Question

Alguém me explica como resolver essa questão da EPCAr ?

Answer 1

Lembrando:
$0,25 = \frac{1}{4} \quad 1,5 = \frac{3}{2} \quad \: 0,5 = \frac{1}{2} \\ \sqrt[b]{ {m}^{a} } = {m}^{ \frac{a}{b} } \quad {m}^{ - a} = \frac{1}{ {m}^{a} } \quad { ({m}^{a}) }^{b} = {m}^{a \cdot b}$
Resolvendo:
$n = {(0,5 \cdot {4}^{0,25} + {4}^{0,25})}^{2} - {4}^{1,5} \cdot (1 + {4}^{ - 0,5} )$
$n = {\left ( \frac{1}{2} \cdot ({2}^{2})^{ \frac{1}{4} } + {({2}^{2})}^{\frac{1}{4}} \right )}^{2} - {{(2}^{2})}^{ \frac{3}{2} } \left (1 + { ({2}^{2}) }^{ - \frac{1}{2} } \right ) \\ n = {\left ( \frac{1}{2} \cdot {2}^{ \frac{1}{2} } + {2}^{ \frac{1}{2} } \right ) }^{2} - {2}^{3} (1 + {2}^{ - 1} )$
$n = {\left ( \frac{ \sqrt{2} }{2} + \sqrt{2} \right ) }^{2} - 8 \left (1 + \frac{1}{2} \right ) \\ n = \frac{2}{4} + 2 \cdot \frac{ \sqrt{2} }{2} \cdot \sqrt{2} + 2 - 8 \cdot \frac{3}{2} \\ n = \frac{1}{2} + 2 + 2 - 12 = \frac{1}{2} - 8 = \frac{1}{2} - \frac{16}{2} = - \frac{15}{2} \\ n = -7,5$