Вычислить:
128 = 2*64 = 2*4³ 1/4 = 1/2² Мы привели выражение в скобках к общему знаменателю, за затем выполнили деление.
Решение приложено_____
![\displaystyle (\sqrt[3]{128} + \sqrt[3]{ \frac{1}{4} }): \sqrt[3]{2} = ( \sqrt[3]{2* 4^{3} }+ \frac{ \sqrt[3]{1} }{ \sqrt[3]{ 2^{2} } } ): \sqrt[3]{2} = \\ \frac{( \sqrt[3]{ 2^{2} }* \sqrt[3]{2* 4^{3} }+1) }{ \sqrt[3]{ 2^{2} } }: \sqrt[3]{2} = \frac{ \sqrt[3]{ 2^{3}* 4^{3} }+1 }{ \sqrt[3]{ 2^{2} }* \sqrt[3]{2} } = \frac{2*4+1}{2} =4,5](https://tex.z-dn.net/?f=%5Cdisplaystyle+%28%5Csqrt%5B3%5D%7B128%7D+%2B+%5Csqrt%5B3%5D%7B+%5Cfrac%7B1%7D%7B4%7D+%7D%29%3A+%5Csqrt%5B3%5D%7B2%7D+%3D+%28+%5Csqrt%5B3%5D%7B2%2A+4%5E%7B3%7D+%7D%2B+%5Cfrac%7B+%5Csqrt%5B3%5D%7B1%7D+%7D%7B+%5Csqrt%5B3%5D%7B+2%5E%7B2%7D+%7D+%7D+%29%3A+%5Csqrt%5B3%5D%7B2%7D+%3D+%5C%5C++%5Cfrac%7B%28+%5Csqrt%5B3%5D%7B+2%5E%7B2%7D+%7D%2A+%5Csqrt%5B3%5D%7B2%2A+4%5E%7B3%7D+%7D%2B1%29++%7D%7B+%5Csqrt%5B3%5D%7B+2%5E%7B2%7D+%7D+%7D%3A+%5Csqrt%5B3%5D%7B2%7D++%3D+%5Cfrac%7B+%5Csqrt%5B3%5D%7B+2%5E%7B3%7D%2A+4%5E%7B3%7D+%7D%2B1+%7D%7B+%5Csqrt%5B3%5D%7B+2%5E%7B2%7D+%7D%2A++%5Csqrt%5B3%5D%7B2%7D++%7D+%3D+%5Cfrac%7B2%2A4%2B1%7D%7B2%7D+%3D4%2C5+ "\displaystyle (\sqrt[3]{128} + \sqrt[3]{ \frac{1}{4} }): \sqrt[3]{2} = ( \sqrt[3]{2* 4^{3} }+ \frac{ \sqrt[3]{1} }{ \sqrt[3]{ 2^{2} } } ): \sqrt[3]{2} = \\ \frac{( \sqrt[3]{ 2^{2} }* \sqrt[3]{2* 4^{3} }+1) }{ \sqrt[3]{ 2^{2} } }: \sqrt[3]{2} = \frac{ \sqrt[3]{ 2^{3}* 4^{3} }+1 }{ \sqrt[3]{ 2^{2} }* \sqrt[3]{2} } = \frac{2*4+1}{2} =4,5")