Question

alguém me ajuda
$\sqrt{16 + \frac{1}{9} } + \frac{1}{4} + {7}^{1} - {12}^{0}$

Answer 1

$\sqrt{16} = 4 \\ \sqrt{ \frac{1}{9} } = \frac{ \sqrt{ 1} }{ \sqrt{9} } = \frac{1}{3} \\ \sqrt{ \frac{1}{4} } = \frac{1}{2} \\ {7}^{1} = 7 \\ {12}^{0} = 1$

$4 + \frac{1}{3} + \frac{1}{2} + 7 + 1 = \frac{24}{6} + \frac{3}{6} + \frac{42}{6} + \frac{6}{6} = \frac{75}{6}$