Soal UTBK-SNBT Paket 16
Soal Nomor 76 Diberikan \( p = 216^{-\frac{1}{3}} + 243^{-\frac{2}{3}} + 256^{-\frac{1}{4}} \), mana yang merupakan bilangan bulat? A. \( \frac{p}{19} \) B. \( \frac{p}{36} \) C. \( p \) D. \( \frac{19}{p} \) E. \( \frac{36}{p} \) Pembahasan \[ p = \frac{1}{216^{\frac{1}{3}}} + \frac{1}{243^{\frac{2}{3}}} + \frac{1}{256^{\frac{1}{4}}} \] \[ p = \frac{1}{\left(6^3\right)^{\frac{1}{3}}} + \frac{1}{\left(3^5\right)^{\frac{2}{5}}} + \frac{1}{\left(2^8\right)^{\frac{1}{4}}} \] \[ p = \frac{1}{6} + \frac{1}{3^2} + \frac{1}{2^2} \] \[ p = \frac{1}{6} + \frac{1}{9} + \frac{1}{4} = \frac{19}{36} \] ...