Question

алгебра номер 4,5 помогите просьба не знаете не пишите​

Answer 1

Ответ:

$4)\ \ \Big(k^{\frac{1}{3}}+q^{\frac{1}{3}}\Big)\Big(k^{\frac{2}{3}}+q^{\frac{2}{3}}-(kq)^{\frac{1}{3}}\Big)= \Big(k^{\frac{1}{3}}\Big)^3+\Big(q^{\frac{1}{3}}\Big)^3=k+q$

$5)\ \ \dfrac{27^{\frac{1}{3}}\cdot 4^{-\frac{1}{2}}-2^{-1}}{625^{-\frac{1}{4}}}=\dfrac{3\cdot \dfrac{1}{2}-\dfrac{1}{2}}{\dfrac{1}{5}}=\dfrac{1}{\dfrac{1}{5}}=5$  

Answer 2

$\displaystyle\bf\\4)\\\\\Big(k^{\frac{1}{3} } +q^{\frac{1}{3} } \Big)\Big(k^{\frac{2}{3} } +q^{\frac{2}{3} }-( kq)^{\frac{1}{3} } \Big)=\Big(k^{\frac{1}{3} } +q^{\frac{1}{3} } \Big)\Big(k^{\frac{2}{3} }-(kq)^{\frac{1}{3} } +q^{\frac{2}{3} }} \Big)=\\\\\\=\Big(k^{\frac{1}{3} } \Big)^{3} +\Big(q^{\frac{1}{3} }\Big)^{3} =k+q$

$\displaystyle\bf\\5)\\\\\frac{27^{\frac{1}{3} }\cdot4^{-\frac{1}{2} } -2^{-1} }{625^{-\frac{1}{4} } } =\frac{(3^{3} )^{\frac{1}{3} }\cdot(2^{2} )^{-\frac{1}{2} } -\frac{1}{2} }{(5^{4} )^{-\frac{1}{4} } } =\frac{3\cdot 2^{-1}-0,5 }{5^{-1} } =\\\\\\=\frac{3\cdot 0,5-0,5}{0,2} =\frac{1,5-0,5}{0,2}=\frac{1}{0,2} =5$