Question

Решите что можете, 1)(27^2/3 )^-2+(3 3/8)^-1/3 2)(2^5/3*5^1-/3-5^5/3*2^-1/3)*³√10 3)(³√а+³√b)*(а^2/3+b^2/3-³√ав)

Answer 1

$1)\quad 27^{\frac{2}{3}}-(-2)^{-2}+(3\frac{3}{8})^{-\frac{1}{3}}=(3^3)^{\frac{2}{3}}-\frac{1}{2^2}+(\frac{27}{8})^{-\frac{1}{3}}=\\\\=3^2-\frac{1}{4}+(\frac{2^3}{3^3})^{\frac{1}{3}}=9-\frac{1}{4}+\frac{2}{3}=\frac{9\cdot 12-3+2\cdot 4 }{12}=\frac{113}{12}=9\frac{5}{12}$

$2)\; \; (2^{\frac{5}{3}}\cdot 5^{-\frac{1}{3}}-5^{\frac{5}{3}}\cdot 2^{-\frac{1}{3}})\cdot \sqrt[3]{10}=\\\\=2^{-\frac{1}{3}}\cdot 5^{-\frac{1}{3}}\cdot \left (\frac{2^{\frac{5}{3}}\cdot 5^{-\frac{1}{3}}}{2^{-\frac{1}{3}}\cdot 5^{-\frac{1}{3}}} - \frac{5^{\frac{5}{3}}\cdot 2^{-\frac{1}{3}}}{2^{-\frac{1}{3}}\cdot 5^{-\frac{1}{3}}} \right )\cdot 10^{\frac{1}{3}}=\\\\=(2\cdot 5)^{-\frac{1}{3}}}\cdot \left (2^{\frac{5}{3}+\frac{1}{3}}-5^{\frac{5}{3}+\frac{1}{3}}\right )\cdot 10^{\frac{1}{3}}=$

$=10^{-\frac{1}{3}}\cdot 10^{\frac{1}{3}}\cdot (2^2-5^2)=10^{-\frac{1}{3}+\frac{1}{3}}\cdot (4-25)=\\\\=10^0\cdot (-21)=1\cdot (-21)=-21\\\\3)\quad (\sqrt[3]{a}+\sqrt[3]{b})\cdot (a^{\frac{2}{3}}+b^{\frac{2}{3}}-\sqrt[3]{ab})=\\\\=(a^{\frac{1}{3}}+b^{\frac{1}{3}})\cdot (\, (a^{\frac{1}{3}})^2-a^{\frac{1}{3}}\cdot b^{\frac{1}{3}}+(a^\frac{1}{3})^2\, )=(a^{\frac{1}{3}})^3+(b^{\frac{1}{3}})^3=a+b$