Question

Resolva as expressões numéricas​

Answer 1

a) cálculos de auxiliares:

$\frac{1}{16}. \sqrt{16}=\frac{1}{\cancel{64}}.\cancel4\\=\frac{1}{16}$

${8}^{-2}={(\frac{1}{8}) }^{2}=\frac{1}{64}$

${16}^{-\frac{1}{2}}=\frac{1}{\sqrt{16}}=\frac{1}{4}$

${2}^{4}.{2}^{-1}={2}^{3}=8$

$\frac{ \frac{1}{64}.\sqrt{16} }{ {8}^{ - 2} } . {16}^{ - \frac{1}{2} } + \frac{ \sqrt{64} }{ {2}^{4}. {2}^{ - 1} } \\ = \frac{ \frac{1}{16} }{ \frac{1}{64} } . \frac{1}{4} + \frac{8}{8}$

$\frac{1}{16}.64. \frac{1}{4} + 1 = 1 + 1 = 2$

b) cálculos auxiliares:

$\frac{6}{9}. (10+\sqrt[3]{125}).0,5\\=\frac{\cancel2}{3}.(10+5).\frac{1}{\cancel2}=\frac{15}{3}=5$

$\frac{1}{2} +\frac{5}{3}+\frac{5</p><p>}{6}+5\\=\frac{3++10+5+30}{6}=\frac{48}{6}=8$

$\sqrt{{2}^{4}-{2}^{3}}=\sqrt{16-8}=\sqrt{8}\\=\sqrt{4.2}=2\sqrt{2}$

$8\div{2}^{3}=8\div8=1$

$\sqrt{2}.\sqrt{2,5}=\sqrt{2.2,5}=\sqrt{5}$

$\frac{ [{ \frac{6}{9}(10 + \sqrt[3]{125}). 0,5 }] ^{ \frac{1}{2} } \div[ {( \frac{1}{2} + \frac{5}{3} + \frac{5}{6} + 5 )}^{ \frac{1}{2} } + \sqrt{2 + 3}] }{ \frac{ \sqrt{ {2}^{4} - {2}^{3} } }{8 \div {2}^{3} } - \sqrt{2} . \sqrt{2,5} }$

$\frac{ {5}^{ \frac{1}{2}} \div[{8}^{\frac{1}{2}}+\sqrt{5}] }{2 \sqrt{2} - \sqrt{5} } = \frac{ \sqrt{5} \div [2 \sqrt{2} +\sqrt{5}] }{2 \sqrt{2} - \sqrt{5} }$

$\frac{ \sqrt{5} }{(2 \sqrt{2} + \sqrt{5})(2 \sqrt{2} - \sqrt{5} )} \\ = \frac{ \sqrt{5} }{ {(2 \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\frac{ \sqrt{5} }{8 - 5} = \boxed{\boxed{\mathsf{ \dfrac{ \sqrt{5} }{3} }}}$