Question

решите пж срочно даю 20б $(-2\frac{1}{3} )^{2} +(\frac{1}{2} )^{3}-(\frac{3}{19}) ^{0} +(-1)^{12} =$

Answer 1

$\displaystyle\bf\\\Big(-2\frac{1}{3} \Big)^{2} +\Big(\frac{1}{2}\Big)^{3} -\underbrace{\Big(\frac{3}{19} \Big)^\circ}_{1}+\underbrace{\Big(-1\Big)^{12}}_{1}=\Big(\frac{7}{3}\Big)^{2} +\frac{1}{8}-1+1=\\\\\\=\frac{49}{9} +\frac{1}{8} =\frac{392+9}{72}=\frac{401}{72} =5\frac{41}{72}$