Question

Resolva as expressões numéricas ​

Answer 1

1)

cálculos auxiliares:

$3.{(-\dfrac{3}{4})}^{-2}+6.(\dfrac{{3}^{-1}}{4}) -4$

$3.\dfrac{16}{9}+6.\dfrac{1}{12}-4$

$\dfrac{16}{3}+\dfrac{1}{2}-4=\dfrac{32+3-24}{6}=\dfrac{11}{6}$

$7.{(\dfrac{-3}{4})}^{-1}+2\\7.(-\dfrac{4}{3}) +2=\dfrac{-28+6}{3}=-\dfrac{22}{3}$

${[\dfrac{3.{(-\dfrac{3}{4})}^{-2}+6.(\dfrac{{3}^{-1}}{4}) -4}{7.{(\dfrac{-3}{4})}^{-1}+2}]}^{-1}+4$

${[\dfrac{\dfrac{11}{6}}{-\dfrac{22}{3}}]}^{-1}+4$

${[\dfrac{\cancel{11}}{\cancel6}.-\dfrac{\cancel3}{\cancel{22}}]}^{-1}+4$

${[-\dfrac{1}{4}]}^{-1}+4=-4+4=0$

$\boxed{\mathsf{{[ \frac{3.{(-\frac{3}{4})}^{-2}+6.(\frac{{3}^{-1}}{4}) -4}{7.{(\frac{-3}{4})}^{-1}+2}]}^{-1}+4=0}}$

cálculos auxiliares:

$\frac{{({2}^{2})}^{2}+{3}^{2}}{{2}^{2}.{3}^{2}} \\=\frac{16+9}{36}=\frac{25}{36}$

$0,5+{3}^{-1}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

${(\dfrac{1}{2+3})}^{-\frac{1}{2}}={(\frac{1}{5})}^{-\frac{1}{2}}=\sqrt{5}$

$\frac{{2}^{\frac{1}{2}}}{2-3}=-\sqrt{2}$

$\frac{ {[ \frac{ { ({2}^{2} ) }^{2} + {3}^{2} }{ {2}^{2}. {3}^{2} } ] }^{ - \frac{1}{2} } .[0,5+{3}^{-1}] }{ { (\frac{1}{2 + 3}) }^{ - \frac{1}{2} } + \frac{ {2}^{ \frac{1}{2} } }{2 - 3} }$

$\dfrac{ {(\frac{25}{36})}^{ - \frac{1}{2} } . \frac{5}{6} }{ \sqrt{5} + (-\sqrt{2})} \\ \frac{ \frac{6}{5}. \frac{5}{6} }{ \sqrt{5} - \sqrt{2} } = \frac{ \sqrt{5} + \sqrt{2} }{5 - 2}$

$\frac{ \sqrt{5} + \sqrt{2}}{3}$