Question

resolva as expressões

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$\frac{\sqrt{(-1)^2}-0,1222... }{(1,2)^{-1}} =\frac{\sqrt{1}-(\frac{1}{10}+\frac{2}{90} ) }{(\frac{6}{5})^{-1} } =\frac{1-\frac{11}{90} }{\frac{5}{6} } =\frac{\frac{79}{90} }{\frac{5}{6} } =\frac{79}{90} *\frac{6}{5} =\frac{474}{450}$

b)

$(\frac{2}{3} -0,333...)^2+\sqrt{0,111...} =(\frac{2}{3} -\frac{3}{9} )^2+\sqrt{\frac{1}{9} }=(\frac{1}{3} )^2+\sqrt{\frac{1}{9} }=\frac{1}{9} +\frac{1}{3} =\frac{4}{9}$

c)

$(\frac{1}{5} )^{-2}+(\frac{1}{5} )^2+\sqrt[3]{-27} =(\frac{5}{1} )^{2}+(\frac{1}{5} )^2+-3=25+\frac{1}{25} -3=22+\frac{1}{25}=\frac{551}{25}$

d)

$\frac{\sqrt{0,999...} +\sqrt{0,444...} }{1+0,424242...} =\frac{\sqrt{1} +\sqrt{\frac{4}{9}} }{1+\frac{42}{99} }=\frac{1 +\frac{2}{3} }{1+\frac{42}{99}}=\frac{\frac{5}{3} }{\frac{141}{99} }=\frac{5}{3}* \frac{99}{141} =\frac{495}{423}$