Преобразование выражений

Question 1

Question 2

Решите задачу:

$\sqrt[3]{18} \cdot \sqrt[3]{-12} = \sqrt[3]{18 \cdot(-12)} =-\sqrt[3]{2\cdot9 \cdot4\cdot3} =-\sqrt[3]{2^3\cdot3^3} =-2\cdot3=-6\\\\ 0.5\cdot \sqrt[3]{ \frac{120}{15} } =0.5\cdot \sqrt[3]{8 } =0.5\cdot \sqrt[3]{2^3 } =0.5\cdot2=1\\\\ 4\cdot \sqrt[3]{ \frac{540}{20} } =4\cdot \sqrt[3]{27} =4\cdot \sqrt[3]{3^3} = 4\cdot 3 = 12\\\\ \sqrt[3]{36\cdot24\cdot54} = \sqrt[3]{9\cdot4\cdot3\cdot8\cdot2\cdot27} = \sqrt[3]{3^2\cdot2^2\cdot3\cdot2^3\cdot2\cdot3^3} = \sqrt[3]{3^6\cdot2^6} =\\ =3^2\cdot2^2=9\cdot4=12$

$\sqrt[3]{12\cdot32\cdot36}= \sqrt[3]{4\cdot3\cdot32\cdot9\cdot4}= \sqrt[3]{2^2\cdot3\cdot2^5\cdot3^2\cdot2^2}= \sqrt[3]{2^9\cdot3^3} =\\ =2^3\cdot3=8\cdot3=24\\\\ \frac{20 \sqrt[3]{3} }{ \sqrt[3]{375} } =20 \sqrt[3]{ \frac{3}{375} } =20 \sqrt[3]{ \frac{1}{125} } = \frac{20}{ \sqrt[3]{5^3} } = \frac{20}{5}=4\\\\ \frac{ \sqrt[3]{320} }{4 \sqrt[3]{5} } = \frac{1}{4} \sqrt[3]{ \frac{320}{5} } = \frac{1}{4} \sqrt[3]{ 64 } =\frac{1}{4} \sqrt[3]{ 2^6 } = \frac{2^2}{4} =1\\$

$(2 \sqrt[4]{25} )^2=(2 \sqrt[4]{5^2} )^2=2^2\cdot\sqrt[4]{5^4}=4\cdot5=20\\\\ (3\cdot \sqrt[3]{64} )^2=3^2\cdot (\sqrt[3]{2^6})^2=9\cdot2^4=9\cdot16=144\\\\ \sqrt[3]{54} \cdot \sqrt[3]{4} - \sqrt{2}\cdot \sqrt{8} = \sqrt[3]{54\cdot4} - \sqrt{2\cdot8} =\sqrt[3]{27\cdot2\cdot2^2} - \sqrt{2\cdot2^3} =\\ \sqrt[3]{3^3\cdot2^3} - \sqrt{2^4} =6-2^2=6-4=2$

$\sqrt[3]{16} \cdot \sqrt[3]{4} -2 \sqrt[3]{27} = \sqrt[3]{4^3} -2 \sqrt[3]{3^3} =4-2\cdot3=4-6=-2\\\\ \sqrt[3]{25\cdot60\cdot18} = \sqrt[3]{5^2\cdot5\cdot4\cdot3\cdot9\cdot2} = \sqrt[3]{5^3\cdot2^3\cdot3^3} =5\cdot2\cdot3=30\\\\ \frac{ \sqrt{22}\cdot \sqrt{18} }{3 \sqrt{11} } = \frac{1}{3} \sqrt{ \frac{396}{11} } =\frac{1}{3} \sqrt{36} =\frac{1}{3} \sqrt{6^2} =2$

$\frac{ \sqrt{21}\cdot \sqrt{12} }{2 \sqrt{7} } = \frac{1}{2} \sqrt{ \frac{252}{7} } =\frac{1}{2} \sqrt{36} =\frac{1}{2} \sqrt{6^2} =3\\\\ \frac{ \sqrt{6}\cdot \sqrt{30} }{2 \sqrt{5} } = \frac{1}{2} \sqrt{ \frac{180}{5} } =\frac{1}{2} \sqrt{36} =\frac{1}{2} \sqrt{6^2} =3\\\\$