqrt{6} }{3}.\\b)~\dfrac{6}{\sqrt[3]{3}}=\dfrac{6\sqrt[3]{3^2} }{\sqrt[3]{3}*\sqrt[3]{3^2}} =\dfrac{6\sqrt[3]{9} }{\sqrt[3]{3^3} }=\dfrac{6\sqrt[3]{9} }{3}=2\sqrt[3]{9}.\\c)~\dfrac{1}{\sqrt{7}-\sqrt{5}} =\frac{1*(\sqrt{7}+\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})} =\dfrac{\sqrt{7}+\sqrt{5}}{(\sqrt{7})^2-(\sqrt{5})^2} =\dfrac{\sqrt{7}+\sqrt{5}}{7-5}= \dfrac{\sqrt{7}+\sqrt{5}}{2}.$

$d)~\dfrac{1}{\sqrt[3]{5}-\sqrt[3]{4}}=\dfrac{1*(\sqrt[3]{5^2}+\sqrt[3]{5}*\sqrt[3]{4}+\sqrt[3]{4^2}}{(\sqrt[3]{5}-\sqrt[3]{4})(\sqrt[3]{5^2}+\sqrt[3]{5}*\sqrt[3]{4}+\sqrt[3]{4^2})}=\dfrac{\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16}}{(\sqrt[3]{5})^3-\sqrt[3]{4})^3} =\\= \dfrac{\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16}}{5-4}= \dfrac{\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16}}{1}= \sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{16}.\\Ex~3)~5^{\frac{3}{2}}*125^{\frac{1}{2}}*5^{\frac{4}{3}}*25^{-\frac{1}{6}}=$

$=5^{\frac{3}{2}}*(5^{3})^{\frac{1}{2}}*5^{\frac{4}{3}}*(5^{2})^{-\frac{1}{6}}=5^{\frac{3}{2}}*5^{3*\frac{1}{2}}*5^{\frac{4}{3}}*5^{2*(-\frac{1}{6})}=5^{\frac{3}{2}}*5^{\frac{3}{2}}*5^{\frac{4}{3}}*5^{-\frac{1}{3}}=\\=5^{\frac{3}{2}+\frac{3}{2}+\frac{4}{3}-\frac{1}{3}}=5^{\frac{6}{2}+\frac{3}{3}}=5^{3+1}=5^4=625.$