Mathematic Problems

107. példa

243^{\frac{1}{5}} - {(\frac{1}{81})}^{\frac{1}{4}} + {(92 + 7, 5)}^{0} + {(9^{\frac{7}{8}} : 3^{\frac{5}{4}})}^{4}

243^{\frac{1}{5}} - {(\frac{1}{81})}^{\frac{1}{4}} + {(92 + 7, 5)}^{0} + {(9^{\frac{7}{8}} : 3^{\frac{5}{4}})}^{4}

$= {(3^{5})}^{\frac{1}{5}} - {(\frac{1}{3^{4}})}^{\frac{1}{4}} + {(92 + 7, 5)}^{0} + {[{(3^{2})}^{\frac{7}{8}} : 3^{\frac{5}{4}}]}^{4}$

$= 3^{\frac{5}{5}} - \frac{1^{\frac{1}{4}}}{3^{\frac{4}{4}}} + 1 + {(3^{\frac{14}{8}} \cdot 3^{- \frac{5}{4}})}^{4}$

$= 3 - \frac{1}{3} + 1 + {[3^{\frac{14}{8} - \frac{5}{4}}]}^{4}$

$= 3 - \frac{1}{3} + 1 + {(3^{\frac{14 - 2 \cdot 5}{8}})}^{4}$

$= 4 - \frac{1}{3} + {(3^{\frac{4}{8}})}^{4}$

$= 3 - \frac{1}{3} + 1 + 3^{\frac{16}{8}}$

= 3 - \frac{1}{3} + 1 + 3^{2}

= 3 - \frac{1}{3} + 1 + 9

= \frac{9 - 1 + 3 + 27}{3}

= \frac{38}{3}

Matematika - Képlettár